CBSE Class 12 Physics Ray Optics Optical Instruments Worksheet

Introduction and Overview

Optics is the branch of physics that studies the behavior and properties of light. It also studies how light interacts with matter and how optical instruments are designed. In Class XII, optics is divided into Ray Optics (dealing with light as rays) and Wave Optics (dealing with light as waves). This chapter covers:

• Ray Optics: Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and its applications, optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lensmaker's formula, magnification, power of a lens, combination of thin lenses in contact, and refraction of light through a prism.
• Scattering of light: Blue color of the sky and the reddish appearance of the sun at sunrise and sunset.
• Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.

Question 601. When a wave is propagating from a rarer to a denser medium, which characteristic of the wave does not change and why ?
OR
When monochromatic light travels from one medium to another, its wavelength changes but its frequency remains same. Why ?
Answer: When a wave propagates from one medium to another, its frequency remains unchanged. This is because frequency is a fundamental characteristic of the source of the waves, representing the rate at which the source produces oscillations. Wavelength and velocity, on the other hand, are properties of the medium and will change depending on the optical density of the medium.
In simple words: The frequency of a wave is determined solely by the source that created it, so it does not change when the wave enters a new medium. Only the wave's speed and wavelength change.

Exam Tip: Always emphasize that frequency is a "source-dependent" property while speed and wavelength are "medium-dependent" properties to secure full marks on this conceptual question.

Question 602. When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Why ?
Answer: Reflection and refraction occur when incident light waves interact with the atoms and molecules of the medium. These atomic constituents act as oscillators, absorbing the energy of the incident wave and vibrating at the exact same frequency. They then re-emit light waves of this identical frequency as reflected or refracted rays. Consequently, the frequency of light remains unaffected during reflection and refraction.
In simple words: When light hits a surface, it forces the atoms in that surface to vibrate back and forth at its own frequency. These vibrating atoms then send out the reflected and refracted light at that same frequency.

Exam Tip: Use the term "atomic oscillators" or "atomic constituents" vibrating in resonance with the incident light to explain why the frequency remains unchanged during these boundaries.

Question 603. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave ?
Answer: No, the reduction in speed does not imply a decrease in the energy carried by the wave. The energy carried by a wave is determined by its amplitude (energy is directly proportional to the square of the amplitude of the wave) and its frequency, not by the speed of wave propagation. Therefore, the energy remains constant as the wave enters a denser medium.
In simple words: The speed of a wave only tells us how fast it travels, not how much energy it carries. Wave energy depends on its frequency and amplitude, which do not drop when the wave slows down in a denser medium.

Exam Tip: State the relation \( E \propto A^2 \) (where \( A \) is amplitude) explicitly to show the mathematical justification of your answer.

Question 604. In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determine the intensity in the photon picture of light ?
Answer: In the photon picture of light, the intensity of light is determined by the number of photons incident normally on a unit area per unit time. Each individual photon carries a discrete quantum of energy \( E = h\nu \), so the total intensity is proportional to the photon flux.
In simple words: While wave theory defines intensity using the square of wave amplitude, photon theory defines it by counting how many packets of light (photons) strike a unit area every second.

Exam Tip: Be sure to include "per unit area per unit time" in your definition, as the rate of incident photons is area-dependent.

Question 605. When light comes from air to glass, the refracted ray is bent towards the normal. Why ?
Answer: According to Snell's law:
\( \mu = \frac{\sin i}{\sin r} \)
Since glass is optically denser than air, the refractive index \( \mu \) is greater than 1:
\( \implies \frac{\sin i}{\sin r} > 1 \)
\( \implies \sin i > \sin r \)
Since the sine function is monotonically increasing for acute angles:
\( \implies i > r \)
Since the angle of refraction \( r \) is less than the angle of incidence \( i \), the refracted ray must bend toward the normal.
In simple words: Because glass is denser than air, it has a higher refractive index, which makes the light bend. Since the angle inside the glass is smaller than the angle in air, the ray bends closer to the normal line.

Exam Tip: Always show the step-by-step inequality \( \sin i > \sin r \implies i > r \) to demonstrate a mathematically complete proof.

Question 606. For the same angle of incidence, the angle of refraction in to two media A and B are \( 25^\circ \) and \( 35^\circ \) respectively. In which medium is the speed of light less ?
Answer: The speed of light is less in medium A.
From the definition of refractive index:
\( \mu = \frac{\sin i}{\sin r} = \frac{c}{v} \)
Since the angle of incidence \( i \) and the speed of light in vacuum \( c \) are constant:
\( v \propto \sin r \)
Given that the angle of refraction in medium A is smaller than in medium B (\( r_A = 25^\circ < r_B = 35^\circ \)):
\( \implies \sin r_A < \sin r_B \)
\( \implies v_A < v_B \)
Thus, the speed of light is less in medium A than in medium B.
In simple words: The smaller the angle of refraction, the more the light has been slowed down. Since the light bends more sharply in medium A (25°), it travels slower in medium A than in medium B.

Exam Tip: Use the proportionality \( v \propto \sin r \) to show a clear and elegant mathematical proof, which helps the examiner grade your answer quickly.

Question 607. Define refractive index of a transparent medium. What is the minimum and maximum value of refractive index ?
Answer: The refractive index (\( \mu \)) of a transparent medium is defined as the ratio of the speed of light in a vacuum (\( c \)) to the speed of light in that specific medium (\( v \)):
\( \mu = \frac{c}{v} \)
The minimum value of refractive index is \( 1 \) (for vacuum/air), and the maximum known value of refractive index for a natural transparent medium is \( 2.42 \) (for diamond).
In simple words: Refractive index is a number that tells us how much a medium slows down light compared to its speed in empty space. It is at least 1 for air, and can go up to 2.42 for diamond.

Exam Tip: Write down the formula \( \mu = \frac{c}{v} \) and state that it is a dimensionless quantity with no units.

Question 608. What is the ratio of the velocity of the wave in the two media of refractive indices \( \mu_1 \) and \( \mu_2 \)?
Answer: The refractive index of a medium is inversely proportional to the velocity of light in that medium:
\( \mu = \frac{c}{v} \implies v \propto \frac{1}{\mu} \)
Therefore, the ratio of the velocities \( v_1 \) and \( v_2 \) in the two media having refractive indices \( \mu_1 \) and \( \mu_2 \) is:
\( \frac{v_1}{v_2} = \frac{\mu_2}{\mu_1} \).
In simple words: Light travels slower in materials with higher refractive indices. Because of this inverse relationship, the ratio of the speeds is the flipped ratio of their refractive indices.

Exam Tip: Make sure the indices are flipped in the final ratio (\( 1 \) corresponds to \( 2 \) and vice versa) to represent the inverse relationship correctly.

Question 609. How does the refractive index of a transparent medium depend on wavelength of light used ?
Answer: The refractive index (\( \mu \)) of a transparent medium is inversely proportional to the wavelength of the incident light (\( \lambda \)). This relationship is mathematically defined by Cauchy's dispersion formula:
\( \mu = a + \frac{b}{\lambda^2} \)
where \( a \) and \( b \) are constants for a given medium. As a result, the refractive index decreases as the wavelength of light increases (light with a longer wavelength, like red, experiences a lower refractive index than light with a shorter wavelength, like violet).
In simple words: Materials bend light of different colors by different amounts. Since red light has a longer wavelength, it experiences a lower refractive index and bends less than violet light.

Exam Tip: Writing Cauchy's formula \( \mu = a + \frac{b}{\lambda^2} \) is highly recommended, as it provides a rigorous, textbook-aligned explanation.

Question 610. When a glass slab is placed on an ink dot, ink dot appears to be raised. Why ?
Answer: When light rays from the ink dot pass from the optically denser glass slab into the optically rarer air, they bend away from the normal due to refraction. To an observer viewing from above, these refracted rays appear to diverge from a point higher than the actual position of the dot. This apparent shift makes the ink dot appear raised.
In simple words: When light travels out of the glass slab into the air, it bends. This bending of light tricks our eyes into seeing the image of the dot higher up than its actual physical location.

Exam Tip: Use the term "apparent shift" and explain that it is caused by the bending of light rays as they transition from a denser to a rarer medium.

Question 611. By how much would an ink dot appear to be raised, when covered by a glass plate of thickness 6.0 cm. Refractive index of glass is 1.5.
Answer: We are given the thickness of the glass plate \( t = 6.0 \text{ cm} \) and the refractive index of glass \( \mu = 1.5 \).
The apparent shift (height by which the ink dot appears to be raised) is given by the formula:
\( \Delta y = t \left(1 - \frac{1}{\mu}\right) \)
Substituting the given values:
\( \Delta y = 6.0 \left(1 - \frac{1}{1.5}\right) \)
\( \implies \Delta y = 6.0 \left(1 - \frac{2}{3}\right) \)
\( \implies \Delta y = 6.0 \left(\frac{1}{3}\right) = 2.0 \text{ cm} \).
\( \implies \) The ink dot appears to be raised by \( 2.0 \text{ cm} \).
In simple words: When you look through a glass slab, objects underneath appear closer than they are. Using the apparent shift formula, we calculate that a 6 cm glass plate raises the image of the dot by 2 cm.

Exam Tip: Double check your fraction conversion: \( \frac{1}{1.5} \) is equivalent to \( \frac{2}{3} \), which simplifies the calculation significantly.

Question 612. The line AB in the ray diagram represents a lens. State whether the lens is convex or concave ?
(i)

(ii)

Answer: Based on the path of the refracted rays in the diagrams:
(i) Convex lens: The lens represented by AB is a convex (converging) lens because the incident parallel rays bend toward the principal axis after passing through it.
(ii) Concave lens: The lens represented by AB is a concave (diverging) lens because the incident parallel rays bend away from the principal axis after passing through it.
In simple words:
(i) Since the lens bends the incoming rays inward toward the center, it is a convex lens.
(ii) Since the lens spreads the incoming rays outward away from the center, it is a concave lens.

Exam Tip: Clearly state the reason—whether the ray bends toward or away from the principal axis—to support your identification of the lens type.

Question 613. What is total internal reflection of light ?
Answer: Whenever light propagating from a medium of higher optical density to one of lower optical density strikes the boundary at an incident angle larger than the critical angle, it experiences complete reflection back inside the denser medium. This physical occurrence is known as total internal reflection of light.
In simple words: When light hits a boundary at a very flat angle while trying to exit a denser material, it cannot escape and acts like it hit a mirror, bouncing straight back inside.

Exam Tip: Ensure you explicitly state both conditions (denser to rarer travel, and angle of incidence exceeding the critical angle) to secure maximum points in descriptive questions.

Question 614. State the conditions for the phenomenon of total internal reflection to occur.
Answer: To observe this optical behavior, two conditions are required:
(i) The path of the light must lead from an optically denser medium into a rarer medium.
(ii) The angle of incidence has to exceed the critical angle (\( i > i_c \)).
In simple words: Light must travel from a heavier medium to a lighter one, and it must strike the boundary at an angle wider than the critical angle.

Exam Tip: Always write the mathematical inequality \( i > i_c \) alongside your description of the conditions to show complete conceptual understanding.

Question 615. Name one phenomenon which is based on total internal reflection.
Answer: This principle is demonstrated by the brilliant sparkle of cut diamonds, the operation of optical fibers, and the formation of mirages in deserts.
In simple words: The sparkling of a diamond or the way optical fibers carry internet data are real-world examples of light bouncing internally.

Exam Tip: Optical fibers or mirages are highly recommended examples because they are frequently asked about in subsequent sections of the exam.

Question 616. Can total internal reflection occur when light goes from rarer to a denser medium ?
Answer: No, this phenomenon cannot take place because when light moves into a denser medium, it refracts toward the normal line rather than away from it.
In simple words: No, it is impossible for light to reflect internally if it is entering a denser material from a lighter one.

Exam Tip: This is a classic trick question. Total internal reflection is strictly restricted to light attempting to exit into an optically rarer medium.

Question 617. Define critical angle.What is the relationbetween refractive index & critical angle for a given pair of optical media ?
Answer: The critical angle is defined as the incidence angle within the denser medium that results in a refraction angle of exactly \( 90^\circ \) in the surrounding rarer medium. The mathematical connection between the refractive index and this angle is:
\[ \mu = \frac{1}{\sin i_c} \]
In simple words: The critical angle is the tipping point where light bends so much that it travels flat along the surface. The refractive index is just one divided by the sine of this angle.

Exam Tip: Draw a small, neat diagram showing the refracted ray skimming the boundary surface at \( 90^\circ \) to secure full marks.

Question 618. When light travels from an optically denser medium to a rarer medium, why does the critical angle of incidence depend on the colour/wavelength of light ?
Answer: Because the critical angle relates to the refractive index through \( i_c = \sin^{-1}\left(\frac{1}{\mu}\right) \), and the refractive index changes with wavelength as described by Cauchy's formula \( \mu = a + \frac{b}{\lambda^2} \), we find that different colors of light experience distinct refractive indices. Consequently, the critical angle varies for different colors and wavelengths of light.
In simple words: Different colors of light have different wavelengths, which changes how much they bend. Because of this, each color has its own unique critical angle.

Exam Tip: State Cauchy's formula clearly to justify the relation between wavelength and refractive index, which directly links to the critical angle.

Question 619. What is the critical angle for a material of refractive index \( \sqrt{2} \) ?
Answer: Using the standard formula relating refractive index and critical angle, we have:
\( \sin i_c = \frac{1}{\mu} \)
Substituting the given value:
\( \sin i_c = \frac{1}{\sqrt{2}} \)

\( \implies i_c = 45^\circ \)
This gives a critical angle of \( 45^\circ \) for the material.
In simple words: By plugging the index value into the sine formula, we find the critical angle is exactly 45 degrees.

Exam Tip: Write down the general formula first before substituting values to ensure step-wise marks even if you make a calculation error.

Question 620. Velocity of light in glass is \( 2 \times 10^8\text{ m/s} \) and in air is \( 3 \times 10^8\text{ m/s} \). If the ray of light passes from glass to air, calculate the value of critical angle.
Answer: First, determine the optical medium's refractive index:
\( \mu = \frac{v_{\text{air}}}{v_{\text{glass}}} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 \)
Using this index, find the sine of the critical angle:
\( \sin i_c = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3} \)

\( \implies i_c = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.8^\circ \)
The critical angle for this transition is approximately \( 41.8^\circ \).
In simple words: We find the refractive index by dividing the speed of light in air by its speed in glass, then use that to calculate the critical angle.

Exam Tip: Make sure to show the ratio of the speed of light in a vacuum/air to the speed of light in the medium to define the refractive index.

Question 621. Calculate the speed of light in a medium whose critical angle is \( 30^\circ \).
Answer: Start by finding the refractive index from the critical angle:
\( \mu = \frac{1}{\sin i_c} = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2 \)
Next, find the propagation speed of light in this medium:
\( \mu = \frac{c}{v} \)

\( \implies v = \frac{c}{\mu} \)
\( v = \frac{3 \times 10^8}{2} = 1.5 \times 10^8\text{ m/s} \)
Hence, light travels at a speed of \( 1.5 \times 10^8\text{ m/s} \) in this medium.
In simple words: A critical angle of 30 degrees means the refractive index is 2. Since the index is 2, light travels at half its speed in air.

Exam Tip: Always include the correct unit (\(\text{m/s}\)) in your final numerical answer to avoid losing half a mark.

Question 622. In the following ray diagram, calculate the speed of light in the liquid of unknown refractive index.
Answer: By analyzing the provided geometry, the hypotenuse is calculated as:
\( \sqrt{30^2 + 40^2} = 50\text{ cm} \)
Since the light ray undergoes critical refraction, we find:
\( \sin i_c = \frac{30}{50} = \frac{3}{5} \)
Using the relationship between speed and critical angle:
\( \frac{v}{c} = \sin i_c \)

\( \implies v = c \sin i_c = 3 \times 10^8 \times \frac{3}{5} = 1.8 \times 10^8\text{ m/s} \)
Therefore, the speed of light in this liquid is \( 1.8 \times 10^8\text{ m/s} \).

In simple words: Using the dimensions 30 cm and 40 cm, we find the diagonal is 50 cm. The sine of the critical angle is 30/50, which helps us find that light travels at 1.8 × 10^8 m/s in this liquid.

Exam Tip: Clearly show the calculation of the hypotenuse using the Pythagorean theorem before finding the value of \( \sin i_c \).

Question 623. Draw a ray diagram to show how a right angled isosceles prism can be used to- (i) deviate a light ray through 90°, (ii) deviate a light ray through 180°/ to obtain the inverted image (iii) to invert an image without the deviation of the rays ?
Answer: An isosceles right-angled prism can alter light paths by utilizing internal reflection, because the critical angle for crown glass (~ \( 42^\circ \)) is smaller than the \( 45^\circ \) angle of incidence.
(i) **To bend light by \( 90^\circ \)**:
Let light enter perpendicular to one of the perpendicular faces. It strikes the hypotenuse face at \( 45^\circ \), where it is totally reflected internally, exiting perpendicular to the second face. (ii) **To bend light by \( 180^\circ \)**:
Let the rays enter perpendicular to the hypotenuse face. They undergo total internal reflection twice (at both perpendicular faces) and emerge opposite to their original path, inverting the image. (iii) **To invert an image without any path deviation**:
Allow light to enter slanted through one perpendicular face. The rays refract, reflect once off the hypotenuse face, and refract again upon leaving the opposite face, emerging parallel but upside down.
In simple words: A 45-90-45 prism can redirect light because glass reflects light completely at 45 degrees. Depending on which side light enters, it can bend 90 degrees, do a full U-turn (180 degrees), or flip upside down while traveling straight.

Exam Tip: Make sure to mark the \( 45^\circ \) and \( 90^\circ \) angles clearly on the prism, and use arrows on your light rays to show the exact direction of travel.

Question 624. Why does a diamond sparkle ?
Answer: The glittering appearance of a diamond is primarily caused by total internal reflection. With a very high refractive index of \( 2.42 \), the diamond's critical angle is exceptionally small, close to \( 24.4^\circ \). The facets are intentionally cut at specific angles so that light entering the gemstone experiences multiple internal reflections before escaping through a limited number of faces, causing a concentrated sparkle.
In simple words: Diamond bends light so strongly that its critical angle is very small. Once light goes inside, it gets trapped and bounces around many times before exiting in bright bursts through a few faces.

Exam Tip: Do not forget to state the exact refractive index of diamond (\( 2.42 \)) and its critical angle (\( 24.4^\circ \)) as these key facts carry specific marks in the grading scheme.

Question 625. Find the relation between critical angle and refractive index.
Answer: According to Snell's law, we have:
\( \frac{\sin i}{\sin r} = \frac{1}{\mu} \)
When the angle of incidence equals the critical angle (\( i = i_c \)), the refracted ray emerges at \( 90^\circ \) (\( r = 90^\circ \)). Substituting these conditions gives:
\( \frac{\sin i_c}{\sin 90^\circ} = \frac{1}{\mu} \)
Since \( \sin 90^\circ = 1 \), this becomes:
\( \sin i_c = \frac{1}{\mu} \)

\( \implies \mu = \frac{1}{\sin i_c} \)
This represents the relation between refractive index and critical angle.
In simple words: Snell's law states how light bends at a boundary. When the incoming light is at the critical angle, the outgoing light bends at exactly 90 degrees, leaving us with a simple formula relating the index and the sine of that angle.

Exam Tip: Always specify that this relation holds true when light travels from a denser medium to a rarer medium.

Question 626. What is an optical fibre ? Name the phenomenon on which working of an optical fibre is based. Give any two uses of optical fibres.
Answer: An optical fiber consists of a very thin strand of high-quality quartz or glass, encased in an outer layer with a slightly lower refractive index, known as the cladding.
**Underlying Phenomenon**: Total internal reflection of light.
**Applications**:
(a) Transporting optical signals for communication networks.
(b) Serving as a light pipe for medical endoscopy procedures.
In simple words: An optical fiber is a tiny glass hair coated in another glass layer that guides light along its path using continuous internal bounces. It is used for fast internet and medical endoscopes.

Exam Tip: Be sure to mention the cladding layer and explain that its refractive index must be less than that of the core strand.

Question 627. Draw a labelled diagram of an optical fibre. Explain how light propagates through the optical fiber.
Answer: **Phenomenon**: Total internal reflection of light.
**Principle of Operation**: Once a light signal is introduced into one end of the core at an appropriate angle, it meets the core-cladding boundary at an angle greater than the critical angle. Consequently, it experiences continuous internal reflections along the fiber's length and eventually exits at the opposite end.
In simple words: Light is fed into the fiber at an angle. It bounces off the inner walls over and over again because it cannot cross into the lower-index coating, traveling all the way to the other end.

Exam Tip: In your diagram, clearly label the 'Core' (high refractive index) and 'Cladding' (low refractive index) to show how light remains trapped.

Question 628. What is scattering light ? What is the condition for Rayleigh scattering to occur ?
Answer: **Scattering of light**: This is the process where light rays are deflected from their straight paths when they interact with particles present in the medium.
**Condition for Rayleigh scattering**: The physical dimension of the scattering particle (\( a \)) must be significantly smaller than the light's wavelength (\( \lambda \)), which is represented as \( a \ll \lambda \).
In simple words: Scattering is when light hits tiny particles and gets splashed in all directions. Rayleigh scattering only happens when the particles are much smaller than the wavelength of the light.

Exam Tip: Always write the mathematical inequality \( a \ll \lambda \) to explicitly show the condition required for Rayleigh's law to be valid.

Question 629. Why cannot we see clearly through fog ? Name the phenomenon responsible for it.
Answer: Water droplets in the fog cause light rays to be scattered, absorbed, and deviated from their original directions, which obscures clear vision.
**Responsible Phenomenon**: Scattering of light.
In simple words: Fog drops scatter the light in different directions, which blurs everything and makes it hard for a clear image to reach our eyes.

Exam Tip: Name both "scattering" and "deviation of light" as the main reasons why visibility drops in foggy conditions.

Question 630. Why does bluish colour predominate in the sky ?
Answer: This is due to the intense scattering of blue light in the atmosphere. Because blue light has a short wavelength, and according to Rayleigh's law, the intensity of scattered light is inversely proportional to the fourth power of the wavelength:
\[ I \propto \frac{1}{\lambda^4} \]
the shorter wavelengths of blue are scattered far more than other colors, dominating the sky's appearance.
In simple words: Blue light has a very short wavelength, so it gets scattered in all directions by the air much more than other colors, making the whole sky look blue.

Exam Tip: Write down the proportional relationship \( I \propto \frac{1}{\lambda^4} \) to demonstrate the mathematical basis of Rayleigh scattering.

Question 631. Why does Sun appears red at sunrise and sunset ?
Answer: This happens because red light experiences the least scattering. Red light has the longest wavelength in the visible spectrum. Since Rayleigh's scattering law states that intensity is inversely proportional to the fourth power of wavelength:
\[ I \propto \frac{1}{\lambda^4} \]
most of the shorter wavelengths (like blue) are scattered away over the long atmospheric path, allowing the unscattered red light to reach our eyes.
In simple words: At sunrise and sunset, sunlight travels a long way through the air. The blue light gets scattered away early, while the longer red waves pass straight through to our eyes.

Exam Tip: Emphasize the long distance that light travels through the atmosphere during dawn and dusk compared to noon.

Question 632. Clouds appear white. Why ?
Answer: This is due to the uniform scattering of all visible colors. Since clouds consist of larger particles like water droplets, dust, and ice, they do not follow Rayleigh's scattering law. Instead, they scatter all wavelengths of light with equal efficiency, producing a white appearance.
In simple words: The water droplets in clouds are relatively large, so they scatter all colors of light equally. When all colors mix together, we see white.

Exam Tip: Contrast this with Rayleigh scattering by explaining that the particle size in clouds is much larger than the wavelength of light.

Question 633. Give reasons for the following observations on the surface of the moon : (a) Sunrise and sunset are abrupt. (b) Sky appears dark (c) a rainbow is never formed.
Answer:
(a) **Abrupt sunrise and sunset**: Since the moon lacks an atmosphere, there is no air to scatter or diffuse light gradually. Sunlight travels directly to the surface, making transitions between light and dark instantaneous.
(b) **Black sky**: The absence of an atmosphere on the moon means there are no particles to scatter sunlight. No scattered light fills the sky, so it remains as dark as night even during the day.
(c) **No rainbow formation**: Rainbows require water droplets to refract and reflect light. Since there is no water vapor, clouds, or rain on the lunar surface, a rainbow can never form.
In simple words: The moon has no air and no water. Because of this, light does not scatter, keeping the sky black and making day-night transitions instant, and there are no water drops to make rainbows.

Exam Tip: For atmospheric questions on the moon, always start by stating that "the moon has no atmosphere" as the primary scientific cause.

Question 634. Why is aperture of objective lens of a telescope is taken large ?
Answer: Using a larger aperture for a telescope's objective lens helps to maximize its light-collecting surface, which makes the observed image significantly brighter.
In simple words: A larger front lens collects more light, which helps make faint space objects look much brighter.

Exam Tip: Mention "light gathering capacity" and "brightness of image" together, as they are key terms evaluated in the grading scheme.

Question 635. State two main considerations taken into account while choosing the objective in optical telescopes with large diameters.
Answer: The primary factors to consider when designing a large-aperture objective lens are:
(i) Maximizing the light gathering capacity to observe faint details.
(ii) Achieving superior resolving power to distinguish close, distinct astronomical objects.
In simple words: When picking a main lens, you want it to capture as much light as possible and keep the final image sharp and clear.

Exam Tip: Structure your answer using clear bullet points specifying "resolving power" and "light gathering power" to gain full marks.

Question 636. The objective of a telescope is of larger focal length and of larger aperture (as compared to eye piece). Why ?
Answer: Both dimensions are selected for distinct optical reasons:
(i) Having a greater focal length for the objective increases the magnifying power, which is calculated as \( m = -\frac{f_o}{f_e} \).
(ii) Having a wider aperture increases the lens's light collection ability, leading to a much brighter image and offering a far higher resolution.
In simple words: A longer front lens makes the view bigger, while a wider front lens makes the view brighter and sharper.

Exam Tip: State the magnifying power formula \( m = -\frac{f_o}{f_e} \) to provide a mathematical basis for your first point.

Question 637. Why is eye piece of a telescope is of short focal length, while objective of large focal length ? Explain.
Answer: The angular magnification of an astronomical telescope is governed by the relation:
\[ m = -\frac{f_o}{f_e} \]
To achieve a high degree of magnification, the focal length of the objective lens (\( f_o \)) must be significantly larger than that of the eyepiece (\( f_e \)), meaning \( f_o \gg f_e \). Therefore, a long focal length is selected for the objective, and a short focal length is chosen for the eyepiece.
In simple words: To zoom in closely, the main lens needs a long focal length and the eyepiece needs a short one, because magnification is the first divided by the second.

Exam Tip: Always show the mathematical condition \( f_o \gg f_e \) when explaining the relative focal lengths.

Question 638. State the condition under which a large magnification can be achieved in an astronomical telescope.
Answer: Based on the telescope magnification formula \( m = -\frac{f_o}{f_e} \), high magnification is obtained through the following conditions:
(i) Selecting an objective with an extremely long focal length (\( f_o \)) and an eyepiece with a short focal length (\( f_e \)), such that \( f_o \gg f_e \).
(ii) Maintaining a separation distance between the two lenses that is greater than or equal to the sum of their focal lengths, i.e., \( L \ge f_o + f_e \).
In simple words: You get the best zoom when the main lens has a very long focal length, the eyepiece has a short one, and they are spaced correctly inside the tube.

Exam Tip: Write both conditions (the focal length ratio and the minimum tube length) to answer this question comprehensively.

Question 639. You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope ? Give reason.
Answer: To build a high-performing astronomical telescope, we should choose:
**Objective lens**: Lens \( L_1 \)
**Eyepiece**: Lens \( L_3 \)
**Reason**: The magnification of a telescope is \( m = -\frac{f_o}{f_e} \). Because focal length is inversely proportional to power (\( f = \frac{1}{P} \)), we have:
\( f_1 = \frac{1}{3}\text{ m} \approx 33.3\text{ cm} \)
\( f_2 = \frac{1}{6}\text{ m} \approx 16.7\text{ cm} \)
\( f_3 = \frac{1}{10}\text{ m} = 10\text{ cm} \ encampment \)
For maximum magnification and light collection, the objective must have the longest focal length and the widest aperture (Lens \( L_1 \), which has \( f = 33.3\text{ cm} \) and aperture \( 8\text{ cm} \)). Conversely, the eyepiece must have the smallest focal length and aperture to focus the light into the eye (Lens \( L_3 \), which has \( f = 10\text{ cm} \) and aperture \( 1\text{ cm} \)).
In simple words: We pick the lens with the lowest power and widest opening for the front because it gathers the most light. We use the highest power lens for the eyepiece to get the highest zoom.

Exam Tip: Always show the step-by-step conversion from lens power to focal length using \( f = \frac{1}{P} \) to validate your selection.

Question 640. You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope. Which lenses should you use as an objective and eyepiece of an astronomical telescope ? Justify your answer.
Answer: To construct the telescope, the lenses should be selected as follows:
**Objective lens**: \( 0.5\text{ D} \) lens
**Eyepiece**: \( 10\text{ D} \) lens
**Justification**: Since magnifying power is defined as \( m = -\frac{f_o}{f_e} \), and focal length is the reciprocal of optical power (\( f = \frac{1}{P} \)), the lens with the lowest power (\( 0.5\text{ D} \)) will have the longest focal length (\( f_o = 2\text{ m} \)). The lens with the highest power (\( 10\text{ D} \)) will have the shortest focal length (\( f_e = 0.1\text{ m} \)). This combination achieves the maximum possible magnification.
In simple words: The lowest power lens gives the longest focal length, so we put it at the front. The highest power lens has the shortest focal length, making it ideal for the eyepiece.

Exam Tip: When justifying, relate power directly to focal length and state the formula for magnifying power.

Question 641. Write two main limitations of refracting telescopes. Explain how these can be minimized in a reflecting telescope.
Answer: Two major issues associated with refracting telescopes are:
(i) **Spherical Aberration**: Light rays passing near the edges of a spherical lens focus at a different point than those passing through the center. This is resolved in reflecting telescopes by using a parabolic mirror instead of a spherical lens.
(ii) **Chromatic Aberration**: Different wavelengths of light refract at slightly different angles through glass, creating colored fringes around images. Reflecting telescopes eliminate this completely by using a primary mirror as the objective, as reflection is independent of color wavelength.
Additionally, reflecting telescopes can be built with much wider apertures, which greatly improves resolving power and image brightness compared to heavy refracting lenses.
In simple words: Lens telescopes suffer from blurry edges (spherical aberration) and rainbow halos (chromatic aberration). Mirror telescopes fix these problems by using curved mirrors instead of glass lenses.

Exam Tip: Clearly pair each refracting limitation with its corresponding correction in reflecting systems to show structured knowledge.

Question 642. Give two reasons to explain why a reflecting telescope is preferred over a refracting telescope.
OR
State the advantages of reflecting telescope over refracting telescope.
Answer: Reflecting telescopes are preferred because of several optical benefits:
(i) **No Aberrations**: Since a primary mirror is used as the objective rather than a lens, both chromatic and spherical aberrations are completely eliminated.
(ii) **Better Image Quality**: Large-aperture parabolic mirrors are easier to manufacture and support than massive lenses. This allows reflecting telescopes to collect more light, producing brighter images and offering much higher resolution.
In simple words: Mirror-based telescopes do not create blurry or rainbow-edged images, and they can be made much larger to capture faint, distant objects with great clarity.

Exam Tip: Highlighting the elimination of chromatic aberration is crucial for scoring full marks in comparison questions.

Question 643. (i) Draw a schematic diagram of a reflecting telescope. State the advantages of reflecting telescope over refracting telescope.
(ii) What is its magnifying power ?
Answer: Reflecting telescopes offer several advantages over refracting ones:
(a) They do not suffer from chromatic aberration because they use mirrors rather than lenses.
(b) Spherical aberration is minimized by employing paraboloidal primary mirrors.
(c) They produce brighter, higher-resolution images as primary mirrors can be made with massive, well-supported apertures.
(d) They offer excellent magnification.
**Magnifying Power**: The magnifying power (\( m \)) of a reflecting telescope is the ratio of the angle subtended at the eye by the final image to the angle subtended by the object directly:
\[ m = \frac{f_o}{f_e} \]
In simple words: A mirror-based telescope produces clear, color-accurate images. Its magnifying power is determined by dividing the focal length of the main mirror by that of the eyepiece.

Exam Tip: Make sure your diagram clearly labels the primary paraboloidal mirror, secondary mirror, and eyepiece to get full marks on the schematic.

Question 644. Does the magnifying power of a microscope depend on the colour of the light used ? Justify your answer.
Answer: Yes, magnifying power depends on the color of the light. The magnification of a compound microscope is inversely proportional to the product of the focal lengths of its lenses (\( m \propto \frac{1}{f_o f_e} \)). Since the refractive index of glass is higher for shorter wavelengths (like violet) and lower for longer wavelengths (like red), the focal length of a lens is shorter for violet light and longer for red light. Therefore, using blue or violet light yields higher magnifying power than red light.
In simple words: Yes. Different colors of light bend differently through lenses, which changes their focal lengths. Shorter wavelengths like blue or violet result in higher magnification.

Exam Tip: Mention the inverse relationship between magnifying power and focal length, and how refractive index depends on the wavelength of light.

Question 645. Explain, why must both the objective and the eye piece of a compound microscope have short focal lengths ?
Answer: The magnifying power of a compound microscope is given by the formula:
\[ m \approx \frac{L}{f_o} \times \frac{D}{f_e} \]
Here, \( L \) is the tube length, and \( D \) is the least distance of distinct vision. To maximize the magnifying power \( m \), both the focal length of the objective (\( f_o \)) and the focal length of the eyepiece (\( f_e \)) must be kept as small as possible.
In simple words: To make small things look as big as possible, we need lenses that bend light very strongly, which means they must have very short focal lengths.

Exam Tip: Use the mathematical expression for magnifying power to demonstrate how reducing the denominators (\( f_o \) and \( f_e \)) directly increases the product.

Question 646. Explain, why is the objective of a compound microscope be of short aperture ?
Answer: The objective lens is designed with a small aperture for two primary reasons:
(i) **Minimizing Aberration**: A small opening keeps light rays close to the optical axis, which reduces spherical aberration and produces a sharper image.
(ii) **Concentrating Light**: Since the specimen is placed extremely close to the objective lens, a small aperture is sufficient to collect the scattered light rays and form a highly focused, bright image.
In simple words: A small front lens on a microscope helps prevent blurry edges and keeps the image sharp and clear.

Exam Tip: Emphasize "spherical aberration control" as it is the key technical phrase examiners look for.

Question 647. Explain, While viewing through a compound microscope, why should our eyes be positioned not on the eye piece but a short distance away from it for best viewing ?
Answer: When using a microscope, our eyes should be positioned slightly behind the eyepiece at the "eye ring" (or exit pupil). Placing the eye at this location allows it to collect all the light rays refracted by the optical system, maximizing both the brightness of the image and the width of the field of view.
In simple words: Looking from a tiny distance back from the eyepiece lets you see the entire circular view with maximum brightness instead of just a small, dim spot in the middle.

Exam Tip: Mention "collecting all refracted rays" and "maximizing the field of view" to construct a complete, high-scoring answer.

Question 648. You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope ? Give reason.
Answer: To build a compound microscope, we should select:
**Objective lens**: Lens \( L_3 \)
**Eyepiece**: Lens \( L_2 \)
**Reason**: In a compound microscope, both lenses must have short focal lengths to ensure high magnification (\( m \approx \frac{L \cdot D}{f_o \cdot f_e} \)). Additionally, the objective must have the absolute shortest focal length and a very small aperture to minimize aberrations:
\( f_1 \approx 33.3\text{ cm} \), \( f_2 \approx 16.7\text{ cm} \), and \( f_3 = 10\text{ cm} \).
Therefore, we select Lens \( L_3 \) (with the smallest focal length of \( 10\text{ cm} \) and aperture \( 1\text{ cm} \)) as the objective, and Lens \( L_2 \) (with the next shortest focal length of \( 16.7\text{ cm} \) and aperture \( 1\text{ cm} \)) as the eyepiece.
In simple words: For a microscope, both lenses should be high-power (short focal length). We use the highest power lens (\( L_3 \)) as the front objective, and the next highest power lens (\( L_2 \)) as the eyepiece.

Exam Tip: Clearly explain that for a microscope, the objective's focal length must be smaller than that of the eyepiece (\( f_o < f_e \)), which is the opposite of a telescope.

Question 649. What is dispersion of light ? What is its cause ?
Answer: **Dispersion**: This refers to the splitting of a beam of composite white light into its seven individual constituent colors when it passes through a refracting medium like a glass prism.
**Cause**: The refractive index of glass (\( \mu \)) varies for different colors of light. According to the deviation formula \( \delta = (\mu - 1)A \ ), each color deviates by a unique angle upon entering the prism, causing the colors to separate into a visible spectrum.
In simple words: Dispersion is when white light gets split into a rainbow after passing through a prism. It happens because each color of light bends by a slightly different angle inside glass.

Exam Tip: Mention the formula \( \delta = (\mu - 1)A \) to show how the deviation angle relates to the refractive index of each color.

Question 650. How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light ? Give reason.
Answer: The angle of minimum deviation will **decrease**.
**Reason**: According to the relation \( \delta = (\mu - 1)A \), the deviation is directly proportional to the refractive index. Since the refractive index of glass is smaller for red light than for violet light (\( \mu_{\text{red}} < \mu_{\text{violet}} \)) because of red's longer wavelength (\( \lambda_{\text{red}} > \lambda_{\text{violet}} \)), red light bends less than violet light.
In simple words: The deviation decreases. Red light has a longer wavelength and bends less than violet light, so it takes a straighter path through the prism.

Exam Tip: State the inequality \( \mu_{\text{red}} < \mu_{\text{violet}} \) to provide a rigorous, clear scientific explanation.

Question 651. Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism. Give reason
Answer: Based on the expression \( \delta = (\mu - 1)A \), the angle of deviation depends directly on the refractive index of the medium. Since glass has a higher refractive index for violet light than for any other color in the visible spectrum (\( \mu_{\text{violet}} > \mu_{\text{red}} \)), violet light experiences the greatest deviation (\( \delta_{\text{violet}} > \delta_{\text{red}} \)) and is bent the most, placing it at the very bottom of the dispersed spectrum.
In simple words: Violet light bends the most of all colors when passing through glass, which is why it always ends up at the bottom of the rainbow spectrum.

Exam Tip: Connect refractive index directly to the deviation angle using the inequality \( \delta_{\text{violet}} > \delta_{\text{red}} \).

Question 652. Out of blue and red light which is more deviate by prism ? Give reason.
Answer: Blue light deviates more than red light when passing through a prism.
**Reason**: According to the formula \( \delta = (\mu - 1)A \), deviation is determined by the refractive index. Since the refractive index of the prism is greater for blue light than for red light (\( \mu_{\text{blue}} > \mu_{\text{red}} \)), blue light undergoes a larger angle of deviation (\( \delta_{\text{blue}} > \delta_{\text{red}} \)).
In simple words: Blue light bends more than red light because glass has a higher refractive index for blue than for red.

Exam Tip: Always relate the comparison directly to the wavelength and refractive index of the two specified colors.

Question 653. For which colour the refractive index of prism material is maximum and minimum ?
Answer: According to Cauchy's equation, the refractive index is related to the wavelength by:
\[ \mu = a + \frac{b}{\lambda^2} \]
Since violet light has the shortest wavelength in the visible spectrum (\( \lambda_{\text{violet}} < \lambda_{\text{red}} \)), the refractive index is **maximum for violet**. Conversely, because red light has the longest wavelength, the refractive index is **minimum for red**.
In simple words: The refractive index is highest for violet light and lowest for red light because violet has the shortest wavelength and red has the longest.

Exam Tip: State Cauchy's formula clearly to show the mathematical relationship between refractive index and wavelength.

Question 654. How is the focal length of a spherical mirror affected, when the wavelength of light used is increased ?
Answer: There is no effect on the focal length. The focal length of a spherical mirror is determined solely by its radius of curvature (\( f = \frac{R}{2} \)) and is entirely independent of the wavelength or color of light used.
In simple words: The focal length of a mirror does not change at all, because mirrors reflect all colors of light in the exact same way.

Exam Tip: Emphasize that the formula \( f = \frac{R}{2} \) depends only on the physical geometry of the mirror, not on the properties of light.

Question 655. How is the focal length of a spherical mirror is affected, when it is immersed in water/Glycerin ?
Answer: The focal length remains completely unchanged. This is because reflection occurs at the silvered surface of the mirror, meaning its focal length (\( f = \frac{R}{2} \)) depends only on its physical shape, not on the surrounding refractive medium.
In simple words: Immersing a mirror in water or glycerin has no effect on its focal length, as mirrors do not rely on refraction to focus light.

Exam Tip: Contrast mirrors with lenses by pointing out that mirror focal lengths are independent of the surrounding medium.

Question 657. How is the focal length of a spherical lens affected, when the wavelength of light used is increased ?
Answer: The focal length of the lens will **increase**.
**Reason**: According to the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
And according to Cauchy's relation, the refractive index decreases as the wavelength increases (\( \mu = a + \frac{b}{\lambda^2} \)). Since increasing the wavelength of light lowers the refractive index (\( \mu \)), the value of \( (\mu - 1) \) decreases, which consequently increases the focal length (\( f \)).
In simple words: The focal length increases. Longer wavelengths of light bend less when passing through glass, so they focus at a point further away from the lens.

Exam Tip: Clearly link the change in wavelength to the refractive index first, then show how that affects focal length through the lens maker's formula.

Question 658. How does focal length of a convex lens change, if violet light is used instead of red light ?
Answer: The focal length of the lens will **decrease**.
**Reason**: From the lens maker's formula, the focal length is inversely proportional to \( (\mu - 1) \). Since violet light has a shorter wavelength than red light (\( \lambda_{\text{violet}} < \lambda_{\text{red}} \)), the refractive index of the lens is higher for violet than for red (\( \mu_{\text{violet}} > \mu_{\text{red}} \)). This higher refractive index causes violet light to bend more sharply, resulting in a shorter focal length (\( f_{\text{violet}} < f_{\text{red}} \)).
In simple words: The focal length gets shorter because violet light bends more strongly than red light, focusing closer to the lens.

Exam Tip: State the final inequality \( f_{\text{violet}} < f_{\text{red}} \) to summarize your explanation clearly.

Question 659. Explain with reason, how the power of a diverging lens changes when incident red light is replaced by violet light.
Answer: The optical power of the diverging lens will **increase** (becoming more negative).
**Reason**: The power of a lens is given by \( P = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \). Since violet light has a shorter wavelength than red light (\( \lambda_{\text{violet}} < \lambda_{\text{red}} \)), the refractive index is greater for violet light (\( \mu_{\text{violet}} > \mu_{\text{red}} \)). This increased refractive index results in stronger divergence, which corresponds to a greater optical power (\( P_{\text{violet}} > P_{\text{red}} \)).
In simple words: The bending power of the lens increases because violet light has a shorter wavelength and gets scattered or bent outward more strongly than red light.

Exam Tip: Remember that for a diverging lens, an increase in power means it becomes more strongly divergent (more negative).

Question 660. What happens to the focal length of a convex lens when it is immersed in water ? Refractive index of the material of lens is greater than that of water.
Answer: The focal length of the lens will **increase**, which means its optical power will **decrease**.
**Reason**: The lens maker's formula in a surrounding medium is:
\[ \frac{1}{f_m} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
When immersed in water, the surrounding refractive index (\( \mu_m \)) increases from \( 1 \) (air) to \( 1.33 \) (water). This reduces the relative refractive index ratio \( \frac{\mu_g}{\mu_m} \), making the term \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) smaller, which increases the focal length \( f_m \).
In simple words: The focal length increases and its power drops. Water bends light less than air does, so light passing through the glass lens in water doesn't bend as sharply and takes longer to focus.

Exam Tip: Clearly write down the relative refractive index term \( \frac{\mu_g}{\mu_m} \) to show how the surrounding medium alters the lens's focusing power.

Question 661. A lens of glass is immersed in water. What will be its effect on the power of lens ?
Answer: The power of the lens will **decrease**.
**Reason**: The power of a lens immersed in a medium is given by:
\[ P_m = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Because the refractive index of water (\( \mu_m \)) is greater than that of air, the factor \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) becomes smaller, directly reducing the overall optical power of the lens.
In simple words: The lens loses some of its power to bend light when placed in water, because the difference in optical density between glass and water is smaller than between glass and air.

Exam Tip: State that power is directly proportional to \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) to clearly justify the reduction.

Question 662. Draw a plot showing the variation of power of a lens with the wavelength of incident ligh
Answer: The power of a lens decreases as the wavelength of the incident light increases.
**Reason**: Power is given by \( P = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \), and the refractive index varies with wavelength as \( \mu = a + \frac{b}{\lambda^2} \). As wavelength (\( \lambda \)) increases, the refractive index (\( \mu \)) decreases, which directly reduces the optical power (\( P \)).
In simple words: As the wavelength of light increases (moving from violet to red), the lens bends the light less, so its optical power decreases.

Exam Tip: Draw the curve sloping downwards from left to right, clearly labeling both axes.

Question 663. A glass lens of refractive index 1.45 disappears when immersed in a liquid. What is the value of refractive index of the liquid ?
Answer: For the glass lens to become invisible in the liquid, the refractive index of the liquid must be exactly equal to that of the glass lens, which is \( 1.45 \). At this value, light passes through the boundary without undergoing any refraction or reflection.
In simple words: The liquid must have a refractive index of 1.45. Since both materials bend light by the exact same amount, light travels straight through without bending, making the lens invisible.

Exam Tip: State that the disappearance occurs when there is no refraction at the liquid-lens interface, meaning \( \mu_{\text{liquid}} = \mu_{\text{lens}} \).

Question 664. What should be the value of the refractive index of the medium in which the lens should be placed so that it acts as a plane sheet of glass ?
OR
Under what condition does a biconvex lens of glass having a certain refractive index acts as a plane glass sheet when immersed in a liquid ?
Answer: The lens will behave as a simple flat glass sheet when the refractive index of the surrounding medium is identical to the refractive index of the lens material (\( \mu_{\text{medium}} = \mu_{\text{lens}} \)). Under this condition, the focal length becomes infinite (\( f \to \infty \)) and the power becomes zero (\( P = 0 \)).
In simple words: The surrounding liquid must have the exact same refractive index as the lens. When they match, light passes straight through without bending, behaving as if the lens were a flat glass sheet.

Exam Tip: Mention that when the refractive indices match, the focal length of the lens becomes infinite, which is characteristic of a plane glass sheet.

Question 665. Explain with reason, how the power of a diverging lens changes when it is kept in a medium of refractive index greater than that of the lens.
Answer: The optical power of the lens will change sign and become positive, meaning the diverging lens will begin to behave as a converging lens.
**Reason**: According to the lens maker's formula:
\[ P = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
For a diverging lens, the curvature term is negative. If the surrounding medium's refractive index is greater than that of the lens (\( \mu_m > \mu_g \)), the ratio \( \frac{\mu_g}{\mu_m} \) is less than \( 1 \), making the term \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) negative. The product of these two negative factors results in a positive power, changing its nature to converging.
In simple words: The lens changes its behavior completely and starts converging light. When the surrounding liquid is optically denser than the glass, the lens acts in the opposite way to how it does in air.

Exam Tip: Clearly explain that the sign of the focal length/power reverses when \( \mu_{\text{medium}} > \mu_{\text{lens}} \).

Question 666. A biconcave lens made of transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave a converging or diverging lens ? Give reason.
Answer: The lens will behave as a **converging lens**.
**Reason**: Using the lens maker's formula for a biconcave lens in a medium:
\[ \frac{1}{f_m} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(-\frac{1}{R_1} - \frac{1}{R_2}\right) = -\left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) \]
Since the refractive index of the lens (\( \mu_g = 1.25 \)) is less than that of water (\( \mu_m = 1.33 \)), the term \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) is negative. This negative value multiplies by the negative sign outside the brackets, resulting in a positive focal length (\( f_m > 0 \)). Consequently, the lens converges light.
In simple words: It behaves as a converging lens. Because water is denser than the lens material, the biconcave lens flips its usual behavior and bends light inward instead of outward.

Exam Tip: Show the numerical comparison \( \mu_g < \mu_m \) (\( 1.25 < 1.33 \)) to prove why the focal length becomes positive.

Question 667. A biconvex lens made of transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave a converging or diverging lens ? Give reason.
Answer: The lens will act as a **diverging lens**.
**Reason**: From the lens maker's formula for a biconvex lens:
\[ \frac{1}{f_m} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) \]
Since the refractive index of the surrounding medium (water, \( \mu_m = 1.33 \)) is greater than the refractive index of the lens glass (\( \mu_g = 1.25 \)), the relative term \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) becomes negative. This negative value makes the focal length negative (\( f_m < 0 \)), causing the lens to diverge light.
In simple words: The lens acts as a diverging lens because water is optically denser than the glass. This reverses the normal focusing behavior, causing light to spread out instead of coming together.

Exam Tip: Clearly write down the comparison \( \mu_m > \mu_g \) to justify the sign change of the focal length.

Question 668. A biconvex lens made of transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave a converging or diverging lens ? Give reason.
Answer: The lens will act as a **converging lens**.
**Reason**: The lens maker's equation gives:
\[ \frac{1}{f_m} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) \]
Since the refractive index of the lens (\( \mu_g = 1.5 \)) is larger than that of water (\( \mu_m = 1.33 \)), the relative refractive term \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) remains positive. This keeps the focal length positive (\( f_m > 0 \)), meaning the lens continues to converge light.
In simple words: Since the glass is still denser than the water, the lens keeps its positive focal length and continues to bring light rays together.

Exam Tip: Note that while the lens remains converging, its focal length will increase compared to its focal length in air.

Question 669. A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in
(i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33
Will the lens behave a converging or diverging lens in the two cases ? Give reason.
Answer:
(i) **Diverging lens**: When dipped in the first medium (\( \mu_m = 1.65 \)), the surrounding medium is optically denser than the glass (\( \mu_m > \mu_g \)). This causes the relative term \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) to become negative, resulting in a negative focal length (\( f_m < 0 \)).
(ii) **Converging lens**: When dipped in the second medium (\( \mu_m = 1.33 \)), the glass is optically denser than the medium (\( \mu_m < \mu_g \)). This keeps the relative term \( \left(\frac{\mu_g}{\mu_m} - 1\right) \) positive, resulting in a positive focal length (\( f_m > 0 \)).
In simple words: (i) When the liquid is denser than glass, the lens acts as a diverging lens. (ii) When the glass is denser than the liquid, it behaves normally as a converging lens.

Exam Tip: This is a common multi-part conceptual question. Solve each sub-case systematically by comparing the refractive index of the lens to that of the medium.

Question 670. A converging lens is kept coaxially in contact with a diverging lens, both the lenses being of equal focal length. What is the focal length of the combination ?
Answer: The focal length of the combined system will be **infinite** (\( F = \infty \)).
**Reason**: The equivalent focal length \( F \) of two thin lenses in contact is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
For a converging lens of focal length \( +f \) and a diverging lens of focal length \( -f \), this becomes:
\[ \frac{1}{F} = \frac{1}{f} - \frac{1}{f} = 0 \]

\( \implies F = \infty \)
Consequently, the lens combination behaves like a flat, plain glass plate.
In simple words: Since the two lenses bend light in opposite directions with the exact same strength, their effects cancel out completely, and light passes straight through like a flat window pane.

Exam Tip: Be sure to state that the focal length is infinite and describe the physical behavior (acting as a plane glass plate) for full credit.

Question 671. Two thin lenses of power +6 D and -2 D are in contact. What is the focal length of this combination ?
Answer: The focal length of the combination is **\( 25\text{ cm} \)** (or \( 0.25\text{ m} \)).
**Reason**: First, calculate the total power \( P \) of the combined lenses in contact:
\( P = P_1 + P_2 = +6\text{ D} + (-2\text{ D}) = +4\text{ D} \)
Next, find the effective focal length \( F \):
\( F = \frac{1}{P} = \frac{1}{4}\text{ m} = 0.25\text{ m} \)

\( \implies F = 25\text{ cm} \)
The positive sign indicates that the system acts as a converging lens.
In simple words: We add the two powers together to get a total power of +4 D, which corresponds to a focal length of 25 cm.

Exam Tip: Always specify focal length in both meters and centimeters to ensure your answer complies with any particular unit requested by the examiner.

Question 672. A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm . Determine the power of the combination will the system be converging or diverging in nature ?
Answer: The power of the combination is **\( -1\text{ D} \)**, and the system is **diverging** in nature.
**Reason**: The focal lengths of the convex and concave lenses are \( f_1 = +25\text{ cm} = +0.25\text{ m} \) and \( f_2 = -20\text{ cm} = -0.20\text{ m} \), respectively. The total power of the combination is:
\( P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} \)
\( P = \frac{1}{0.25} + \frac{1}{-0.20} = +4\text{ D} - 5\text{ D} = -1\text{ D} \)
The equivalent focal length is:
\( F = \frac{1}{P} = -1\text{ m} = -100\text{ cm} \)
Since the net power and focal length are negative, the combination behaves as a diverging system.
In simple words: The concave lens is stronger than the convex lens, so the combined system ends up behaving like a diverging lens with a power of -1 D.

Exam Tip: Convert focal lengths to meters before calculating power directly to avoid decimal errors.

Question 673. The focal length of a convex lens made of glass(\( \mu_g = 1.5 \)) is 22 cm. What will be its new focal length when placed in a medium of refractive index 4/3 ?
Answer: The new focal length in the medium will be **\( 88\text{ cm} \)**.
**Reason**: The relation between focal lengths in a medium and air is given by:
\[ f_m = \left[\frac{\mu_g - 1}{\frac{\mu_g}{\mu_m} - 1}\right] f_a \]
Substitute the given values (\( \mu_g = 1.5 = \frac{3}{2} \), \( \mu_m = \frac{4}{3} \), and \( f_a = 22\text{ cm} \)):
\[ f_m = \left[\frac{\frac{3}{2} - 1}{\frac{3/2}{4/3} - 1}\right] \times 22 \]
\[ f_m = \left[\frac{1/2}{9/8 - 1}\right] \times 22 = \left[\frac{1/2}{1/8}\right] \times 22 = 4 \times 22 = 88\text{ cm} \]
The focal length increases by a factor of 4.
In simple words: When the glass lens is placed in water, it bends light less, which quadruples its focal length from 22 cm to 88 cm.

Exam Tip: Remember that immersing a glass lens (\( \mu = 1.5 \)) in water (\( \mu = 1.33 \)) always increases its focal length by exactly four times.

Question 674. A double convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm.
Answer: The required radius of curvature for each face is **\( 22\text{ cm} \)**.
**Reason**: According to the lens maker's formula for a biconvex lens with equal radii of curvature (\( R_1 = R \), \( R_2 = -R \)):
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R} - \left(-\frac{1}{R}\right)\right) = (\mu - 1)\left(\frac{2}{R}\right) \]
Given \( \mu = 1.55 \) and \( f = 20\text{ cm} \), substitute these values:
\[ \frac{1}{20} = (1.55 - 1)\left(\frac{2}{R}\right) \]
\[ \frac{1}{20} = 0.55 \times \frac{2}{R} = \frac{1.10}{R} \]

\( \implies R = 1.10 \times 20 = 22\text{ cm} \)
Thus, the radius of curvature of each surface must be \( 22\text{ cm} \).
In simple words: By plugging the focal length of 20 cm and refractive index of 1.55 into the lens formula, we calculate that each curved face must have a radius of 22 cm.

Exam Tip: Pay careful attention to the sign convention for the radii of curvature (\( R_1 = +R \) and \( R_2 = -R \)) during your derivations.

Question 675. The focal length of an equiconvex lens is equal to the radius of curvature of either face. What is the refractive index of the material of the lens ?
Answer: The refractive index of the lens material is **\( 1.5 \)**.
**Reason**: For an equiconvex lens, the lens maker's formula simplifies to:
\[ \frac{1}{f} = (\mu - 1)\left(\frac{2}{R}\right) \]
We are given that the focal length equals the radius of curvature (\( f = R \)). Substituting this into the formula:
\[ \frac{1}{R} = (\mu - 1)\left(\frac{2}{R}\right) \]
Dividing both sides by \( \frac{1}{R} \):
\[ 1 = 2(\mu - 1) \implies \mu - 1 = \frac{1}{2} = 0.5 \]

\( \implies \mu = 1.5 \)
Hence, the refractive index of the lens material is \( 1.5 \).
In simple words: If the focal length is the same as the radius of curvature, the math shows that the lens material must have a refractive index of exactly 1.5.

Exam Tip: This is a classic short-answer derivation. Simplify the formula before performing algebraic operations to avoid mistakes.

Question 676. The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If the focal length of the lens is 12 cm, find the refractive index of the material of the lens ?
Answer: The refractive index of the lens material is **\( 1.5 \)**.
**Reason**: Using the lens maker's formula with sign conventions (\( R_1 = +10\text{ cm} \), \( R_2 = -15\text{ cm} \), and \( f = 12\text{ cm} \)):
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
\[ \frac{1}{12} = (\mu - 1)\left(\frac{1}{10} - \left(-\frac{1}{15}\right)\right) \]
\[ \frac{1}{12} = (\mu - 1)\left(\frac{1}{10} + \frac{1}{15}\right) \]
Find a common denominator for the fractions:
\[ \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} \]
Substitute this back into the equation:
\[ \frac{1}{12} = (\mu - 1)\left(\frac{1}{6}\right) \]
\[ \mu - 1 = \frac{6}{12} = 0.5 \]

\( \implies \mu = 1.5 \)
Thus, the refractive index of the material is \( 1.5 \).
In simple words: Substituting the curved face sizes and focal length into the lens maker's equation shows that the glass has a refractive index of 1.5.

Exam Tip: Ensure you correctly assign a positive sign to \( R_1 \) and a negative sign to \( R_2 \) when plugging values into the formula.

Question 678. A concave mirror produces a real and magnified image of an object kept in front of it. Draw a ray diagram to show The image formation and use it to derive the mirror equation.
Answer: Let's derive the mirror formula for a concave mirror forming a real, magnified image:
**Derivation**: Object \( AB \) is placed between \( C \) and \( F \). An incident ray parallel to the principal axis reflects through the focus \( F \). Another ray heading towards the pole \( P \) reflects at an equal angle. They intersect to form a real, inverted, magnified image \( A'B' \) beyond \( C \).
From the geometry of the similar triangles:
1. \( \triangle ABC \) and \( \triangle A'B'C \) are similar:
\[ \frac{A'B'}{AB} = \frac{CB'}{CB} = \frac{PC - PB'}{PB - PC} \quad \text{--- (1)} \]
2. \( \triangle ABP \) and \( \triangle A'B'P \) are similar:
\[ \frac{A'B'}{AB} = \frac{PB'}{PB} \quad \text{--- (2)} \]
Equating (1) and (2):
\[ \frac{PC - PB'}{PB - PC} = \frac{PB'}{PB} \]
Applying Cartesian sign conventions: let \( PB = -u \) (object distance), \( PB' = -v \) (image distance), and \( PC = -R = -2f \) (radius of curvature).
Substitute these values:
\[ \frac{-2f - (-v)}{-u - (-2f)} = \frac{-v}{-u} \]
\[ \frac{v - 2f}{2f - u} = \frac{v}{u} \]
Cross-multiplying gives:
\[ u(v - 2f) = v(2f - u) \implies uv - 2uf = 2vf - uv \]
Rearranging terms:
\[ 2uv = 2vf + 2uf \]
Dividing the entire equation by \( 2uvf \) yields:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
This is the mirror equation.
In simple words: By using the geometry of similar triangles formed by the object, the mirror, and its image, we can mathematically prove that the reciprocals of the object and image distances add up to the reciprocal of the focal length.

Exam Tip: Make sure to clearly state that the distances are measured from the pole \( P \) and apply the correct signs (all values \( u \), \( v \), and \( f \) are negative in this case) to get full derivation marks.

Question 679. A point object O on the principal axis of a spherical surface of radius separating two media of refractive indices \( \mu_1 \) and \( \mu_2 \) forms an image as shown in the figure. Prove that \( \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \)
Answer: Let us derive the relation for refraction at a spherical surface separating two media of indices \( \mu_1 \) and \( \mu_2 \):
**Derivation**:
According to Snell's law, we have:
\[ \frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} \]
For paraxial rays, the angles of incidence \( i \) and refraction \( r \) are very small, so \( \sin i \approx i \) and \( \sin r \approx r \). This simplifies the relation to:
\[ \mu_1 i = \mu_2 r \quad \text{--- (1)} \]
From the geometry of the triangle \( \triangle OAC \) (exterior angle theorem):
\[ i = \alpha + \gamma \]
Similarly, from \( \triangle IAC \):
\[ \gamma = r + \beta \implies r = \gamma - \beta \]
Substitute these expressions for \( i \) and \( r \) into equation (1):
\[ \mu_1(\alpha + \gamma) = \mu_2(\gamma - \beta) \]
\[ \mu_1 \alpha + \mu_2 \beta = (\mu_2 - \mu_1)\gamma \quad \text{--- (2)} \]
Since the aperture of the refracting surface is extremely small, the angles \( \alpha, \beta, \gamma \) are also very small. Therefore, we can approximate them by their tangents:
\[ \alpha \approx \tan \alpha = \frac{AM}{OM} \approx \frac{AM}{PO} = \frac{AM}{-u} \]
\[ \beta \approx \tan \beta = \frac{AM}{MI} \approx \frac{AM}{PI} = \frac{AM}{+v} \]
\[ \gamma \approx \tan \gamma = \frac{AM}{MC} \approx \frac{AM}{PC} = \frac{AM}{+R} \]
Now, substitute these approximations into equation (2):
\[ \mu_1 \left(\frac{AM}{-u}\right) + \mu_2 \left(\frac{AM}{+v}\right) = (\mu_2 - \mu_1)\left(\frac{AM}{+R}\right) \]
Dividing the entire equation by \( AM \) gives:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
This proves the relation.
In simple words: We can find how light bends through a curved surface by linking the incoming and outgoing angles to the slope of the curve using Snell's law.

Exam Tip: Be extremely careful with sign conventions: \( u \) is measured against the direction of incident light and is negative, while \( v \) and \( R \) are measured in the direction of light and are positive.

Question 680. Derive expression for the lens maker’s formula using necessary ray diagrams.
\[ \frac{1}{f} = (\mu_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Also state the assumptions in deriving the above relation and the sign conventions used.
Answer: Let us derive the lens maker's formula step-by-step:
**Derivation**:
Consider a thin lens of refractive index \( \mu_2 \) placed in a medium of refractive index \( \mu_1 \). The lens has two spherical surfaces with radii of curvature \( R_1 \) and \( R_2 \).
1. **Refraction at the first surface (ABC)**: The light ray from point object \( O \) in medium \( \mu_1 \) refracts at surface ABC to form a virtual image \( I' \) at distance \( v' \):
\[ \frac{\mu_2}{v'} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1} \quad \text{--- (1)} \]
2. **Refraction at the second surface (ADC)**: The image \( I' \) acts as a virtual object for the second surface ADC, which forms the final image \( I \) at distance \( v \). Since the lens is very thin, the distance of this virtual object from the second surface is also taken as \( v' \). Refraction occurs from medium \( \mu_2 \) to \( \mu_1 \):
\[ \frac{\mu_1}{v} - \frac{\mu_2}{v'} = \frac{\mu_1 - \mu_2}{R_2} = -\frac{\mu_2 - \mu_1}{R_2} \quad \text{--- (2)} \]
Adding equations (1) and (2) eliminates the intermediate distance \( v' \):
\[ \frac{\mu_1}{v} - \frac{\mu_1}{u} = (\mu_2 - \mu_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Divide both sides by \( \mu_1 \):
\[ \frac{1}{v} - \frac{1}{u} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Let \( \frac{\mu_2}{\mu_1} = \mu_{21} \) (relative refractive index of the lens with respect to the medium):
\[ \frac{1}{v} - \frac{1}{u} = (\mu_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
By definition, if the object is at infinity (\( u = -\infty \)), the rays focus at the focal point (\( v = f \)). Substituting these into the equation:
\[ \frac{1}{f} - 0 = (\mu_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
\[ \frac{1}{f} = (\mu_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
**Assumptions used**:
(i) The lens is extremely thin.
(ii) The lens has a very small aperture.
(iii) The object is a point object positioned on the principal axis.
(iv) All incident rays are paraxial (close to the principal axis).
**New Cartesian sign conventions used**:
(i) All distances are measured from the optical center of the lens.
(ii) Distances measured in the direction of the incident ray are taken as positive.
(iii) Distances measured in the direction opposite to the incident ray are taken as negative.
In simple words: We find how a lens focuses light by looking at how light refracts at the front surface and then refracts again at the back surface.

Exam Tip: To get full marks, make sure you write down all four assumptions and the three sign conventions exactly as shown in standard textbooks.

Question 681. Two thin convex lenses \( L_1 \) and \( L_2 \) of focal lengths \( f_1 \) and \( f_2 \) respectively, are placed coaxially in contact. An object is placed at a point beyond the focus of lens \( L_1 \). Draw a ray diagram to show the image formation and hence derive the expression for the focal length of the combined system.
Answer: Let us derive the expression for the combined focal length of two thin lenses in contact:
**Derivation**:
1. **Refraction by the first lens \( L_1 \)**: Let an object be placed at point \( O \) at distance \( u \). In the absence of the second lens, the first lens would form an image \( I' \) at distance \( v' \):
\[ \frac{1}{v'} - \frac{1}{u} = \frac{1}{f_1} \quad \text{--- (1)} \]
2. **Refraction by the second lens \( L_2 \)**: The second lens is placed coaxially in contact with the first. The intermediate image \( I' \) acts as a virtual object for \( L_2 \), which forms the final real image \( I \) at distance \( v \):
\[ \frac{1}{v} - \frac{1}{v'} = \frac{1}{f_2} \quad \text{--- (2)} \]
Adding equations (1) and (2) cancels out the virtual image term \( v' \):
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \quad \text{--- (3)} \]
If we replace the two-lens system with a single equivalent lens of focal length \( F \) that forms the same image \( I \) for the object \( O \), we have:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \quad \text{--- (4)} \]
Comparing equations (3) and (4) yields the equivalent focal length of the combination:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
This is the required expression.
In simple words: When lenses are placed side-by-side, the first lens bends the light, and the second lens bends it even more. Adding their individual focal length reciprocals gives the reciprocal of the total focal length.

Exam Tip: Clearly describe the role of the intermediate image \( I' \) as a virtual object for the second lens, as this is the core physical argument of the derivation.

Question 682. Draw a ray diagram to show the refraction of light through a glass prism. Hence derive the relation
\[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Answer: Let us derive the prism formula:
**Derivation**:
Let a ray of light be incident on face AB of a glass prism ABC with refracting angle \( A \).
1. **Geometry of Quadrilateral \( AQNR \)**: The normals drawn at the points of incidence \( Q \) and emergence \( R \) meet at \( N \). Since the angles at \( Q \) and \( R \) are right angles:
\[ \angle A + \angle QNR = 180^\circ \quad \text{--- (1)} \]
2. **Geometry of Triangle \( \triangle QNR \)**: In this triangle:
\[ r_1 + r_2 + \angle QNR = 180^\circ \quad \text{--- (2)} \]
Comparing equations (1) and (2) yields:
\[ r_1 + r_2 = A \quad \text{--- (3)} \]
3. **Total Angle of Deviation (\( \delta \))**: The total deviation is the sum of deviations at both refracting surfaces:
\[ \delta = (i - r_1) + (e - r_2) = (i + e) - (r_1 + r_2) \ ]
Substitute \( r_1 + r_2 = A \) from (3):
\[ \delta = i + e - A \quad \text{--- (4)} \]
4. **Condition of Minimum Deviation (\( \delta = \delta_m \))**: At the position of minimum deviation, the ray passes symmetrically through the prism, meaning the angle of incidence equals the angle of emergence (\( i = e \)), and the internal refraction angles are equal (\( r_1 = r_2 = r \)).
Substituting these into equation (3):
\[ r + r = A \implies 2r = A \implies r = \frac{A}{2} \]
Substituting \( i = e \) and \( \delta = \delta_m \) into equation (4):
\[ \delta_m = 2i - A \implies 2i = A + \delta_m \implies i = \frac{A + \delta_m}{2} \]
5. **Applying Snell's Law**: The refractive index \( \mu \) of the prism material is:
\[ \mu = \frac{\sin i}{\sin r} \]
Substituting our derived expressions for \( i \) and \( r \) gives:
\[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
This is the prism formula.
In simple words: We track the path of a light ray as it enters one face of a prism and leaves another. By analyzing the angles of the prism and the total bending, we derive a direct formula for the glass's refractive index.

Exam Tip: Make sure to clearly state the conditions under which minimum deviation occurs (\( i = e \) and \( r_1 = r_2 \)), as this step is essential to complete the proof.

Question 683. A ray of light incident on an equilateral glass prism propagates parallel to the base line of the prism inside it. Find the angle of incidence of this ray. Given refractive index of material of glass prism is \( \sqrt{3} \).
Answer: The angle of incidence is **\( 60^\circ \)**.
**Reason**: For an equilateral prism, the angle of the prism is \( A = 60^\circ \). Since the refracted ray inside the prism travels parallel to the base, the path is symmetric, meaning the angles of refraction at both faces are equal (\( r_1 = r_2 = r \)).
Using the relation:
\( r_1 + r_2 = A \implies 2r = 60^\circ \)

\( \implies r = 30^\circ \)
Now, apply Snell's law at the first refracting surface:
\( \mu = \frac{\sin i}{\sin r} \)
Given \( \mu = \sqrt{3} \) and \( r = 30^\circ \):
\( \sqrt{3} = \frac{\sin i}{\sin 30^\circ} \)
\( \sin i = \sqrt{3} \times \sin 30^\circ = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \)

\( \implies i = 60^\circ \)
Thus, the angle of incidence of the ray is \( 60^\circ \).
In simple words: Since the prism is equilateral, its main angle is 60 degrees. The parallel ray path inside means the bending angle is exactly 30 degrees, which gives an entry angle of 60 degrees.

Exam Tip: Mention that the parallel propagation of the ray inside the prism represents the condition of minimum deviation, where \( r = A/2 \).

Question 684. Determine the value of the angle of incidence for a ray of light travelling from a medium of refractive index \( \mu_1 = \sqrt{2} \) into the medium of refractive index \( \mu_2 = 1 \), so that it just grazes along the surface of separation.
Answer: The required angle of incidence is **\( 45^\circ \)**.
**Reason**: For the refracted ray to graze along the boundary of separation, the angle of refraction must be \( r = 90^\circ \). This means the angle of incidence is equal to the critical angle (\( i = i_c \)).
Applying Snell's law for light going from medium 1 to medium 2:
\( \frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} \)
Substitute \( \mu_1 = \sqrt{2} \), \( \mu_2 = 1 \), and \( r = 90^\circ \):
\( \frac{\sin i_c}{\sin 90^\circ} = \frac{1}{\sqrt{2}} \)
Since \( \sin 90^\circ = 1 \):
\( \sin i_c = \frac{1}{\sqrt{2}} \)

\( \implies i_c = 45^\circ \)
Hence, the angle of incidence must be \( 45^\circ \).
In simple words: Grazing along the surface means the refracted light is bent by exactly 90 degrees. This requires the incoming light to strike at the critical angle, which is 45 degrees.

Exam Tip: Recognize that the phrase "just grazes along the surface of separation" is synonymous with total internal reflection starting, meaning the angle of incidence equals the critical angle.

Question 685. A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is 3/4 th of the angle of prism. Calculate the speed of light in the prism.
Answer: The speed of light in the prism is **\( 2.12 \times 10^8\text{ m/s} \)** (or \( 2.1 \times 10^8\text{ m/s} \)).
**Reason**: For an equilateral prism, the angle of the prism is \( A = 60^\circ \). The angle of incidence is given as:
\( i = \frac{3}{4} A = \frac{3}{4} \times 60^\circ = 45^\circ \)
At the minimum deviation condition, the angle of refraction \( r \) is:
\( r = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \)
Find the refractive index \( \mu \) of the prism material:
\( \mu = \frac{\sin i}{\sin r} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx 1.414 \)
The speed of light \( v \) in the prism is calculated using:
\( \mu = \frac{c}{v} \)

\( \implies v = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{2}}\text{ m/s} \approx 2.12 \times 10^8\text{ m/s} \)
Thus, the speed of light in the prism is approximately \( 2.12 \times 10^8\text{ m/s} \).
In simple words: An angle of incidence equal to 3/4 of the prism's angle gives 45 degrees, which yields a refractive index of 1.414. This slows the speed of light down to about 2.12 × 10^8 m/s inside the glass.

Exam Tip: Show the calculation of both \( i \) and \( r \) clearly before applying Snell's law to prevent intermediate calculation errors.

Question 686. (i) Draw a labelled ray diagram to show the image formation by an astronomical telescope in normal adjustment.
(ii) Define magnifying power of an astronomical telescope in normal adjustment (i,e, when the final image is formed at infinity).
(iii) Derive the expression for its magnifying power in normal adjustment.
Answer: Let us discuss the image formation and magnifying power of an astronomical telescope in normal adjustment:
(i) **Ray Diagram**:
An objective lens with a long focal length focuses light from a distant source, forming an intermediate inverted image at its focus. The eyepiece then magnifies this image and forms the final image at infinity. (ii) **Definition**: The magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye by the final image (formed at infinity) to the angle subtended at the unaided eye by the object (also at infinity).
(iii) **Derivation**:
Let \( \alpha \) be the angle subtended by the distant object at the objective lens, and \( \beta \) be the angle subtended by the final image at the eyepiece.
For small angles:
\[ \alpha \approx \tan \alpha = \frac{A'B'}{OB'} = \frac{A'B'}{f_o} \]
\[ \beta \approx \tan \beta = \frac{A'B'}{EB'} = \frac{A'B'}{-f_e} \]
The magnifying power is:
\[ m = \frac{\beta}{\alpha} = \frac{\frac{A'B'}{-f_e}}{\frac{A'B'}{f_o}} \]

\( \implies m = -\frac{f_o}{f_e} \]
This is the expression for magnifying power in normal adjustment.
In simple words: In normal adjustment, a telescope brings parallel light from infinity to a focus, and the eyepiece bends it back into parallel rays for our eyes. The magnifying power is simply the focal length of the objective divided by that of the eyepiece.

Exam Tip: Remember that the negative sign in \( m = -\frac{f_o}{f_e} \) indicates that the final image is inverted.

Question 687. (i) Draw a labelled ray diagram of an astronomical telescope when the final image is formed at least distance of distinct vision.
(ii) Define its magnifying power and deduce the expression for the magnifying power of telescope.
Answer: Let us examine the telescope system when the final image is formed at the least distance of distinct vision (\( D \)):
(i) **Ray Diagram**:
An intermediate image \( A'B' \) is formed inside the focus of the eyepiece, and divergent rays refracted by the eyepiece appear to diverge from a virtual image at distance \( D \) on the left side. (ii) **Definition**: The magnifying power is the ratio of the angle subtended at the eye by the final virtual image (located at the least distance of distinct vision) to the angle subtended by the object at the unaided eye.
(iii) **Derivation**:
The magnifying power is given by:
\[ m = \frac{\beta}{\alpha} \approx \frac{\tan \beta}{\tan \alpha} = \frac{A'B'/u_e}{A'B'/f_o} = -\frac{f_o}{u_e} \quad \text{--- (1)} \]
Apply the lens equation for the eyepiece:
\[ \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \]
Using sign conventions: \( v_e = -D \), \( u_e = -u_e \), and focal length is \( +f_e \):
\[ \frac{1}{-D} - \frac{1}{-u_e} = \frac{1}{f_e} \]
\[ \frac{1}{u_e} = \frac{1}{f_e} + \frac{1}{D} = \frac{1}{f_e}\left(1 + \frac{f_e}{D}\right) \]
Substitute this back into equation (1):
\[ m = -f_o \left[\frac{1}{f_e}\left(1 + \frac{f_e}{D}\right)\right] \]

\( \implies m = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) \]
This is the magnifying power when the image is formed at the near point.
In simple words: If we adjust the telescope so that the image is formed as close as possible to our eye (25 cm away), we get slightly higher magnification than normal adjustment.

Exam Tip: Notice that the magnification in this case is higher than the normal adjustment value by a factor of \( \left(1 + \frac{f_e}{D}\right) \).

Question 688. Write the main considerations required in selecting the objective and eye piece lenses in order to have large magnifying power and high resolution of the telescope
Answer: To optimize an astronomical telescope for high magnification and resolving power, the following optical design criteria must be met:
(i) **For High Magnification**: Since magnification is given by \( m = -\frac{f_o}{f_e} \), we must select an objective lens with a very long focal length (\( f_o \)) and an eyepiece with a short focal length (\( f_e \)), such that \( f_o \gg f_e \).
(ii) **For High Resolving Power**: Resolving power is given by \( \text{R.P.} = \frac{D}{1.22 \lambda} \), where \( D \) is the diameter (aperture) of the objective lens. Therefore, the telescope must have a primary objective with a very large aperture to resolve fine details.
In simple words: For high zoom, the front lens needs a long focal length and the eyepiece needs a short one. For sharp details, the front lens must be as wide as possible.

Exam Tip: Write the mathematical formulas for both magnification and resolving power to back up your structural recommendations.

Question 689. Draw a labelled ray diagram of a compound microscope when image is formed at least distance of distinct vision. Define its magnifying power and deduce the expression for the magnifying power of the microscope.
Answer: Let us study the compound microscope when the final image is formed at the least distance of distinct vision \( D \):
(i) **Ray Diagram**:
A small objective lens forms a real, inverted, magnified intermediate image \( A'B' \) inside the tube. This image lies within the focus of the eyepiece, which forms a highly magnified virtual final image. (ii) **Definition**: The magnifying power of a compound microscope at the near-point adjustment is the ratio of the angle subtended at the eye by the final virtual image (at the near point) to the angle subtended by the object at the unaided eye, with both placed at the least distance of distinct vision (\( D \)).
(iii) **Derivation**:
The angular magnification is:
\[ m = \frac{\beta}{\alpha} \approx \frac{\tan \beta}{\tan \alpha} = \frac{A'B'/u_e}{AB/D} = \left(\frac{A'B'}{AB}\right)\left(\frac{D}{u_e}\right) \]
The linear magnification of the objective lens is \( m_o = \frac{A'B'}{AB} = -\frac{v_o}{u_o} \). This gives:
\[ m = -\frac{v_o}{u_o}\left(\frac{D}{u_e}\right) \quad \text{--- (1)} \]
Using the lens formula for the eyepiece (\( v_e = -D \), \( u_e = -u_e \)):
\[ \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \implies \frac{1}{-D} - \frac{1}{-u_e} = \frac{1}{f_e} \]
\[ \frac{1}{u_e} = \frac{1}{f_e} + \frac{1}{D} \]
Multiplying both sides by \( D \):
\[ \frac{D}{u_e} = 1 + \frac{D}{f_e} \]
Substituting this back into equation (1) yields the final magnifying power expression:
\[ m = -\frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right) \]
Or in terms of focal lengths when the object is very close to \( F_o \):
\[ m \approx -\frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \]
This is the required expression.
In simple words: A microscope uses two lenses: a front one to create a real, inverted image inside the tube, and an eyepiece that acts like a magnifying glass to blow up that image as big as possible for our eyes.

Exam Tip: Explicitly state the intermediate step relating linear magnification \( m_o = -\frac{v_o}{u_o} \) to secure partial credit if any algebraic step goes wrong.

Question 690. (i) Draw a labelled ray diagram for the formation of image by a compound microscope in normal adjustment.
(ii) Define magnifying power of a compound microscope in normal adjustment and derive an expression for it.
Answer: Let us derive the relations for a compound microscope in normal adjustment (when the final image is formed at infinity):
(i) **Ray Diagram**:
The objective lens forms an intermediate image \( A'B' \) exactly at the principal focus \( F_e \) of the eyepiece, which then produces parallel light rays that focus at infinity. (ii) **Definition**: The magnifying power of a compound microscope in normal adjustment is defined as the ratio of the angle subtended at the eye by the final virtual image (at infinity) to the angle subtended by the object at the unaided eye (placed at the least distance of distinct vision \( D \)).
(iii) **Derivation**:
The total magnification \( m \) of the microscope is the product of the linear magnification of the objective (\( m_o \)) and the angular magnification of the eyepiece (\( m_e \)):
\[ m = m_o \times m_e \]
The objective lens produces a real image with magnification:
\[ m_o = -\frac{v_o}{u_o} \]
Since the final image is formed at infinity, the eyepiece acts as a simple magnifier with its focal point at the intermediate image, yielding:
\[ m_e = \frac{D}{f_e} \]
Thus, the total magnification is:
\[ m = -\frac{v_o}{u_o} \times \frac{D}{f_e} \]
When the object is placed very close to the principal focus of the objective lens (\( u_o \approx f_o \)) and the intermediate image is formed very close to the eyepiece (\( v_o \approx L \), the tube length):
Substitute these approximations:
\[ m \approx -\frac{L}{f_o} \times \frac{D}{f_e} \]
This represents the magnifying power in normal adjustment.
In simple words: In normal adjustment, the intermediate image is formed exactly at the focal point of the eyepiece. This allows the eyepiece to send parallel rays of light to our eyes, giving a relaxed, comfortable view of the specimen at infinity.

Exam Tip: Clearly state the paraxial approximations \( u_o \approx f_o \) and \( v_o \approx L \) when deriving the simplified form of the formula.

Question 691. Three rays (1,2,3) of different colours fall normally on one of the sides of an isosceles right angled prism as shown.The refractive index of prism for these rays is 1.39, 1.47 and 1.52 respectively. Find which of these rays get internally reflected and which get only refracted from AC. Trace the path of rays. Justify your answer.
Answer: For total internal reflection (TIR) to occur at the boundary face AC, the angle of incidence \( i \) must be greater than the critical angle \( i_c \):
\[ i > i_c \implies \sin i > \sin i_c \]
Since \( \sin i_c = \frac{1}{\mu} \), this requires:
\[ \sin i > \frac{1}{\mu} \implies \mu > \frac{1}{\sin i} \]
For an isosceles right-angled prism, a ray incident normally on one perpendicular face strikes the hypotenuse face AC at an angle of incidence \( i = 45^\circ \). Substituting this value:
\[ \mu > \frac{1}{\sin 45^\circ} = \sqrt{2} \approx 1.414 \]
Thus, only rays with a refractive index greater than \( 1.414 \) will undergo total internal reflection:
- For Ray 1 (\( \mu = 1.39 \)): Since \( 1.39 < 1.414 \), it undergoes refraction and exits the face AC.
- For Ray 2 (\( \mu = 1.47 \)): Since \( 1.47 > 1.414 \), it undergoes total internal reflection.
- For Ray 3 (\( \mu = 1.52 \)): Since \( 1.52 > 1.414 \), it also undergoes total internal reflection.
Therefore, Ray 1 refracts out of face AC, while Ray 2 and Ray 3 are totally internally reflected inside the prism.
In simple words: Light rays must have a refractive index of at least 1.414 to undergo total internal reflection inside a 45-degree prism. Rays 2 and 3 meet this condition and reflect downward, while Ray 1 has a lower index and simply bends out of the prism.

Exam Tip: Show the calculation of the limiting refractive index (\( \mu > \sqrt{2} = 1.414 \)) clearly before comparing individual values to secure full marks.

Question 692. A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown. What must be the minimum value of refractive index of glass ? Give relevant calculations.
Answer: For total internal reflection to happen at the hypotenuse face of the right isosceles prism, the angle of incidence \( i \) must be greater than or equal to the critical angle \( i_c \):
\[ i \ge i_c \implies \sin i \ge \sin i_c \]
Since \( \sin i_c = \frac{1}{\mu} \) and the angle of incidence on the internal boundary is \( i = 45^\circ \), substituting these values gives:
\[ \sin 45^\circ \ge \frac{1}{\mu} \]
\[ \frac{1}{\sqrt{2}} \ge \frac{1}{\mu} \]

\( \implies \mu \ge \sqrt{2} \approx 1.414 \)
Therefore, the minimum required refractive index for the glass is \( 1.414 \).
In simple words: To make light reflect completely inside a 45-degree prism, the glass must bend light well enough to have a refractive index of at least 1.414.

Exam Tip: Start your calculation with the fundamental inequality \( \sin i \ge \frac{1}{\mu} \) to establish a strong theoretical foundation for your solution.

Question 693. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
(i) a convex lens of focal length 20 cm,
(ii) a concave lens of focal length 16 cm ?
Answer: In this scenario, the virtual object distance \( u \) is positive because the beam converges toward point P behind the lens, so \( u = +12\text{ cm} \).
(i) **For a convex lens of focal length \( f = +20\text{ cm} \)**:
Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
\[ \frac{1}{v} - \frac{1}{12} = \frac{1}{20} \]
\[ \frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{3 + 5}{60} = \frac{8}{60} = \frac{2}{15} \]

\( \implies v = +7.5\text{ cm} \)
Therefore, the beam converges at a point \( 7.5\text{ cm} \) behind the convex lens.

(ii) **For a concave lens of focal length \( f = -16\text{ cm} \)**:
Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
\[ \frac{1}{v} - \frac{1}{12} = -\frac{1}{16} \]
\[ \frac{1}{v} = -\frac{1}{16} + \frac{1}{12} = \frac{-3 + 4}{48} = \frac{1}{48} \]

\( \implies v = +48\text{ cm} \)
Therefore, the beam converges at a point \( 48\text{ cm} \) behind the concave lens.
In simple words: (i) Placing a converging lens in the path of the converging light pinches the beam closer, making it focus at 7.5 cm. (ii) Placing a diverging lens fights the convergence slightly, shifting the focus further out to 48 cm.

Exam Tip: Note that when dealing with a convergent incident beam, the object distance is positive (\( u = +12\text{ cm} \)), which is a common point where students lose marks.

Question 694. A ray of light incident on one of the faces of a glass prism of angle ‘A’ has angle of incidence 2A. The refracted ray in the prism strikes the opposite face which is silvered, the reflected ray from it retracing its path. Trace the ray diagram and find the relation between the refractive index of the material of the prism and the angle of the prism.
Answer: For a prism, the angles of refraction are related to the prism angle \( A \) by:
\[ r_1 + r_2 = A \]
Since the ray of light strikes the silvered second face normally, it reflects straight back along its incident path. This normal incidence means the angle of refraction at the second surface is \( r_2 = 0 \). Substituting this:
\[ r_1 + 0 = A \implies r_1 = A \]
Given that the angle of incidence on the first face is \( i = 2A \), we apply Snell's law at the first boundary:
\[ \mu = \frac{\sin i}{\sin r_1} = \frac{\sin 2A}{\sin A} \]
Using the trigonometric identity \( \sin 2A = 2 \sin A \cos A \):
\[ \mu = \frac{2 \sin A \cos A}{\sin A} \]

\( \implies \mu = 2 \cos A \)
This represents the relationship between the refractive index and the prism angle.
In simple words: Since the silvered face reflects the light straight back along its path, the light must hit that second face at exactly 90 degrees. This simplifies our angles so we can relate the refractive index directly to twice the cosine of the prism's angle.

Exam Tip: Draw the double-headed arrow on the refracted ray to indicate the retracing of its path as specified in the problem statement.

Question 695. Using mirror formula, explain why does a convex mirror always produce a virtual image ?
Answer: According to the mirror equation:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
Rearranging this to solve for the image position \( v \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
Applying the Cartesian sign conventions for a convex mirror:
- The focal length \( f \) is always positive because the focus lies behind the reflecting surface (\( f > 0 \)).
- The object is positioned in front of the mirror, so the object distance \( u \) is always negative (\( u < 0 \)).
Since \( u \) is negative, the term \( -\frac{1}{u} \) becomes positive:
\[ -\frac{1}{u} = -\frac{1}{-\lvert u \rvert} = +\frac{1}{\lvert u \rvert} > 0 \]
Because both \( \frac{1}{f} \) and \( -\frac{1}{u} \) are positive quantities, their sum \( \frac{1}{v} \) must also be positive. Consequently, the image distance \( v \) remains positive under all conditions, showing that a convex mirror always produces a virtual image located behind the mirror surface.
In simple words: Because a convex mirror's focal length is always positive and the object distance is always negative, the mirror formula forces the image distance to always be positive. A positive distance means the image forms behind the mirror, which is virtual.

Exam Tip: Clearly state that both components of the equation are positive, leading to a mathematically positive value of \( v \).

Question 696. You are given two converging lenses of focal lengths 1.25 cm and 5 cm to design a compound microscope. If it is desired to have a magnification of 30, find out the separation between the objective and the eyepiece.
Answer: The magnification of a compound microscope for a final image at the near point is given by:
\[ m = -\frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \]
Here, \( f_o = 1.25\text{ cm} \) (objective), \( f_e = 5\text{ cm} \) (eyepiece), \( D = 25\text{ cm} \) (least distance of distinct vision), and \( m = -30 \). Substitute these values into the formula:
\[ -30 = -\frac{L}{1.25}\left(1 + \frac{25}{5}\right) \]
\[ 30 = \frac{L}{1.25}(1 + 5) \]
\[ 30 = \frac{L}{1.25} \times 6 \]
\[ 5 = \frac{L}{1.25} \]

\( \implies L = 5 \times 1.25 = 6.25\text{ cm} \)
Thus, the separation (tube length) between the objective and the eyepiece is \( 6.25\text{ cm} \).
In simple words: Using the microscope magnification formula with the given focal lengths, we calculate that the two lenses must be placed 6.25 cm apart to achieve a zoom of 30.

Exam Tip: Ensure you use the negative sign for magnification (\( m = -30 \)) as the final image in a compound microscope is always inverted.

Question 697. (i) A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment ?
(ii) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens ?
Answer:
(i) **Magnifying Power in Normal Adjustment**:
The magnifying power of a telescope in normal adjustment is:
\[ m = -\frac{f_o}{f_e} \]
Given \( f_o = 150\text{ cm} \) and \( f_e = 5\text{ cm} \):
\[ m = -\frac{150}{5} = -30 \]
The magnifying power is \( 30 \) (the negative sign indicates an inverted image).

(ii) **Height of the Image Formed by the Objective Lens**:
Here, the tower acts as the object at a distance of \( u = -3\text{ km} = -3000\text{ m} \), and its height is \( h_1 = 100\text{ m} \). The focal length of the objective lens is \( f_o = 150\text{ cm} = 1.5\text{ m} \).
Using the lens formula for the objective:
\[ \frac{1}{v_o} - \frac{1}{u} = \frac{1}{f_o} \]
\[ \frac{1}{v_o} - \frac{1}{-3000} = \frac{1}{1.5} \]
\[ \frac{1}{v_o} = \frac{1}{1.5} - \frac{1}{3000} = \frac{2000 - 1}{3000} = \frac{1999}{3000} \]

\( \implies v_o \approx 1.5\text{ m} \)
Now, using the linear magnification formula:
\[ m = \frac{h_2}{h_1} = \frac{v_o}{u} \]
Taking magnitudes:
\[ h_2 = h_1 \times \frac{v_o}{\lvert u \rvert} = 100 \times \frac{1.5}{3000} = 100 \times 0.0005 = 0.05\text{ m} = 5\text{ cm} \]
Thus, the height of the image formed by the objective lens is \( 5\text{ cm} \).
In simple words: (i) The telescope has a magnifying power of 30. (ii) The objective lens forms a tiny, 5 cm tall inverted image of the massive 100 m tall tower.

Exam Tip: Ensure all units are consistent (convert all values either to meters or centimeters) before performing calculations for the second part.

Question 698. (i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope ?
(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 × 10^6 m and the radius of lunar orbit is 3.8 × 10^8 m.
Answer:
(i) **Angular Magnification**:
Given the focal length of the objective lens \( f_o = 15\text{ m} \) and the focal length of the eyepiece \( f_e = 1.0\text{ cm} = 1.0 \times 10^{-2}\text{ m} \).
The angular magnification is:
\[ \lvert m \rvert = \frac{f_o}{f_e} = \frac{15}{1.0 \times 10^{-2}} = 1500 \]

(ii) **Diameter of the Image of the Moon**:
Let \( d \) be the diameter of the moon, \( r \) be the radius of the lunar orbit, and \( D \) be the diameter of the image formed by the objective lens at its focal plane.
The angle subtended by the moon at the objective is:
\[ \alpha = \frac{\text{Diameter of moon}}{\text{Radius of lunar orbit}} = \frac{3.48 \times 10^6}{3.8 \times 10^8}\text{ rad} \]
The angle subtended by the image at the optical center of the objective is:
\[ \alpha = \frac{\text{Diameter of image}}{f_o} = \frac{D}{15} \]
Equating the two angular expressions:
\[ \frac{D}{15} = \frac{3.48 \times 10^6}{3.8 \times 10^8} \]
\[ D = 15 \times \frac{3.48}{3.8} \times 10^{-2}\text{ m} = 0.1373\text{ m} \approx 13.73\text{ cm} \]
Thus, the diameter of the image formed is \( 13.73\text{ cm} \).
In simple words: (i) The telescope zooms in by 1500 times. (ii) When looking at the moon, the primary lens forms a physical image of the moon that is about 13.73 cm wide on its focal plane.

Exam Tip: Use the fact that the angle subtended by the object equals the angle subtended by the image at the lens center to simplify astronomical image calculations.

Question 699. Monochromatic light of wavelength 589 nm is incident from air on a water surface. If μ for water is 1.33, find the wavelength, frequency and speed of the refracted light.
Answer: When light refracts into a different medium:
(i) **Frequency (\( \nu \))**: The frequency of light depends only on the source and remains constant during refraction. Using the properties in air:
\[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8\text{ m/s}}{589 \times 10^{-9}\text{ m}} \approx 5.09 \times 10^{14}\text{ Hz} \]
Thus, the frequency of refracted light is \( 5.09 \times 10^{14}\text{ Hz} \).

(ii) **Speed (\( v' \))**: The speed of light in water is:
\[ v' = \frac{c}{\mu} = \frac{3 \times 10^8\text{ m/s}}{1.33} \approx 2.25 \times 10^8\text{ m/s} \]

(iii) **Wavelength (\( \lambda' \))**: The wavelength in water is:
\[ \lambda' = \frac{\lambda}{\mu} = \frac{589\text{ nm}}{1.33} \approx 442.86\text{ nm} \]
Thus, the wavelength of the refracted light in water is \( 442.86\text{ nm} \).
In simple words: When light enters water, its color-frequency stays exactly the same, but its speed slows down to 2.25 × 10^8 m/s and its wavelength shrinks to 442.86 nm.

Exam Tip: Explicitly state that frequency is an intrinsic property of the source and does not change when the wave crosses into another medium.

Question 699*. Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Also find the location of the image.
Answer: We are given the radius of curvature of the concave mirror is \( R = -20\text{ cm} \). This gives a focal length of:
\[ f = \frac{R}{2} = \frac{-20}{2} = -10\text{ cm} \]
Since the image is real, the magnification \( m \) is negative, so \( m = -2 \).
Using the magnification formula for spherical mirrors:
\[ m = -\frac{v}{u} \]
\[ -2 = -\frac{v}{u} \implies v = 2u \quad \text{--- (1)} \]
Now, substitute \( f = -10\text{ cm} \) and \( v = 2u \) into the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
\[ \frac{1}{-10} = \frac{1}{2u} + \frac{1}{u} \]
\[ -\frac{1}{10} = \frac{1 + 2}{2u} = \frac{3}{2u} \]
\[ -2u = 30 \]

\( \implies u = -15\text{ cm} \)
Using equation (1), find the image location:
\[ v = 2 \times (-15) = -30\text{ cm} \]
Thus, the object must be placed at a distance of \( 15\text{ cm} \) in front of the mirror, and the real image is formed at a distance of \( 30\text{ cm} \) in front of the mirror.
In simple words: To double the image size of the object, we place it 15 cm in front of the mirror, which forms the real image at a distance of 30 cm on the same side.

Exam Tip: For real images formed by mirrors, always assign a negative sign to the magnification (\( m = -2 \)), as a positive value would represent a virtual, erect image.