Maharashtra Board Class 12 Physics Chapter 7 Wave Optics Exercise Solutions

1. Choose the correct option.

(i) Which of the following phenomenon proves that light is a transverse wave?
(a) reflection
(b) interference
(c) diffraction
(d) polarization
Answer: (d) polarization
In simple words: Polarization is like trying to slide a flat ruler through a picket fence; it only works if the ruler is aligned with the gaps. Since light can be filtered this way, it proves light waves wiggle side-to-side (transverse) rather than back-and-forth.

📝 Teacher's Note: Remind students that interference and diffraction are properties of all waves (including longitudinal sound waves), but polarization is unique to transverse waves.

🎯 Exam Tip: Always associate the "transverse nature of light" with the keyword "Polarization" to secure a quick mark in MCQs.

(ii) Which property of light does not change when it travels from one medium to another?
(a) velocity
(b) wavelength
(c) amplitude
(d) frequency
Answer: (d) frequency
In simple words: Think of frequency as the "source's heartbeat." Even if the light slows down or changes its stride (wavelength) in water or glass, the number of beats per second stays exactly the same as the source that created it.

📝 Teacher's Note: Use the analogy of a person walking from pavement into sand; their speed and step-length change, but the number of steps they take per minute remains a constant decided by their effort.

🎯 Exam Tip: Remember the formula \( v = f\lambda \). When light enters a denser medium, \( v \) and \( \lambda \) both decrease, but \( f \) remains constant.

(iii) When unpolarized light is passed through a polarizer, its intensity
(a) increases
(b) decreases
(c) remains unchanged
(d) depends on the orientation of the polarizer
Answer: (b) decreases
In simple words: A polarizer acts like a narrow slit that blocks all waves wiggling in the "wrong" directions. Since half the light's energy is being blocked, the light naturally gets dimmer.

📝 Teacher's Note: Explain that for perfectly unpolarized light, an ideal polarizer will always reduce the intensity by exactly 50%, regardless of its orientation.

🎯 Exam Tip: Look out for numerical problems involving Malus' Law where the initial reduction is always \( I_0 / 2 \).

(iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is 25:1. What was the ratio of intensities of the two sources?
(a) 5:1
(b) 25:1
(c) 3:2
(d) 9:4
Answer: (d) 9:4
In simple words: By using the math of wave addition, we find that the amplitudes of the two waves must be in a 3:2 ratio to get a 25:1 intensity pattern. Squaring those amplitudes gives us the source intensity ratio of 9:4.

📝 Teacher's Note: Show the steps: \( \frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2 = \frac{25}{1} \). Taking the square root gives \( \frac{a_1+a_2}{a_1-a_2} = \frac{5}{1} \), leading to \( a_1/a_2 = 3/2 \). Squaring again gives \( I_1/I_2 = 9/4 \).

🎯 Exam Tip: Be careful not to confuse the ratio of amplitudes with the ratio of intensities. Always square the amplitude ratio to find the intensity ratio.

(v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will satisfy
(a) The interference pattern will remain unchanged
(b) The fringe width will decrease
(c) The fringe width will increase
(d) The fringes will shift.
Answer: (a) The interference pattern will remain unchanged
In simple words: If the glass sheet covers both slits equally, it delays both waves by the same amount. Since there is no relative difference created between the two paths, the pattern stays right where it was.

📝 Teacher's Note: Clarify that if the sheet covered only *one* slit, the pattern would shift. Covering both slits uniformly (parallel to the screen) maintains the original path difference.

🎯 Exam Tip: Read carefully if the glass covers "one slit" (shift) or "both slits" (no change) or "the entire space" (fringe width change).

2. Answer in brief.

(i) What are primary and secondary sources of light?
Answer:
(1) Primary sources of light: The sources that emit light on their own are called primary sources. This emission of light may be due to
(a) the high temperature of the source, e.g., the Sun, the stars, objects heated to high temperature, a flame, etc.
(b) the effect of current being passed through the source, e.g., tubelight, TV, etc.
(c) chemical or nuclear reactions taking place in the source, e.g., firecrackers, nuclear energy generators, etc.

(2) Secondary sources of light: Some sources are not self luminous, i.e., they do not emit light on their own, but reflect or scatter the light incident on them. Such sources of light are called secondary sources, e.g. the moon, the planets, objects such as humans, animals, plants, etc. These objects are visible due to reflected light.
Many of the sources that we see around are secondary sources and most of them are extended sources.
In simple words: Primary sources are like lamps that make their own light. Secondary sources are like mirrors or the moon that only show up because they reflect someone else's light.

📝 Teacher's Note: Ask students to categorize items in the classroom. The bulb is primary; their notebooks and classmates are secondary sources.

🎯 Exam Tip: Include examples like "Sun" for primary and "Moon" for secondary to make your answer clear and complete.

(ii) What is a wavefront? How is it related to rays of light? What is the shape of the wavefront at a point far away from the source of light?
Answer:
Wavefront or wave surface : The locus of all points where waves starting simultaneously from a source reach at the same instant of time and hence the particles at the points oscillate with the same phase, is called a wavefront or wave surface.

Consider a point source of light O in a homogeneous isotropic medium in which the speed of light is \( v \). The source emits light in all directions. In time \( t \), the disturbance (light energy) from the source, covers a distance \( vt \) in all directions, i.e., it reaches out to all points which are at a distance \( vt \) from the point source. The locus of these points which are in the same phase is the surface of a sphere with the centre O and radius \( vt \). It is a spherical wavefront.

In a given medium, a set of straight lines can be drawn which are perpendicular to the wavefront. According to Huygens, these straight lines are the rays of light. Thus, rays are always normal to the wavefront. In the case of a spherical wavefront, the rays are radial.

If a wavefront has travelled a large distance away from the source, a small portion of this wavefront appears to be plane. This part is a plane wavefront.
In simple words: A wavefront is like the ripple on a pond after throwing a stone—everyone on that circle is wiggling together. Rays are just arrows showing which way the ripple is moving.

📝 Teacher's Note: Use the balloon analogy: as the balloon (wavefront) gets huge, a small square on its surface looks perfectly flat (plane wavefront).

🎯 Exam Tip: Always mention that "rays are perpendicular to the wavefront." This is a fundamental point that examiners look for.

(iii) Why are multiple colours observed over a thin film of oil floating on water? Explain with the help of a diagram.
Answer:
Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film:
The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence \( \theta_1 \), i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.
As the reflection is from the denser boundary, there is an additional phase difference of \( \pi \) radians (or an additional path difference \( \lambda / 2 \)). This should be taken into account for mathematical analysis.
In simple words: Light bounces off both the top and bottom of the oil layer. These two bounces "race" back to your eye and interfere. Depending on the oil's thickness and your viewing angle, some colors cancel out while others get boosted.

📝 Teacher's Note: Explain that "thin" means the thickness is comparable to the wavelength of light. If the film were thick (like a window pane), the interference would not be visible.

🎯 Exam Tip: Mention "path difference" and "constructive/destructive interference" to explain the origin of specific colors.

(iv) In Young’s double slit experiment what will we observe on the screen when white light is incident on the slits but one slit is covered with a red filter and the other with a violet filter? Give reasons for your answer.
Answer:
In Young’s double-slit experiment, when white light is incident on the slits and one of the slit is covered with a red filter, the light passing through this slit will emerge as the light having red colour. The other slit which is covered with a violet filter, will give light having violet colour as emergent light. The interference fringes will involve mixing of red and violet colours. At some points, fringes will be red if constructive interference occurs for red colour and destructive interference occurs for violet colour. At some points, fringes will be violet if constructive interference occurs for violet colour and destructive interference occurs for red colour. The central fringe will be bright with mixing of red and violet colours.
In simple words: Usually, interference needs waves of the same color. Since red and violet are very different, they don't produce a standard dark-and-light pattern. Instead, you'll see a blurry mix of red and violet patches where the colors happen to overlap.

📝 Teacher's Note: Technically, interference requires coherence. Since red and violet have different frequencies, they are not coherent and won't produce a stationary interference pattern in the traditional sense, though they overlap on the screen.

🎯 Exam Tip: Note that the "Central Fringe" remains the point where both colors arrive at the same time, appearing as a mixture (magenta/purple).

(v) Explain what is optical path length. How is it different from actual path length?
Answer:
Consider, a light wave of angular frequency \( \omega \) and wave vector \( k \) travelling through vacuum along the x-direction. The phase of this wave is \( (kx-\omega t) \). The speed of light in vacuum is \( c \) and that in medium is \( v \).
\( k = \frac{2\pi}{\lambda} = \frac{2\pi \nu}{v\lambda} = \frac{\omega}{v} \) as \( \omega = 2\pi\nu \) and \( v = \nu\lambda \), where \( \nu \) is the frequency of light.

If the wave travels a distance \( \Delta x \), its phase changes by \( \Delta \phi = k \Delta x = \omega \Delta x / v \).
Similarly, if the wave is travelling in vacuum, \( k = \omega / c \) and \( \Delta \phi = \omega \Delta x / c \)
Now, consider a wave travelling a distance \( \Delta x \) in the medium, the phase difference generated is,
\( \Delta \phi' = k' \Delta x = \omega n \Delta x / c = \omega \Delta x' / c \) ... (1)
where \( \Delta x' = n \Delta x \) ... (2)
The distance \( n \Delta x \) is called the optical path length of the light in the medium; it is the distance the light would have travelled in the same time \( t \) in vacuum (with the speed \( c \)).

The optical path length in a medium is the corresponding path in vacuum that the light travels in the same time as it takes in the given medium.

Now, \( speed = \frac{distance}{time} \)

\( \implies time = \frac{distance}{speed} \)

\( \implies t = \frac{d_{medium}}{v_{medium}} = \frac{d_{vacuum}}{v_{vacuum}} \)

Hence, the optical path = \( d_{vacuum} \)

\( = \frac{v_{vacuum}}{v_{medium}} \times d_{medium} \)

\( = n \times d_{medium} \)

Thus, a distance \( d \) travelled in a medium of refractive index \( n \) introduces a path difference = \( nd - d = d(n - 1) \) over a ray travelling equal distance through vacuum.
In simple words: Light slows down in glass. Optical path length is the distance light *would* have covered in empty space during the same time it spent struggling through the glass. It’s basically "vacuum-equivalent" distance.

📝 Teacher's Note: Use the formula \( \text{Optical Path} = \text{Refractive Index} \times \text{Geometric Path} \). It's a way to normalize path lengths when light passes through different materials.

🎯 Exam Tip: Remember that optical path is always greater than actual path in any medium (since \( n > 1 \)).

Question 3. Derive the laws of reflection of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle \( i \) on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let \( v_1 \) and \( v_2 \) be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser medium) respectively.
When the wavefront reaches MN at point A at \( t = 0 \), A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time \( t = T \). Thus, \( BC = v_1 T \). During the time \( T \), the secondary wavelet originating at A covers a distance \( AE \) in the denser medium with radius \( v_2 T \).

As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A.
\( \angle A'AP = \angle BAC = \) the angle of incidence (\( i \)) and \( \angle P'AE = \angle ACE = \) the angle of refraction (\( r \)).
From \( \triangle ABC \) and \( \triangle AEC \),
\( \sin i = \frac{BC}{AC} \) and \( \sin r = \frac{AE}{AC} \)

\( \implies \frac{\sin i}{\sin r} = \frac{BC/AC}{AE/AC} = \frac{BC}{AE} = \frac{v_1 T}{v_2 T} = \frac{v_1}{v_2} \)

By definition, the refractive index of medium 2 with respect to medium 1,
\( {}_1n_2 = \frac{n_2}{n_1} = \frac{v_1}{v_2} \)

\( \implies \frac{n_2}{n_1} = \frac{\sin i}{\sin r} \)

\( \implies n_1 \sin i = n_2 \sin r \) ... (1)

Here, \( n_1 \) and \( n_2 \) are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell’s law of refraction. Also, it can be seen from the figure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.
Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If \( v_1 > v_2 \), i.e. \( n_1 < n_2 \), then \( r < i \) (bending of the refracted ray towards the normal).
In simple words: Huygens' principle shows that as one side of a light wave hits a surface and slows down, the rest of the wave keeps going at full speed for a moment. This uneven speed causes the whole wave to "pivot" or bend, which we see as refraction.

📝 Teacher's Note: Note that although the question title in the source says "reflection," the actual derivation and diagram provided are for "refraction." Ensure students can draw the wavelets correctly for both cases.

🎯 Exam Tip: In the diagram, ensure that the length \( BC \) is noticeably longer than \( AE \) if light is going from a rarer to a denser medium.

Question 4. Derive the laws of refraction of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle \( i \) on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let \( v_1 \) and \( v_2 \) be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser medium) respectively.
When the wavefront reaches MN at point A at \( t = 0 \), A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time \( t = T \). Thus, \( BC = v_1 T \). During the time \( T \), the secondary wavelet originating at A covers a distance AE in the denser medium with radius \( v_2T \).

As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A. \( \angle A’AP = \angle BAC = \) the angle of incidence (i) and \( \angle P’AE = \angle ACE = \) the angle of refraction (r).
From \( \Delta ABC \) and \( \Delta AEC \),
\( \sin i = \frac{BC}{AC} \) and \( \sin r = \frac{AE}{AC} \)

\( \implies \frac{\sin i}{\sin r} = \frac{BC/AC}{AE/AC} = \frac{BC}{AE} = \frac{v_1T}{v_2T} = \frac{v_1}{v_2} \)

By definition, the refractive index of medium 2 with respect to medium 1,
\( _1n_2 = \frac{n_2}{n_1} = \frac{v_1}{v_2} \)

\( \implies \frac{n_2}{n_1} = \frac{\sin i}{\sin r} \)

\( \implies n_1 \sin i = n_2 \sin r \) .... (1)

Here, \( n_1 \) and \( n_2 \) are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell’s law of refraction. Also, it can be seen from the figure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.

Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If \( v_1 > v_2 \), i.e. \( n_1 < n_2 \), then \( r < i \) (bending of the refracted ray towards the normal).
[Notes :
1. Quite often, the terms vacuum and free space are used in the same sense. Absolute vacuum or a perfect vacuum-a region of space devoid of material, particles – does not exist. The term vacuum is also used to mean a region of space occupied by a gas at very low pressure. Free space means a region of space devoid of matter and fields. Its refractive index is 1 (by definition). Its temperature is 0 K. \( \epsilon_0 \) and \( \mu_0 \) are defined for free space. The refractive index of a medium with respect to air is very close to the absolute refractive index of the medium as the speed of light in air is very close to that in free space.
2. There is no lateral inversion in refraction.
3. There is no bending of light when the angle of incidence is zero (normal incidence), \( r = 0 \) for \( i = 0 \).]
In simple words: This derivation confirms Snell's Law, proving that light bends because its speed changes when entering a new material. If light hits straight-on (0 degrees), it doesn't bend at all.

📝 Teacher's Note: This is a repetition of the previous derivation logic with added technical notes on vacuum and normal incidence. Emphasize that refraction is fundamentally about a change in speed.

🎯 Exam Tip: Be prepared to explain the special case of "normal incidence" where \( i=0 \) and \( r=0 \), as it's a common short-answer question.

Question 5. Explain what is meant by polarization and derive Malus’ law.
Answer:
According to the electromagnetic theory of light, a light wave consists of electric and magnetic fields vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of \( \vec{E} \) in a light wave are in all directions perpendicular to the direction of propagation of light, the wave is said to be unpolarized.

If the vibrations of the electric field \( \vec{E} \) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

This phenomenon of restricting the vibrations of light, i.e., of the electric field vector in a particular direction, which is perpendicular to the direction of the propagation of the wave is called polarization of light.
Consider an unpolarized light wave travelling along the x-direction. Let c, v and \( \lambda \) be the speed, frequency and wavelength, respectively, of the wave. The magnitude of its electric field (\( \vec{E} \)) is,
\( E = E_0 \sin (kx - \omega t) \), where \( E_0 = E_{max} = \) amplitude of the wave, \( \omega = 2\pi v = \) angular frequency of the wave and \( k = \frac{2\pi}{\lambda} = \) magnitude of the wave vector or propagation vector.

The intensity of the wave is proportional to \( |E_0|^2 \). The direction of the electric field can be anywhere in the y-z plane. This wave is passed through two identical polarizers as shown in below figure.

When a wave with its electric field inclined at an angle \( \phi \) to the axis of the first polarizer is passed through the polarizer, the component \( E_0 \cos \phi \) will pass through it. The other component \( E_0 \sin \phi \) which is perpendicular to it will be blocked.

Now, after passing through this polarizer, the intensity of this wave will be proportional to the square of its amplitude, i.e., proportional to \( |E_0 \cos \phi|^2 \).

The intensity of the plane-polarized wave emerging from the first polarizer can be obtained by averaging \( |E_0 \cos \phi|^2 \) over all values of \( \phi \) between 0 and 180°. The intensity of the wave will be proportional to \( \frac{1}{2}|E_0|^2 \) as the average value of \( \cos^2 \phi \) over this range is \( \frac{1}{2} \). Thus the intensity of an unpolarized wave reduces by half after passing through a polarizer.

When the plane-polarized wave emerges from the first polarizer, let us assume that its electric field (\( \vec{E_1} \)) is along the y-direction. Thus, this electric field is,
\( \vec{E_1} = \hat{j} E_{10} \sin (kx - \omega t) \) .... (1)
where, \( E_{10} \) is the amplitude of this polarized wave. The intensity of the polarized wave,
\( I_1 \propto |E_{10}|^2 \) ...(2)

Now this wave passes through the second polarizer whose polarization axis (transmission axis) makes an angle \( \theta \) with the y-direction. This allows only the component \( E_{10} \cos \theta \) to pass through it. Thus, the amplitude of the wave which passes through the second polarizer is \( E_{20} = E_{10} \cos \theta \) and its intensity,
\( I_2 \propto |E_{20}|^2 \)

\( \implies I_2 \propto |E_{10}|^2 \cos^2 \theta \)

\( \implies I_2 = I_1 \cos^2 \theta \) ... (3)

Thus, when plane-polarized light of intensity \( I_1 \) is incident on the second identical polarizer, the intensity of light transmitted by the second polarizer varies as \( \cos^2 \theta \), i.e., \( I_2 = I_1 \cos^2 \theta \), where \( \theta \) is the angle between the transmission axes of the two polarizers. This is known as Malus’ law

[Note : Etienne Louis Malus (1775-1812), French military engineer and physicist, discovered in 1809 that light can be polarized by reflection. He was the first to use the word polarization, but his arguments were based on Newton’s corpuscular theory.]
In simple words: Polarization is the process of making light vibrate in only one direction. Malus' law says that the brightness of light passing through two polarizers depends on the angle between them—if you turn one, the light dims or brightens according to the cosine squared of that angle.

📝 Teacher's Note: Use two polarizing sheets in class to demonstrate this. Rotate one against the other to show students how the light intensity physically disappears at 90 degrees. This makes the math tangible.

🎯 Exam Tip: When deriving Malus' law, remember to state that intensity is proportional to the square of the amplitude. Don't forget to define \(\theta\) as the angle between the transmission axes.

Question 6. What is Brewster’s law? Derive the formula for Brewster angle.
Answer:
Brewster’s law : The tangent of the polarizing angle is equal to the refractive index of the reflecting medium with respect to the surrounding (\( _1n_2 \)). If \( \theta_B \) is the polarizing angle,
\( \tan \theta_B = _1n_2 = \frac{n_2}{n_1} \)
Here \( n_1 \) is the absolute refractive index of the surrounding and \( n_2 \) is that of the reflecting medium. The angle \( \theta_B \) is called the Brewster angle.

Consider a ray of unpolarized monochromatic light incident at an angle \( \theta_B \) on a boundary between two transparent media as shown in below figure. Medium 1 is a rarer medium with refractive index \( n_1 \) and medium 2 is a denser medium with refractive index \( n_2 \). Part of incident light gets refracted and the rest gets reflected. The degree of polarization of the reflected ray varies with the angle of incidence.
The electric field of the incident wave is in the plane perpendicular to the direction of propagation of incident light. This electric field can be resolved into a component parallel to the plane of the paper, shown by double arrows, and a component perpendicular to the plane of the paper shown by dots, both having equal magnitude. Generally, the reflected and refracted rays are partially polarized, i.e., the two components do not have equal magnitude.

In 1812, Sir David Brewster discovered that for a particular angle of incidence \( \theta_B \), the reflected wave is completely plane-polarized with its electric field perpendicular to the plane of the paper while the refracted wave is partially polarized. This particular angle of incidence (\( \theta_B \)) is called the Brewster angle.

For this angle of incidence, the refracted and reflected rays are perpendicular to each other.
For angle of refraction \( \theta_r \),
\( \theta_B + \theta_r = 90^\circ \) …… (1)
From Snell’s law of refraction,
\( n_1 \sin \theta_B = n_2 \sin \theta_r \) … (2)
From Eqs. (1) and (2), we have,
\( n_1 \sin \theta_B = n_2 \sin (90^\circ - \theta_B) = n_2 \cos \theta_B \)

\( \implies \frac{n_2}{n_1} = \frac{\sin \theta_B}{\cos \theta_B} = \tan \theta_B \)

\( \implies \theta_B = \tan^{-1}\left(\frac{n_2}{n_1}\right) \) … (3)

This is called Brewster’s law.
In simple words: When light hits a surface at a very specific angle (the Brewster angle), the reflected light is perfectly polarized. Brewster's law simply states that the tangent of this special angle equals the refractive index of the surface.

📝 Teacher's Note: Emphasize that at the Brewster angle, the reflected and refracted rays are exactly 90 degrees apart. Drawing this right angle on the board helps students remember why the sine becomes a cosine in the derivation.

🎯 Exam Tip: Always draw the ray diagram showing the reflected and refracted rays as perpendicular. Use the relation \(\theta_B + \theta_r = 90^\circ\) as the starting point for your mathematical proof to get full marks.

Question 7. Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.
Answer:
Description of Young’s double-slit interference experiment:
1. A plane wavefront is obtained by placing a linear source S of monochromatic light at the focus of a convex lens. It is then made to pass through an opaque screen AB having two narrow and similar slits \( S_1 \) and \( S_2 \). \( S_1 \) and \( S_2 \) are equidistant from S so that the wavefronts starting simultaneously from S and reaching \( S_1 \) and \( S_2 \) at the same time are in phase. A screen PQ is placed at some distance from screen AB as shown in below figure
2. \( S_1 \) and \( S_2 \) act as secondary sources. The crests/-troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in above figure. The point where these lines meet the screen have high intensity and are bright.
3. Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference.
4. These dark and bright regions are called fringes or bands and the whole pattern is called interference pattern.

Conditions for occurence of dark and bright fringes on the screen :
Consider Young’s double-slit experimental set up. Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength \( \lambda \) emerges out of two narrow and closely spaced, parallel slits \( S_1 \) and \( S_2 \) of equal widths. The separation \( S_1S_2 = d \) is very small. The interference pattern is observed on a screen placed parallel to the plane of and at considerable distance D (D >> d) from the slits. OO’ is the perpendicular bisector of segment \( S_1S_2 \).
Consider, a point P on the screen at a distance y from O’ (y << D). The two light waves from \( S_1 \) and \( S_2 \) reach P along paths \( S_1P \) and \( S_2P \), respectively. If the path difference (\( \Delta l \)) between \( S_1P \) and \( S_2P \) is an integral multiple of \( \lambda \), the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between \( S_1P \) and \( S_2P \) is half integral multiple of \( \lambda \), there will be destructive interference and a dark fringe will be produced at P.
From above figure,
\( (S_2P)^2 = (S_2S_{2’})^2 + (PS_{2’})^2 \)
\( = (S_2S_{2’})^2 + (PO’ + O’S_{2’})^2 \)
\( = D^2 + \left(y + \frac{d}{2}\right)^2 \) .... (1)
and
\( (S_1P)^2 = (S_1S_{1’})^2 + (PS_{1’})^2 \)
\( = (S_1S_{1’})^2 + (PO’ - O’S_{1’})^2 \)
\( = D^2 + \left(y - \frac{d}{2}\right)^2 \) .... (2)

\( \implies (S_2P)^2 - (S_1P)^2 = \left\{D^2 + \left(y + \frac{d}{2}\right)^2\right\} - \left\{D^2 + \left(y - \frac{d}{2}\right)^2\right\} \)

\( \implies (S_2P + S_1P)(S_2P - S_1P) \)
\( = [D^2 + y^2 + \frac{d^2}{4} + yd] - [D^2 + y^2 + \frac{d^2}{4} - yd] = 2yd \)

\( \implies S_2P - S_1P = \Delta l = \frac{2yd}{S_2P + S_1P} \)
In practice, \( D \gg y \) and \( D \gg d \),

\( \implies S_2P + S_1P \approx 2D \)

\( \implies \text{Path difference, } \Delta l = (S_2P - S_1P) \approx \frac{2yd}{2D} = y\frac{d}{D} \) .... (3)

Expression for the fringe width (or band width) : The distance between consecutive bright (or dark) fringes is called the fringe width (or band width) W. Point P will be bright (maximum intensity), if the path difference, \( \Delta l = y\frac{d}{D} = n\lambda \) where n = 0, 1, 2, 3…, Point P will be dark (minimum intensity equal to zero), if \( y\frac{d}{D} = (2m - 1)\frac{\lambda}{2} \), where, m = 1, 2, 3…,
Thus, for bright fringes (or bands),
\( y_n = 0, \frac{\lambda D}{d}, \frac{2\lambda D}{d} \dots \)
and for dark fringes (or bands),
\( y_m = \frac{\lambda D}{2d}, 3\frac{\lambda D}{2d}, 5\frac{\lambda D}{2d} \dots \)

These conditions show that the bright and dark fringes (or bands) occur alternately and are equally – spaced. For Point O’, the path difference \( (S_2O’ - S_1O’) = 0 \). Hence, point O’ will be bright. It corresponds to the centre of the central bright fringe (or band). On both sides of O’, the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit.

Let \( y_n \) and \( y_{n+1} \), be the distances of the \( n^{th} \) and \( (n + 1)^{th} \) bright fringes from the central bright fringe.

\[ \therefore \frac{y_n d}{D} = n\lambda \quad \therefore y_n = \frac{n\lambda D}{d} \quad \dots (4) \]

\[ \text{and } \frac{y_{n+1} d}{D} = (n + 1)\lambda \quad \therefore y_{n+1} = \frac{(n + 1)\lambda D}{d} \quad \dots (5) \]

The distance between consecutive bright fringes
\[ = y_{n+1} - y_n = \frac{\lambda D}{d} [(n + 1) - n] = \frac{\lambda D}{d} \quad \dots (6) \]

Hence, the fringe width,
\[ \therefore W = \Delta y = y_{n+1} - y_n = \frac{\lambda D}{d} \text{ (for bright fringes)} \quad \dots (7) \]

Alternately, let \( y_m \) and \( y_{m+1} \) be the distances of the \( m^{th} \) and \( (m + 1)^{th} \) dark fringes respectively from the central bright fringe.
\[ \therefore \frac{y_m d}{D} = (2m - 1) \frac{\lambda}{2} \text{ and} \]

\[ \frac{y_{m+1} d}{D} = [2(m + 1) - 1] \frac{\lambda}{2} = (2m + 1) \frac{\lambda}{2} \quad \dots (8) \]

\[ \therefore y_m = (2m - 1) \frac{\lambda D}{2d} \text{ and} \]

\[ y_{m+1} = (2m + 1) \frac{\lambda D}{2d} \quad \dots (9) \]

\(\therefore\) The distance between consecutive dark fringes,
\[ y_{m+1} - y_m = \frac{\lambda D}{2d} [(2m + 1) - (2m - 1)] = \frac{\lambda D}{d} \quad \dots (10) \]

\[ \therefore W = y_{m+1} - y_m = \frac{\lambda D}{d} \text{ (for dark fringes)} \quad \dots (11) \]

Eqs. (7) and (11) show that the fringe width is the same for bright and dark fringes.

[Note : In the first approximation, the path difference is \( d \sin \theta \).]

In simple words: This experiment shows how light waves from two small slits overlap to create a pattern of bright and dark stripes on a screen. Bright stripes happen when waves add up, and dark ones happen when they cancel each other out. The distance between these stripes is the fringe width.

📝 Teacher's Note: When explaining the path difference, use the analogy of two runners starting at slightly different points; the difference in the distance they run determines if they reach the finish line in step (in phase) or out of step (out of phase).

🎯 Exam Tip: The approximation \(S_1P + S_2P \approx 2D\) is crucial. Be sure to explicitly state the assumption \(D \gg d\) and \(D \gg y\) to justify this step in your derivation.

Question 8. What are the conditions for obtaining good interference pattern? Give reasons.
Answer:
The conditions necessary for obtaining well defined and steady interference pattern :
1. The two sources of light should be coherent:
The two sources must maintain their phase relation during the time required for observation. If the phases and phase difference vary with time, the positions of maxima and minima will also change with time and consequently the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed. For coherence, the two secondary sources must be derived from a single original source.
2. The light should be monochromatic :
Otherwise, interference will result in complex coloured bands (fringes) because the separation of successive bright bands (fringes) is different for different colours. It also may produce overlapping bands.
3. The two light sources should be of equal brightness, i.e., the waves must have the same amplitude.
The interfering light waves should have the same amplitude. Then, the points where the waves meet in opposite phase will be completely dark (zero intensity). This will increase the contrast of the interference pattern and make it more distinct.
4. The two light sources should be narrow :
If the source apertures are wide in comparison with the light wavelength, each source will be equivalent to multiple narrow sources and the superimposed pattern will consist of bright and less bright fringes. That is, the interference pattern will not be well defined.
5. The interfering light waves should be in the same state of polarization :
Otherwise, the points where the waves meet in opposite phase will not be completely dark and the interference pattern will not be distinct.
6. The two light sources should be closely spaced and the distance between the screen and the sources should be large : Both these conditions are desirable for appreciable fringe separation. The separation of successive bright or dark fringes is inversely proportional to the closeness of the slits and directly proportional to the screen distance.
In simple words: To see a clear and stable pattern of light and dark bands, the light sources must be exactly in step, be of one single color, have equal brightness, and be very tiny and close together.

📝 Teacher's Note: Use the analogy of two identical water wave generators in a pool; if they are not synchronized (coherent), the waves become a mess rather than a pattern.

🎯 Exam Tip: Listing 'Coherence' and 'Monochromaticity' as the first two points is crucial as they are the primary requirements for any interference.

Question 9. What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?
Answer:
Coherent sources : Two sources of light are said to be coherent if the phase difference between the emitted waves remains constant.
It is not possible to observe interference pattern with light from any two different sources. This is because, no observable interference phenomenon occurs by superposing light from two different sources. This happens due to the fact that different sources emit waves of different frequencies. Even if the two sources emit light of the same frequency, the phase difference between the wave trains from them fluctuates randomly and rapidly, i.e., they are not coherent. Consequently, the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed.
In the laboratory, coherent sources can be obtained by using
(1) Lloyd’s mirror and
(2) Fresnel’s biprism.

(1) Lloyd’s mirror : A plane polished mirror is kept at some distance from the source of monochromatic light and light is made incident on the mirror at a grazing angle.
Some light falls directly on the screen as shown by the black lines in above figure, while some light falls on the screen after reflection from the mirror as shown by red lines. The reflected light appears to come from a virtual source and thus two sources can be obtained. These two sources are coherent as they are derived from a single source. Superposition of the waves coming from these coherent sources, under appropriate conditions, gives rise to interference pattern consisting of alternate bright and dark bands on the screen as shown in the figure.

(2) Fresnel’s biprism : It is a single prism having an obtuse angle of about 178° and the other two angles of about 1° each. The biprism can be considered as made of two thin prisms of very small refracting angle of about 1°. The source, in the form of an illuminated narrow slit, is aligned parallel to the refracting edge of the biprism. Monochromatic light from the source is made to pass through that narrow slit and fall on the biprism.
Two virtual images \( S_1 \) and \( S_2 \) are formed by the two halves of the biprism. These are coherent sources which are obtained from a single secondary source S. The two waves coming from \( S_1 \) and \( S_2 \) interfere under appropriate conditions and form interference fringes, like those obtained in Young’s double-slit experiment, as shown in the figure in the shaded region. The formula for y is the same as in Young’s experiment.
In simple words: Coherent sources are like two speakers playing exactly the same song at the same time. Since two separate lamps can't do this, we use tricks like mirrors or prisms to split light from one lamp into two identical virtual sources.

📝 Teacher's Note: Emphasize that "phase difference remains constant" is the technical definition. Explain that independent sources have random phase jumps because atoms emit light in bursts.

🎯 Exam Tip: When describing Lloyd's mirror or Fresnel Biprism, always mention that the two sources are derived from a "single parent source" to ensure coherence.

Question 10. What is diffraction of light? How does it differ from interference? What are Fraunhoffer and Fresnel diffractions?
Answer:
1. Phenomenon of diffraction of light: When light passes by the edge of an obstacle or through a small opening or a narrow slit and falls on a screen, the principle of rectilinear propagation of light from geometrical optics predicts a sharp shadow. However, it is found that some of the light deviates from its rectilinear path and penetrates into the region of the geometrical shadow. This is a general characteristic of wave phenomena, which occurs whenever a portion of the wavefront is obstructed in some way. This bending of light waves at an edge into the region of geometrical shadow is called diffraction of light.
2. Differences between interference and diffraction :
1. The term interference is used to characterise the superposition of a few coherent waves (say, two). But when the superposition at a point involves a large number of waves coming from different parts of the same wavefront, the effect is referred to as diffraction.
2. Double-slit interference fringes are all of equal width. In single-slit diffraction pattern, only the non-central maxima are of equal width which is half of that of the central maximum.
3. In double-slit interference, the bright and dark fringes are equally spaced. In diffraction, only the non-central maxima lie approximately halfway between the minima.
4. In double-slit interference, bright fringes are of equal intensity. In diffraction, successive non-central maxima decrease rapidly in intensity.
[Note : Interference and diffraction both have their origin in the principle of superposition of waves. There is no physical difference between them. It is just a question of usage. When there are only a few sources, say two, the phenomenon is usually called interference. But, if there is a large number of sources the word diffraction is used.]
3. Diffraction can be classified into two types depending on the distances involved in the experimental setup :
(A) Fraunhofer diffraction : In this class of diffraction, both the source and the screen are at infinite distances from the aperture. This is achieved by placing the source at the focus of a convex lens and the screen at the focal plane of another convex lens.
(B) Fresnel diffraction : In this class of diffraction, either the source of light or the screen or both are at finite distances from the diffracting aperture. The incident wavefront is either cylindrical or spherical depending on the source. A lens is not needed to observe the diffraction pattern on the screen.
In simple words: Diffraction is light bending around corners. Interference happens when light from two slits overlaps, while diffraction happens when light from different parts of the same single slit overlaps.

📝 Teacher's Note: Use the example of hearing sound from around a corner to explain diffraction, as sound waves have larger wavelengths and diffract more easily than light.

🎯 Exam Tip: In difference questions, highlighting that interference fringes are of equal intensity while diffraction maxima decrease in intensity is a key point for full marks.

Question 11. Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.
Answer:
When a parallel beam of monochromatic light of wavelength \( \lambda \) illuminates a single slit of finite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called ‘ the diffraction pattern of a single slit.
Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D (\( \gg a \)) from the slit and at the focal plane of the convex lens.
We can imagine the single slit as being made up of a large number of Huygens’ sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens’ wavelets.
Now, imagine the single slit as made up of two adjacent slits, each of width a/2. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also be in phase at the point \( P_0 \) on the screen, where \( P_0 \) is equidistant from all the Huygens sources. At \( P_0 \), then, we get the central maximum.
For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or O and B) is \( \lambda / 2 \), which is the condition for destructive interference. Suppose, the nodal line OP for the first minimum subtends an angle \( \theta \) at the slit; \( \theta \) is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, \( D \gg a \), the arc AC can be considered a straight line at right angles to PB. Then, \( \Delta ABC \) is a right-angled triangle similar to \( \Delta OP_0P \).
This means that, \( \angle BAC = \theta \)
\( \therefore BC = a \sin \theta \)
\( \therefore \) Difference in path length,
\( BC = PB - PA = \lambda \)
\( \therefore a \sin \theta = \lambda \)
\( \therefore \sin \theta \simeq \theta = \frac{\lambda}{a} \quad \dots (1) \)
(\( \because \theta \) is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the \( m^{th} \) minimum (\( m = \pm 1, \pm 2, \pm 3, \dots \)).
\( \theta_m = \frac{m\lambda}{a} \text{ (mth minimum)} \quad \dots (2) \)
as \( \theta_m \) is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of \( \frac{\lambda}{2} \):
\( a \sin \theta_m = (2m + 1) \frac{\lambda}{2} = (m + \frac{1}{2})\lambda \)
i.e., at angles given by,
\( \theta_m \simeq \sin \theta_m = (m + \frac{1}{2})\frac{\lambda}{a} \text{ (with secondary maximum)} \quad \dots (3) \)
In simple words: To find where the dark spots are, we imagine splitting the slit into parts. If the light from the top half and bottom half cancels out, we get a dark spot. This happens when the slit width, angle, and wavelength follow the formula \( a \sin \theta = m\lambda \).

📝 Teacher's Note: Draw the slit as a series of 8-10 points. Show how at the first minimum, point 1 cancels point 5, point 2 cancels point 6, etc., to make the concept of "destructive interference" from a single slit more intuitive.

🎯 Exam Tip: Remember that for diffraction, the condition for a dark fringe is \( a \sin \theta = m\lambda \), which looks like the bright fringe condition for interference—don't swap them!

Question 12. Describe what is Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.
Answer:
Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.
The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.
Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.
Resolving power of an optical instrument:
The primary aim of using an optical instrument is to see fine details, whether observing a star system through a telescope or a living cell through a microscope. After passing through an optical system, light from two adjacent parts of the object should produce sharp, distinct (separate) images of those parts. The objective lens or mirror of a telescope or microscope acts like a circular aperture. The diffraction pattern of a circular aperture consists of a central bright spot (called the Airy disc and corresponds to the central maximum) and concentric dark and bright rings.
Light from two close objects or parts of an object after passing through the aperture of an optical system produces overlapping diffraction patterns that tend to obscure the image. If these diffraction patterns are so broad that their central maxima overlap substantially, it is difficult to decide if the intensity distribution is produced by two separate objects or by one.
The resolving power of an optical instrument, e.g., a telescope or microscope, is a measure of its ability to produce detectably separate images of objects that are close together.
Definition : The smallest linear or angular separation between two point objects which appear just resolved when viewed through an optical instrument is called the limit of resolution.
The limit of resolution of the instrument and its reciprocal is called the resolving power of the instrument.
Rayleigh’s Criterion For Minimum Resolution:
Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.
The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.
Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.
The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.
Consider two stars seen through a telescope. The diameter (D) of the objective lens or mirror corresponds to the diffracting aperture. For a distant point source, the first diffraction minimum is at an angle \( \theta \) away from the centre such that
\[ D \sin \theta = 1.22 \lambda \]
where \( \lambda \) is the wavelength of light. The angle \( \theta \) is usually so small that we can substitute \( \sin \theta \approx \theta \) (\( \theta \) in radian). Thus, the Airy disc for each star will be spread out over an angular half-width \( \theta = 1.22 \lambda / D \) about its geometrical image point. The radius of the Airy disc at the focal plane of the objective lens is \( r = f \theta = 1.22 f \lambda / D \), where \( f \) is the focal length of the objective.
When observing two closely-spaced stars, the Rayleigh criterion for just resolving the images as that of two point sources (instead of one) is met when the centre of one Airy disc falls on the first minimum of the other pattern. Thus, the angular limit (or angular separation) of resolution is \[ \theta = \frac{1.22 \lambda}{D} \dots (1) \] and the linear separation between the images at the focal plane of the objective lens is
\[ y = f \theta \dots (2) \]
\( \implies \) Resolving power of a telescope,
\[ R = \frac{1}{\theta} = \frac{D}{1.22 \lambda} \dots (3) \]
It depends
1. directly on the diameter of the objective lens or mirror,
2. inversely on the wavelength of the radiation.

In simple words: Rayleigh's criterion says that to tell two objects apart, the bright center of one object's light pattern shouldn't be closer than the first dark ring of the other object's pattern. It's the rule that tells us how "clear" a telescope or microscope can be.

📝 Teacher's Note: Compare this to pixels on a screen. If the pixels are too big (diffraction patterns too wide), images of two dots merge into one blurry blob.

🎯 Exam Tip: Always include the diagram showing the two overlapping curves to explain "Just resolved" — it's the most effective way to show you understand the criterion.

Question 13. Whitelight consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55? [Ans: 258.1 - 451.6 nm]
Answer:
Let \( \lambda_1 \) and \( \lambda_2 \) be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively. Let \( \lambda_a \) be the wavelength of light in vacuum.
\( \lambda_1 = \frac{\lambda_a}{n} = \frac{400 \times 10^{-9} \text{ m}}{1.55} = 258.06 \times 10^{-9} \text{ m} \)
\( \lambda_2 = \frac{\lambda_a}{n} = \frac{700 \times 10^{-9} \text{ m}}{1.55} = 451.61 \times 10^{-9} \text{ m} \)
The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.
In simple words: When light enters glass, it slows down, and its wavelength gets shorter. By dividing the original wavelengths by the glass's refractive index (1.55), we find the new, smaller range of wavelengths.

📝 Teacher's Note: Remind students that frequency remains constant when light enters a different medium; only the speed and wavelength change. This is a fundamental concept in wave optics.

🎯 Exam Tip: Always use the formula \( \lambda_{medium} = \lambda_{vacuum} / n \). Ensure you convert units to nanometers or meters consistently.

Question 14. The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance x cm through a medium having refractive index 1.25. Determine the value of x. [Ans: 3.84 cm]
Answer:
Let \( d_{fg} \) and \( d_m \) be the distances by the ray of light in the flint glass and the medium respectively. Also, let \( n_{fg} \) and \( n_m \) be the refractive indices of the flint glass and the medium respectively.
Data : \( d_{fg} = 3 \text{ cm}, n_{fg} = 1.6, n_m = 1.25 \),
Optical path = \( n_m \times d_m = n_{fg} \times d_{fg} \)

\( \implies d_m = \frac{n_{fg} \times d_{fg}}{n_m} = \frac{1.6 \times 3}{1.25} = 3.84 \text{ cm} \)
Thus, x cm = 3.84 cm

\( \implies x = 3.84 \)
This is the value of x.
In simple words: "Optical path" is the distance light would travel in a vacuum in the same amount of time. To keep the paths equal, light must travel further in a medium with a lower refractive index.

📝 Teacher's Note: Use the concept of 'equivalent vacuum distance' to help students visualize why the product of refractive index and actual distance is constant for equal optical paths.

🎯 Exam Tip: Simply equate \( n_1d_1 = n_2d_2 \). It's a quick calculation, but check that units for distance are the same on both sides.

Question 15. A double-slit arrangement produces interference fringes for sodium light (\( \lambda = 589 \text{ nm} \)) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)? [Ans: 0.15°]
Answer:
Data : \( \theta_1 = 0.20^\circ, n_w = 1.33 \)
In the first approximation,
\( D \sin \theta_1 = y_1 \) and \( D \sin \theta_2 = y_2 \)

\( \implies \frac{\sin \theta_2}{\sin \theta_1} = \frac{y_2}{y_1} \dots (1) \)
Now, \( y \propto \frac{\lambda D}{d} \)
For given \( d \) and \( D \),
\( y \propto \lambda \)

\( \implies \frac{y_2}{y_1} = \frac{\lambda_2}{\lambda_1} \dots (2) \)
Now, \( n_w = \frac{\lambda_1}{\lambda_2} \dots (3) \)
From Eqs. (1), (2) and (3), we get,
\( \frac{\sin \theta_2}{\sin \theta_1} = \frac{\lambda_2}{\lambda_1} = \frac{1}{n_w} \)

\( \implies \sin \theta_2 = \frac{\sin \theta_1}{n_w} = \frac{\sin 0.2}{1.33} = \frac{0.0035}{1.33} = 0.0026 \)

\( \implies \theta_2 = \sin^{-1}(0.0026) = 0.15^\circ \)
This is the required angular fringe separation.
In simple words: When the apparatus is put underwater, the light's wavelength shrinks. Since the gap between fringes depends on wavelength, the fringes squeeze closer together by the same factor as the refractive index.

📝 Teacher's Note: Explain that since \( \theta \) is small, the relation \( \theta \propto \lambda \) simplifies calculations significantly. Students often forget to account for the medium's effect on wavelength.

🎯 Exam Tip: For small angles, use the shortcut \( \theta_2 = \theta_1 / n \). This saves time and minimizes trigonometric errors.

Question 16. In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits.
(a) What is the angular separation in radians between the central maximum and an adjacent maximum?
(b) What is the distance between these maxima on a screen 50.0 cm from the slits?
[Ans: 0.01 rad, 0.5 cm]
Answer:
Data : \( d = 100\lambda, D = 50.0 \text{ cm} \)
(a) The condition for maximum intensity in Young’s experiment is, \( d \sin \theta = n\lambda, n = 0, 1, 2 \dots \),
The angle between the central maximum and its adjacent maximum can be determined by setting n equal to 1,

\( \implies d \sin \theta = \lambda \)

\( \implies \theta = \sin^{-1} \left( \frac{\lambda}{d} \right) = \sin^{-1} \left( \frac{\lambda}{100\lambda} \right) = \sin^{-1} (0.01) \approx 0.01 \text{ rad} \)
(b) The distance between these maxima on the screen is \( D \sin \theta = D \left( \frac{\lambda}{d} \right) \)
\( = (50.0 \text{ cm}) \left( \frac{\lambda}{100\lambda} \right) \)
\( = 0.50 \text{ cm} \)
In simple words: (a) We find the angle where the next bright spot appears by comparing the slit width to the wavelength. (b) We then project this angle onto the screen 50 cm away to see how many centimeters apart they actually look.

📝 Teacher's Note: Highlight the use of the small angle approximation \( \sin \theta \approx \theta \) (in radians). This is valid here because 0.01 is much smaller than 1.

🎯 Exam Tip: State clearly which order of maximum (\( n=1 \)) you are using. Label your units (rad vs cm) properly to avoid losing marks.

Question 17. Unpolarized light with intensity \( I_0 \) is incident on two polaroids. The axis of the first polaroid makes an angle of 50° with the vertical, and the axis of the second polaroid is horizontal. What is the intensity of the light after it has passed through the second polaroid? [Ans: \( I_0/2 \times (\cos 40^\circ)^2 \)]
Answer:
According to Malus’ law, when the unpolarized light with intensity \( I_0 \) is incident on the first polarizer, the polarizer polarizes this incident light. The intensity of light becomes \( I_1 = I_0/2 \).
Now, \( I_2 = I_1 \cos^2 \theta \)

\( \implies I_2 = \left( \frac{I_0}{2} \right) \cos^2 \theta \)
Also, the angle \( \theta \) between the axes of the two polarizers is \( \theta_2 - \theta_1 \).

\( \implies I_2 = \left( \frac{I_0}{2} \right) \cos^2 (90^\circ - 50^\circ) = \left( \frac{I_0}{2} \right) \cos^2 40^\circ \)
The intensity of light after it has passed through the second polaroid = \( \left( \frac{I_0}{2} \right) \cos^2 40^\circ = \frac{I_0}{2} (0.7660)^2 \)
\( = 0.2934 I_0 \)
In simple words: The first filter cuts the light in half. The second filter cuts it further based on the angle between the two filters. Since one is 50° from vertical and the other is horizontal (90° from vertical), the angle between them is 40°.

📝 Teacher's Note: Students often use the initial angle with the vertical directly. Emphasize that Malus' Law uses the relative angle *between* the two polarizers' axes.

🎯 Exam Tip: Don't forget the factor of 1/2 for the first unpolarized-to-polarized transition. It's the most common mistake in polarization problems.

Question 18. In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used. [Ans: 5000 Å]
Answer:
Data : \( D = 1.2 \text{ m} \)
The distance between the central bright band and the 20th bright band is 0.4 cm.

\( \implies y_{20} = 0.4 \text{ cm} = 0.4 \times 10^{-2} \text{ m} \)
\( W = \frac{y_{20}}{20} = \frac{0.4}{20} \times 10^{-2} \text{ m} = 2 \times 10^{-4} \text{ m} \),
\( d_1 = 0.9 \text{ cm} = 0.9 \times 10^{-2} \text{ m} \), \( v_1 = 90 \text{ cm} = 0.9 \text{ m} \)

\( \implies u_1 = D - v_1 = 1.2 \text{ m} - 0.9 \text{ m} = 0.3 \text{ m} \)
Now, \( \frac{d_1}{d} = \frac{v_1}{u_1} \)

\( \implies d = \frac{d_1 u_1}{v_1} = \frac{(0.9 \times 10^{-2})(0.3)}{0.9} \text{ m} \)
\( = 3 \times 10^{-3} \text{ m} \)
The wavelength of light,
\( \lambda = \frac{Wd}{D} = \frac{2 \times 10^{-4} \times 3 \times 10^{-3}}{1.2} \text{ m} \)
\( = 5 \times 10^{-7} \text{ m} \)
\( = 5 \times 10^{-7} \times 10^{10} \text{ \AA} \)
\( = 5000 \text{ \AA} \)
In simple words: We first find the width of a single fringe. Then, using lens equations, we figure out the actual distance between the two light sources. Finally, we plug these into the standard formula to find the light's wavelength.

📝 Teacher's Note: The biprism experiment is a common lab question. Explain how the lens allows us to measure 'd' (the virtual slit separation) which otherwise cannot be measured directly.

🎯 Exam Tip: Be very careful with units. Converting cm to m before doing any division or multiplication is the safest way to avoid power-of-ten errors.

Question 19. In Fraunhoffer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light. [Ans: 5000 Å]
Answer:
Data : \( D = 2 \text{ m}, y_{1d} = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}, a = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m} = 2 \times 10^{-4} \text{ m} \)
\( y_{md} = m \frac{\lambda D}{a} \)

\( \implies \lambda = \frac{y_{1d} a}{D} \) (for m = 1)
\( \lambda = \frac{5 \times 10^{-3} \times 2 \times 10^{-4}}{2} \)
\( \lambda = 5 \times 10^{-7} \text{ m} = 5 \times 10^{-7} \times 10^{10} \text{ \AA} = 5000 \text{ \AA} \)
This is the wavelength of light.
In simple words: In a diffraction pattern, the first dark spot's position depends on the slit size and wavelength. By measuring how far the dark spot is from the center, we can calculate exactly what kind of light was used.

📝 Teacher's Note: Contrast this with interference. In single-slit diffraction, the formula for minima (\( a \sin \theta = m\lambda \)) looks like the formula for maxima in double-slit interference. Don't let students mix them up.

🎯 Exam Tip: Remember that 'a' represents the width of the single slit. Check if the question gives the total width of the central maximum or just the distance to the first minimum.

Question 20. The intensity of the light coming from one of the slits in Young’s experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern? [Ans: 34]
Answer:
Data : \( I_1 : I_2 = 2 : 1 \)
If \( E_{10} \) and \( E_{20} \) are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is
\( \frac{I_{max}}{I_{min}} = \left( \frac{E_{10} + E_{20}}{E_{10} - E_{20}} \right)^2 = \left( \frac{r + 1}{r - 1} \right)^2 \)
where \( r = \frac{E_{10}}{E_{20}} \)

\( \implies \frac{I_1}{I_2} = \left( \frac{E_{10}}{E_{20}} \right)^2 = r^2 \)

\( \implies r = \sqrt{\frac{I_1}{I_2}} = \sqrt{2} \)

\( \implies \frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)^2 = \left( \frac{2.414}{0.414} \right)^2 = (5.83)^2 \)
\( = 33.99 \approx 34 \).
The ratio of the intensities of the bright and dark fringes in the resulting interference pattern is 34 : 1.
In simple words: Even though one light source is only twice as bright as the other, the way they interfere makes the brightest spots 34 times brighter than the darkest spots! This happens because intensity depends on the square of the added waves.

📝 Teacher's Note: Help students understand that intensities don't simply add; amplitudes do. Use this problem to reinforce the relationship \( I \propto A^2 \).

🎯 Exam Tip: Rationalizing the denominator inside the square or using the decimal values are both acceptable. Be careful with the square at the final step.

Question 21. A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima? [Ans: 1.1 mm]
Answer:
Data : \( \lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m}, a = 0.4 \text{ mm} = 4 \times 10^{-4} \text{ m}, D = 40 \text{ cm} = 40 \times 10^{-2} \text{ m} \)
\( y_{md} = m \frac{\lambda D}{a} \)

\( \implies y_{1d} = 1 \frac{\lambda D}{a} \) and
\( 2y_{1d} = \frac{2 \lambda D}{a} \)
\( = \frac{2 \times 550 \times 10^{-9} \times 40 \times 10^{-2}}{4 \times 10^{-4}} \)
\( = 2 \times 550 \times 10^{-6} = 1100 \times 10^{-6} \)
\( = 1.100 \times 10^{-3} \text{ m} = 1.100 \text{ mm} \)
This is the distance between the two first order minima.
In simple words: We are looking for the total width of the central bright patch. It stretches from the first dark spot on the left to the first dark spot on the right. We calculate this by doubling the distance of one dark spot from the center.

📝 Teacher's Note: "Distance between the two first order minima" is exactly the same thing as the "width of the central maximum". These terms are interchangeable in exams.

🎯 Exam Tip: Watch out for "first order minima" vs "first order maxima". Minima are where the light vanishes. The calculation requires a factor of 2 because there's a minimum on both sides of the center.

Question 22. What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°? [Ans: 1.274]
Answer:
Data : \( \theta = 45^\circ, m = 1 \)
\( a \sin \theta = m\lambda \) for (m = 1, 2, 3... minima)
Here, m = 1 (First minimum)

\( \implies a \sin 45^\circ = (1) \lambda \)

\( \implies \frac{a}{\lambda} = \frac{1}{\sin 45^\circ} = 1.414 \)
This is the required ratio.
In simple words: This ratio tells us how narrow the slit is compared to the light's wavelength. For light to spread out to a wide 45° angle at the first dark spot, the slit must be about 1.4 times wider than the wavelength.

📝 Teacher's Note: Note that the calculation \( 1/\sin(45^\circ) \) gives 1.414, but the provided answer key in the text says 1.274. In exams, show your steps clearly to get partial credit even if the final number differs from a provided answer key.

🎯 Exam Tip: Always state the formula \( a \sin \theta = n\lambda \) first. It shows the examiner you understand the physics even if you make a calculator error.

Question 23. Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is
(a) 500 nm (visible light);
(b) 50 µm (infrared radiation);
(c) 0.500 nm (X-rays)?
[Ans: 0.4167 mm, 41.67 mm, \( 4.167 \times 10^{-4} \text{ mm} \)]
Answer:
Data: \( 2W = 6 \text{ mm} \implies W = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}, D = 2.5 \text{ m} \),
(a) \( \lambda_1 = 500 \text{ nm} = 5 \times 10^{-7} \text{ m} \)
(b) \( \lambda_2 = 50 \text{ \mu m} = 5 \times 10^{-5} \text{ m} \)
(c) \( \lambda_3 = 0.500 \text{ nm} = 5 \times 10^{-10} \text{ m} \)
Let a be the slit width.
Using \( a = \frac{2 \lambda D}{2W} \):
(a) \( a = \frac{2 \times 5 \times 10^{-7} \times 2.5}{6 \times 10^{-3}} = 4.167 \times 10^{-4} \text{ m} = 0.4167 \text{ mm} \)
(b) \( a = \frac{2 \times 5 \times 10^{-5} \times 2.5}{6 \times 10^{-3}} = 4.167 \times 10^{-2} \text{ m} = 41.67 \text{ mm} \)
(c) \( a = \frac{2 \times 5 \times 10^{-10} \times 2.5}{6 \times 10^{-3}} = 4.167 \times 10^{-7} \text{ m} = 4.167 \times 10^{-4} \text{ mm} \)

(a) \( W = \frac{y\lambda_1}{a} \)
\( \therefore a = \frac{y\lambda_1}{W} = \frac{(2.5)(5 \times 10^{-7})}{3 \times 10^{-3}} \)
\( = 4.167 \times 10^{-4} \text{ m} \)
\( = 0.4167 \text{ mm} \)

(b) \( W = \frac{y\lambda_2}{a} \)
\( \therefore a = \frac{y\lambda_2}{W} = \frac{(2.5)(5 \times 10^{-5})}{3 \times 10^{-3}} \)
\( = 4.167 \times 10^{-2} \text{ m} \)
\( = 41.67 \text{ mm} \)

(c) \( W = \frac{y\lambda_3}{a} \)
\( \therefore a = \frac{y\lambda_3}{W} = \frac{(2.5)(5 \times 10^{-10})}{3 \times 10^{-3}} \)
\( = 4.167 \times 10^{-7} \text{ m} \)
\( = 4.167 \times 10^{-4} \text{ mm} \)

In simple words: This problem shows that to get the exact same sized diffraction pattern, different types of light (like X-rays vs Infrared) need very different slit sizes. Longer wavelengths (infrared) need huge slits, while tiny wavelengths (X-rays) need microscopic slits.

📝 Teacher's Note: This is a great problem to show the scale of different EM waves. It helps students understand why we don't see diffraction through a doorway for visible light, but we do for much longer radio waves.

🎯 Exam Tip: When given "width of central maximum", remember it is \( 2y_1 \). You must use the factor of 2 in your numerator or divide the given width by 2 before starting.

Question 24. A star is emitting light at the wavelength of 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter of 200 inch. [Ans: 1.2 × 10⁻⁷ rad]
Answer:
Data: \( \lambda = 5000 \text{ Å} = 5 \times 10^{-7} \text{ m} \),
\( D = 200 \times 2.54 \text{ cm} = 5.08 \text{ m} \)
\( \theta = \frac{1.22\lambda}{D} \)
\( = \frac{1.22 \times 5 \times 10^{-7}}{5.08} \)
\( = 1.2 \times 10^{-7} \text{ rad} \)
This is the required quantity.
In simple words: The limit of resolution is the smallest angle between two stars that allows a telescope to see them as separate objects instead of one blurry spot. A bigger telescope lens or mirror makes this angle smaller, allowing us to see more detail.

📝 Teacher's Note: Remind students to always convert non-SI units like inches and Angstroms to meters before using formulas to avoid calculation errors.

🎯 Exam Tip: Rayleigh's criterion is the key concept here. Ensure you memorize the factor of 1.22 which is specific to circular apertures like telescope lenses.

Question 25. The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used? 
Answer:
Data: \( \lambda_1 = 6000 \text{ Å} = 6 \times 10^{-7} \text{ m}, \lambda_2 = 4800 \text{ Å} = 4.8 \times 10^{-7} \text{ m}, W_1 = 0.32 \text{ mm} = 3.2 \times 10^{-4} \text{ m} \)
Distance between consecutive bright fringes,
\( W = \frac{\lambda D}{d} \)
For \( \lambda_1, W_1 = \frac{\lambda_1 D}{d} \) and .... (1)
for \( \lambda_2, W_2 = \frac{\lambda_2 D}{d} \) .... (2)
\( \frac{W_2}{W_1} = \frac{\lambda_2 D/d}{\lambda_1 D/d} = \frac{\lambda_2}{\lambda_1} \)
\( \therefore W_2 = \left( \frac{\lambda_2}{\lambda_1} \right) W_1 = \left( \frac{4.8 \times 10^{-7}}{6 \times 10^{-7}} \right) (3.2 \times 10^{-4}) \)
\( = (0.8) (3.2 \times 10^{-4}) \text{ m} \)
\( = 2.56 \times 10^{-4} \text{ m} \)
\( \therefore \Delta W = W_1 - W_2 \)
\( = 3.2 \times 10^{-4} \text{ m} - 2.56 \times 10^{-4} \text{ m} \)
\( = 0.64 \times 10^{-4} \text{ m} \)
\( = 6.4 \times 10^{-5} \text{ m} \)
\( = 0.064 \text{ mm} \)
This is the required change in distance.
In simple words: Fringe width is the gap between bright bands. Since this gap depends on the wavelength of light, changing the color (wavelength) causes the bands to move closer together or further apart.

📝 Teacher's Note: Emphasize the direct proportionality between fringe width and wavelength \( (W \propto \lambda) \). If the wavelength decreases, the fringes will naturally get closer together.

🎯 Exam Tip: When the question asks "by how much will the distance change", always calculate the difference (\( \Delta W \)), not just the new value (\( W_2 \)).

Use your brain power (Textbook Page No. 167)

What will you observe if

Question 1. you look at a source of unpolarized light through a polarizer ?
Answer:
When a source of unpolarized light is viewed through a polarizer, the intensity of the light transmitted by the polarizer is reduced and hence the source appears less bright.
In simple words: A polarizer works like a filter that blocks half of the light waves. Because less light reaches your eye, the source just looks dimmer than usual.

📝 Teacher's Note: Use the analogy of a picket fence and a vibrating rope to explain why only specific orientations of light waves pass through.

🎯 Exam Tip: State clearly that the intensity becomes exactly half for an ideal polarizer when viewing perfectly unpolarized light.

Question 2. you look at the source through two polarizers and rotate one of them around the path of light for one full rotation?
Answer:
When a source of unpolarized light is viewed through two polarizers and the second polarizer is rotated gradually, the intensity of the light transmitted by the second polarizer goes on decreasing. When the axes of polarization of the two polarizers are at 90° to each other, light almost disappears depending on the quality of the polarizers. (Ideally the intensity of the transmitted light should be zero.) The light reappears, i.e., its intensity increases, when the second polarizer is rotated further, and the intensity of the light becomes maximum when the axes of polarization are parallel again.
In simple words: Imagine two slotted gates. If both slots are vertical, light passes through. If you turn the second gate sideways (90°), the light that made it through the first gate gets blocked by the second, creating darkness.

📝 Teacher's Note: This is a demonstration of Malus's Law. Point out that the intensity varies as \( \cos^2 \theta \), where \( \theta \) is the angle between the transmission axes.

🎯 Exam Tip: Use terms like "crossed polarizers" for the 90° position and "parallel polarizers" for the 0°/180° positions to sound more technical.

Question 3. instead of rotating only one of the polaroids, you rotate both polaroids simultaneously in the same direction?
Answer:
If both the polaroids are rotated simultaneously in the same direction with the same angular velocity, then there would be no change in the intensity of the transmitted light observed.
In simple words: Since the angle between the two "gates" never changes, the amount of light getting through stays exactly the same throughout the rotation.

📝 Teacher's Note: Focus on the concept of relative angle. Since the relative angle \( \theta \) remains constant, the transmitted intensity \( I = I_0 \cos^2 \theta \) remains constant.

🎯 Exam Tip: The keyword here is "relative orientation". If the relative orientation is fixed, the intensity is constant.

Can you tell? (Textbook Page No. 168)

Question 1. If you look at the sky in a particular direction through a polaroid and rotate the polaroid around that direction what will you see ?
Answer:
As the scattered light is polarized, the sky appears bright and dim alternately.
In simple words: Sunlight scattering in the atmosphere becomes polarized. When you spin a polarizer while looking at the blue sky, you'll see the blue color get darker and lighter because you're blocking and then letting through that polarized light.

📝 Teacher's Note: Mention that polarization by scattering is maximum at a 90° angle from the sun's direction.

🎯 Exam Tip: Mention "scattering of light" as the physical process responsible for this polarization.

Question 2. Why does the sky appear to be blue while the clouds appear white ?
Answer:
The blue colour of the sky is because of the scattering of light by air molecules and dust particles in the atmosphere. As the wavelength of blue light is less than that of red light, the blue light is preferentially scattered than the light corresponding to other colours in the visible region. Clouds are seen due to scattering of light from lower parts of the atmosphere. The clouds contain the dust particles and water droplets which are sometimes large enough to scatter light of all the wavelengths such that the combined effect makes the clouds appear white.
In simple words: Tiny air molecules scatter blue light much better than other colors, making the sky look blue. But clouds have big water drops that scatter all colors equally, so they just look like a mix of everything, which is white light.

📝 Teacher's Note: This distinguishes between Rayleigh scattering (small particles, wavelength dependent) and Mie scattering (large particles, wavelength independent).

🎯 Exam Tip: For the sky, emphasize "wavelength dependent scattering" (Rayleigh scattering). For clouds, mention "large particles scattering all wavelengths equally".

Remember this (Textbook Page No. 171)

Question 1. For the interference pattern to be clearly visible on the screen, the distance (D) between the slits and the screen should be much larger than the distance (d) between the two slits (S₁ and S₂), i.e., D » d.
Answer:
The condition for constructive interference at P is,
\( \Delta l = y_n \frac{d}{D} = n\lambda \) …. (1)
\( y_n \) being the position (y-coordinate) of nth bright fringe \( (n = 0, \pm 1, \pm 2, \dots) \).
\( \therefore y_n = n\lambda \frac{D}{d} \) ….. (2)
Similarly, the position of mth \( (m = +1, \pm 2, \dots) \) dark fringe (destructive interference) is given by,
\( \Delta l = y_m \frac{d}{D} = (2m - 1)\frac{\lambda}{2} \) giving
\( y_m = (2m - 1)\lambda \frac{D}{2d} \) …(3)
The distance between any two consecutive bright or dark fringes, i.e., the fringe width
\( = W = \Delta y = y_{n + 1} - y_n = \lambda \frac{D}{d} \) …(4)
Conditions given by Eqs. (1) to (4) and hence the location of the fringes are derived assuming that the two sources \( S_1 \) and \( S_2 \) are in phase. If there is a non-zero phase difference between them it should be added appropriately. This will shift the entire fringe pattern but will not change the fringe width.
In simple words: To see a clear pattern of light and dark stripes, the screen needs to be far away compared to how close the two slits are. The width of these stripes depends on the wavelength of light and the distances involved.

📝 Teacher's Note: Explain that the approximation \( \sin \theta \approx \tan \theta \approx \theta \) is only valid when \( D \gg d \), which is why this condition is crucial for the standard fringe width formula.

🎯 Exam Tip: Be ready to derive the formula for fringe width \( W = \lambda D/d \). Remember that the fringe width is the same for both bright and dark fringes.

Do you know (Textbook Page No. 172 & 173)

Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.
The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.
The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence \( \theta_1 \) i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.
As the reflection is from the denser boundary, there is an additional phase difference of \( \pi \) radians (or an additional path difference \( \lambda/2 \)). This should be taken into account for mathematical analysis.