Class 10 Science Light Refraction Exam Notes

Class 10 Science Light Refraction Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

LIGHT: REFRACTION

REFRACTION OF LIGHT

When light rays travelling in one medium are incident on a transparent surface of another medium, they are bent as they travel in the second medium. This bending of light is called refraction.

Fig. Refraction of light from a plane transparent denser surface.

SOME ASSOCIATED TERMS

Consider a ray PQ incident at the point Q on a plane transparent surface XY, with the normal N₁QN₂ and the refracted ray QR.

• Transparent surface: the plane surface which refracts light (XY).
• Point of incidence: the point on the transparent surface where the ray of light meets it (Q).
• Normal: the perpendicular drawn to the transparent surface at the point of incidence (N₁QN₂).
• Incident ray: the ray of light which strikes the transparent surface at the point of incidence (PQ).
• Refracted ray: the ray of light which travels from the point of incidence into the other medium (QR).
• Angle of incidence (i): the angle between the incident ray and the normal at the point of incidence (angle PQN₁).
• Angle of refraction (r): the angle between the refracted ray and the normal at the point of incidence (angle RQN₂).
• Plane of incidence: the plane containing the normal and the incident ray.
• Plane of refraction: the plane containing the normal and the refracted ray.

LAWS OF REFRACTION OF LIGHT

• First law: the incident ray, the normal to the transparent surface at the point of incidence and the refracted ray all lie in one and the same plane.
• Second law (Snell’s law): the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media. This constant is called the refractive index of the second medium with respect to the first medium.

\[\dfrac{\sin i}{\sin r}=\text{constant}=\mu\]

REFRACTIVE INDEX

(a) In terms of the speed of light.

\[\text{Refractive index}\ \mu=\dfrac{\text{speed of light in vacuum}\ (c)}{\text{speed of light in the medium}\ (v)}\]

(b) In terms of wavelength. The frequency of light remains unchanged when it passes from one medium to another, so

\[\mu=\dfrac cv=\dfrac{\nu\lambda_{vac}}{\nu\lambda_{med}}=\dfrac{\lambda_{vac}}{\lambda_{med}}\]

(c) Relative refractive index. The refractive index of medium 2 with respect to medium 1 is the ratio of the speed of light (v₁) in medium 1 to the speed of light (v₂) in medium 2. It is denoted by ₁μ₂:

\[{}_1\mu_2=\dfrac{v_1}{v_2}=\dfrac{\lambda_1}{\lambda_2}=\dfrac{\mu_2}{\mu_1}\]

Since the refractive index is the ratio of two similar physical quantities, it has no unit and no dimension.

The refractive index of a medium depends on: (i) the nature of the medium, (ii) the wavelength of the light used, (iii) the temperature, and (iv) the nature of the surrounding medium. Refractive index is a characteristic of the pair of media and also depends on the wavelength of the light, but it is independent of the angle of incidence.

REFRACTION THROUGH A GLASS SLAB

Refraction through a rectangular glass slab and the principle of reversibility of light. A ray AE is incident on the face PQ of a rectangular glass slab at an angle of incidence i. On entering the slab it bends towards the normal and travels along EF at an angle of refraction r. The refracted ray EF is incident on the face SR at an angle of incidence r′, and the emerging ray FD bends away from the normal, with an angle of refraction e. The emergent ray FD is parallel to the incident ray AE, but it is laterally displaced with respect to the incident ray.

Fig. Refraction through a rectangular glass slab: lateral displacement.

Lateral shift is the perpendicular distance between the incident and emergent rays when light is incident obliquely on a refracting slab with parallel faces. It depends on the following factors:

1. It is directly proportional to the thickness of the glass slab.
2. It is directly proportional to the incident angle.
3. It is directly proportional to the refractive index of the glass slab.
4. It is inversely proportional to the wavelength of the incident light.

Proof that i = e. At E (air to glass): \(\mu_a\sin i=\mu_g\sin r\) …(1). At F (glass to air): \(\mu_g\sin r'=\mu_a\sin e\) …(2), where r′ = r because the faces are parallel. From (1) and (2), sin i = sin e, so i = e. Hence the incident and emergent rays are parallel to each other.

SPHERICAL LENS

A piece of transparent medium bounded by at least one spherical surface is called a spherical lens. There are two types:

(i) Convex (converging) lenses: thick in the middle and thin at the edges. The three types are: (a) double convex lens (both surfaces convex), (b) plano-convex lens (one surface plane, the other convex), (c) concavo-convex lens (one surface concave, the other convex).

Fig. Three types of convex lenses.

(ii) Concave (diverging) lenses: thin in the middle and thick at the edges. The three types are: (a) double concave lens (both surfaces concave), (b) plano-concave lens (one surface plane, the other concave), (c) convexo-concave lens (one surface convex, the other concave).

Fig. Three types of concave lenses.

SOME ASSOCIATED TERMS

• Centre of curvature (C): the centre of the sphere of which a lens surface forms a part. A lens has two surfaces, so it has two centres of curvature, C₁ and C₂.
• Radius of curvature (R): the radius of the sphere of which a lens surface forms a part (R₁ and R₂).
• Principal axis: the line passing through the two centres of curvature C₁ and C₂ of the lens.
• Optical centre (O): the point on the principal axis of a lens such that a ray of light passing through it emerges parallel to its incident direction, i.e. without deviation. If the radii of curvature of the two surfaces are equal, the optical centre coincides with the geometric centre of the lens.
• First principal focus (F₁) and first focal length: a fixed point on the principal axis such that rays starting from this point (convex lens), or appearing to go towards it (concave lens), after refraction through the lens become parallel to the principal axis.
• Second principal focus (F₂) and second focal length: a fixed point on the principal axis such that light rays incident parallel to the principal axis, after refraction through the lens, either converge to this point (convex lens) or appear to diverge from it (concave lens). If the medium on both sides of the lens is the same, the numerical values of the first and second focal lengths are equal: f₁ = f₂ = f.
• Aperture: the diameter of the circular boundary of the lens.

Figure: Characteristics of a convex lens (centres of curvature, radii of curvature, principal axis, optical centre).

(a)

(b) Figure: Characteristics of convex and concave lenses.

Optical centre: a ray of light through the optical centre O emerges parallel to its incident direction, undeviated.

First principal focus: rays from F1 (convex lens) or rays directed towards F1 (concave lens) become parallel to the principal axis after refraction.

Figure: Ray diagram showing first principal focus.

Second principal focus: rays parallel to the principal axis converge to F2 (convex lens) or appear to diverge from F2 (concave lens).

Figure: Ray diagram showing second principal focus.

RULES FOR IMAGE FORMATION BY THE RAY DIAGRAM METHOD

Three special rays for a convex lens

1. A ray incident parallel to the principal axis passes through the second principal focus F₂ after refraction.
2. A ray passing through the first principal focus F₁ emerges parallel to the principal axis.
3. A ray passing through the optical centre goes straight without any deviation.

Three special rays for a concave lens

1. A ray incident parallel to the principal axis appears to diverge from the second principal focus F₂ after refraction.
2. A ray directed towards the first principal focus F₁ emerges parallel to the principal axis.
3. A ray passing through the optical centre goes straight without any deviation.

Convex lens: three special rays

Concave lens: three special rays

IMAGE FORMATION BY A LENS

From the lens formula, for a lens of fixed focal length, as the object distance u changes, the image distance v also changes. As u decreases or increases, the position, the nature and the size of the image change.

Convex lens in different cases

1. Object at infinity: a point object on the principal axis sends parallel rays, which after refraction actually meet at the second principal focus F₂. The image is formed at F₂; it is real and point-sized. For a large object at infinity, the image is formed at F₂, real, inverted and highly diminished.
2. Object beyond 2f: the image is formed between f and 2f; it is real, inverted and diminished.
3. Object at 2f: the image is formed at 2f on the other side; it is real, inverted and of the same size as the object.
4. Object between f and 2f: the image is formed beyond 2f; it is real, inverted and enlarged.
5. Object at the focus: the image is formed at infinity; it is real, inverted and highly enlarged.
6. Object between the focus and the optical centre: the image is formed on the same side of the lens as the object; it is virtual, erect and enlarged.

Concave lens in different cases

1. Object at infinity: a point object on the principal axis sends parallel rays, which after refraction appear to come from the second principal focus F₂. The image is formed at F₂; it is virtual and point-sized. For a large object at infinity, the image is formed at F₂, virtual, erect and highly diminished.
2. Object at a finite distance: the image is formed between the second principal focus F₂ and the optical centre; it is virtual, erect and diminished.

(The corresponding ray diagrams are not reproduced here. The source called the image for an object at the focus “imaginary”; it is real, inverted and formed at infinity.)

Convex lens, case 1 (object at infinity): point object

Image at F2: real and point sized.

Convex lens, case 1: big object at infinity, image at focus

Image at F2: real, inverted and diminished.

Convex lens, case 2: object beyond 2f, image between f and 2f

Image: real, inverted and diminished.

Convex lens, case 3: object at 2f, image at 2f

Image: real, inverted and of the same size as the object.

Convex lens, case 4: object between f and 2f, image beyond 2f

Image: real, inverted and enlarged.

Convex lens, case 5: object at focus, image at infinity

Image: at infinity, inverted and very large.

Convex lens, case 6: object between focus and optical centre

Image: virtual, erect and enlarged, on the same side as the object.

Concave lens, case 1: point object at infinity

Image at F2: virtual and point sized.

Concave lens, case 1: big object at infinity, image at focus

Image at F2: virtual, erect and diminished.

NUMERICAL METHOD IN LENSES

(A) Lens formula. The equation relating the object distance (u), the image distance (v) and the focal length (f) of a lens:

\[\dfrac1v-\dfrac1u=\dfrac1f\]

Assumptions: the lens is thin; the lens has a small aperture; the object lies close to the principal axis; the incident rays make small angles with the lens surface or the principal axis.

(B) Linear magnification. The ratio of the size of the image formed by refraction through the lens to the size of the object. If I is the size of the image and O is the size of the object:

\[m=\dfrac IO=\dfrac vu\]

(C) Power of a lens. The capacity or ability of a lens to deviate (converge or diverge) the path of rays passing through it. A lens that converges or diverges the rays more has a higher power.

\[P\ (\text{in dioptre})=\dfrac{1}{f\ (\text{in metre})}\]

TOTAL INTERNAL REFLECTION

When light travels from a denser medium to a rarer medium and is incident at an angle greater than the critical angle for that medium, it is completely returned into the denser medium. This complete return of light is called total internal reflection.

Critical angle (C). The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90° is called the critical angle. For a denser medium (water, w) and a rarer medium (air, a): \(\mu_w\sin C=\mu_a\sin90^\circ\), so

\[\sin C=\dfrac{\mu_a}{\mu_w}=\dfrac1{{}_a\mu_w}\]

The greater the value of μ, the smaller the critical angle.

Fig. Total internal reflection.

Conditions for total internal reflection: (i) the light must travel from a denser to a rarer medium; (ii) the angle of incidence must be greater than the critical angle for the denser medium.

Merit. In total internal reflection nearly 100% of the light is reflected, so the images formed are brighter. In ordinary reflection from mirrors, only about 85% of the light is reflected, and the rest is absorbed by the mirror glass or transmitted because of poor polish, so the images are less bright.

REAL AND APPARENT DEPTH AND HEIGHT

(A) Seeing from air into a liquid. An object in a liquid appears raised. If the refractive index of the liquid is μ:

Apparent depth from surface = d(actual)/μ.

\[\mu=\dfrac{\text{actual depth}}{\text{apparent depth}},\qquad \text{apparent depth}=\dfrac{d_{actual}}{\mu}\]

(B) Seeing from a liquid into air. An object in air, seen from inside a liquid, appears higher than it really is: apparent height from the surface = H_actual × μ.

Apparent height from surface = H(actual) × μ.

SOLVED EXAMPLES

Example 1

The speed of light in water is 2.25 × 10⁸ m/s. Calculate the refractive index of water.

Answer: 1.33.

\[\mu=\dfrac{c}{v}=\dfrac{3\times10^8}{2.25\times10^8}=1.33\]

Example 2

The refractive index of diamond is 2.42. Calculate the speed of light in diamond.

Answer: 1.24 × 10⁸ m/s.

\[v=\dfrac{c}{n}=\dfrac{3\times10^8}{2.42}=1.24\times10^8\ \text{m/s}\]

Example 3

A ray of light travelling in air falls on the surface of water. The angle of incidence is 60° with the normal to the surface. The refractive index of water = 4/3. Calculate the angle of refraction.

Answer: About 40.5°.

\(\dfrac{\sin i}{\sin r}=n\): \(\dfrac{\sin60^\circ}{\sin r}=\dfrac43\), so \(\sin r=\dfrac{\sqrt3/2\times3}{4}=0.65\) and \(r=\sin^{-1}0.65\approx40.5^\circ\).

Example 4

If the refractive index of water is 4/3 and that of glass is 3/2, calculate the refractive index of glass with respect to water.

Answer: 9/8 ≈ 1.1.

\[{}_w\mu_g=\dfrac{\mu_g}{\mu_w}=\dfrac{3/2}{4/3}=\dfrac98\approx1.1\]

Example 5

A ray of light is incident on the plane surface of a transparent medium at an angle of 60° with the normal. The angle of refraction is 30°. Calculate the refractive index of the transparent material.

Answer: √3 ≈ 1.73.

\[n=\dfrac{\sin i}{\sin r}=\dfrac{\sin60^\circ}{\sin30^\circ}=\dfrac{\sqrt3/2}{1/2}=\sqrt3\]

Example 6

A ray of light travelling in air falls on the surface of a glass slab at an angle of 45° with the normal. The refractive index of glass is 1.5. Calculate the angle of refraction.

Answer: About 28°.

\(1.5=\dfrac{\sin45^\circ}{\sin r}\), so \(\sin r=\dfrac{0.707}{1.5}=0.47\) and \(r=\sin^{-1}0.47\approx28^\circ\).

Example 7

The refractive index of diamond is 2.42 and that of carbon disulphide is 1.63. Calculate the refractive index of diamond with respect to carbon disulphide.

Answer: 1.48.

\[{}_1n_2=\dfrac{n_2}{n_1}=\dfrac{2.42}{1.63}=1.48\]

Example 8

A coin is placed in a tumbler, and water is then filled in the tumbler to a height of 20 cm. If the refractive index of water is 4/3, calculate the apparent depth of the coin.

Answer: 15 cm.

\(n=\dfrac{\text{real depth}}{\text{apparent depth}}\): \(\dfrac43=\dfrac{20}{\text{apparent depth}}\), so apparent depth = \(\dfrac{20\times3}{4}=15\) cm.

Example 9

There is a black spot on a table. A glass slab of thickness 6 cm is placed on the table over the spot. The refractive index of glass is 3/2. At what depth from the upper surface will the spot appear when viewed from above?

Answer: 4 cm.

\(\dfrac32=\dfrac{6}{\text{apparent depth}}\), so apparent depth = \(\dfrac{6\times2}{3}=4\) cm.

Example 10

The refractive index of diamond is 2.42 and that of glass is 1.5. Calculate the critical angle for the diamond–glass surface.

Answer: About 38°.

\(\sin i_c=\dfrac{n_2}{n_1}=\dfrac{1.5}{2.42}=0.62\), so \(i_c=\sin^{-1}0.62\approx38^\circ\).

Example 11

The refractive index of glass is 3/2. A ray of light travelling in glass is incident on the glass–water surface at an angle of 30° with the normal. Will it be able to come out into the water? Refractive index of water = 4/3.

Answer: Yes, it will be refracted into the water.

\(\sin i_c=\dfrac{n_2}{n_1}=\dfrac{4/3}{3/2}=\dfrac89=0.89\), so \(i_c\approx63^\circ\). Since the angle of incidence (30°) is less than the critical angle, the ray will be refracted into the water.

(The source gave the critical angle as 62°; sin⁻¹ 0.89 is about 63°.)

Example 12

The refractive index of dense flint glass is 1.65 and that of alcohol is 1.36, both with respect to air. What is the refractive index of flint glass with respect to alcohol?

Answer: 1.21.

\[{}_1n_2=\dfrac{n_2}{n_1}=\dfrac{1.65}{1.36}=1.21\]

Example 13

An object is placed 36 cm from a convex lens. A real image is formed 24 cm from the lens. Calculate the focal length of the lens.

Answer: 14.4 cm.

The object is on the left of the lens, so u = −36 cm; the real image is on the other side, so v = +24 cm. \(\dfrac1f=\dfrac1v-\dfrac1u=\dfrac1{24}+\dfrac1{36}=\dfrac5{72}\), so f = 14.4 cm.

Example 14

A 2 cm long pin is placed perpendicular to the principal axis of a lens of focal length 15 cm at a distance of 25 cm from the lens. Find the position of the image and its size.

Answer: The image is formed 37.5 cm from the lens on the other side; it is 3 cm long and inverted.

u = −25 cm, f = +15 cm. \(\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{15}-\dfrac1{25}=\dfrac2{75}\), so v = +37.5 cm. The positive sign shows that the image is formed on the right-hand side of the lens. \(m=\dfrac vu=\dfrac{37.5}{-25}=-1.5\), so \(h'=-1.5\times2=-3\) cm. The image of the pin is 3 cm long; the negative sign shows it is inverted.

Example 15

A point object is placed at a distance of 18 cm from a convex lens on its principal axis. Its image is formed on the other side of the lens at 27 cm. Calculate the focal length of the lens.

Answer: 10.8 cm.

u = −18 cm, v = +27 cm. \(\dfrac1f=\dfrac1{27}+\dfrac1{18}=\dfrac5{54}\), so f = 10.8 cm.

Example 16

A convex lens forms an image of the same size as the object at a distance of 30 cm from the lens. Find the focal length of the lens. Also find the power of the lens. What is the distance of the object from the lens?

Answer: f = 15 cm; power = 6.7 D; object distance = 30 cm.

A convex lens forms an image of the same size as the object only when the object is at 2f from the lens; the image is then also at 2f. So 2f = 30 cm, f = 15 cm = 0.15 m. Power \(P=\dfrac1f=\dfrac1{0.15}=6.7\) D. The object is at 2f = 30 cm from the lens.

(The source gave the power as 6.6 D; 1/0.15 = 6.67 D.)

Example 17

A 3 cm high object is placed at a distance of 80 cm from a concave lens of focal length 20 cm. Find the position and size of the image.

Answer: The image is 16 cm from the lens on the same side as the object; it is 0.6 cm high and erect.

u = −80 cm, f = −20 cm. \(\dfrac1v=\dfrac1f+\dfrac1u=-\dfrac1{20}-\dfrac1{80}=-\dfrac5{80}=-\dfrac1{16}\), so v = −16 cm. \(m=\dfrac vu=\dfrac{-16}{-80}=+\dfrac15\), so \(h'=\dfrac{3}{5}=0.6\) cm. The positive sign shows that the image is erect.

Example 18

An object is placed on the principal axis of a concave lens at a distance of 40 cm from it. If the focal length of the lens is also 40 cm, find the location of the image and the magnification.

Answer: The image is 20 cm from the lens on the same side as the object; m = +1/2.

f = −40 cm, u = −40 cm. \(\dfrac1v=-\dfrac1{40}-\dfrac1{40}=-\dfrac1{20}\), so v = −20 cm. The minus sign shows that the image is on the same side of the lens as the object. \(m=\dfrac vu=\dfrac{-20}{-40}=+\dfrac12\); the positive sign shows the image is erect.

Example 19

A beam of light travelling parallel to the principal axis of a concave lens appears to diverge from a point 25 cm behind the lens after refraction. Calculate the power of the lens.

Answer: −4 D.

A parallel beam, after refraction through a concave lens, appears to diverge from the focus, so the focal length is 25 cm. For a concave lens f is negative: f = −25 cm = −0.25 m. Power \(P=\dfrac1f=\dfrac1{-0.25}=-4\) D.

Example 20

A convex lens of power 5 D is placed at a distance of 30 cm from a screen. At what distance from the lens should the object be placed so that its image is formed on the screen?

Answer: 60 cm from the lens.

P = +5 D, so f = 1/5 = 0.20 m = 20 cm. The screen is 30 cm from the lens, so v = +30 cm. \(\dfrac1u=\dfrac1v-\dfrac1f=\dfrac1{30}-\dfrac1{20}=-\dfrac1{60}\), so u = −60 cm. The object should be placed 60 cm from the lens.

(The source asked “at what distance should the screen be placed”, but the screen position is given as 30 cm and the answer, 60 cm, is the object distance, so the question was corrected.)

Example 21

A pin 3 cm long is placed at a distance of 24 cm from a convex lens of focal length 18 cm. The pin is placed perpendicular to the principal axis. Find the position, size and nature of the image.

Answer: The image is 72 cm from the lens on the other side; it is real, inverted and 9 cm long.

u = −24 cm, f = +18 cm. \(\dfrac1v=\dfrac1{18}-\dfrac1{24}=\dfrac1{72}\), so v = +72 cm; the image is real. \(m=\dfrac vu=\dfrac{72}{-24}=-3\), so \(h'=-3\times3=-9\) cm. The image is 9 cm in size; the negative sign shows it is inverted.

Example 22

A convex lens of focal length 40 cm and a concave lens of focal length 25 cm are placed in contact in such a way that they have a common principal axis. Find the power of the combination.

Answer: −1.5 D.

Convex lens: f₁ = +0.40 m, \(P_1=\dfrac1{0.40}=+2.5\) D. Concave lens: f₂ = −0.25 m, \(P_2=\dfrac1{-0.25}=-4\) D. Power of the combination P = P₁ + P₂ = 2.5 − 4 = −1.5 D.

Example 23

A concave lens has a focal length of 15 cm. At what distance should the object be placed from the lens so that it forms an image 10 cm from the lens? Also find the magnification.

Answer: The object should be 30 cm from the lens; m = +1/3.

A concave lens always forms a virtual, erect image on the same side as the object. v = −10 cm, f = −15 cm. \(\dfrac1u=\dfrac1v-\dfrac1f=-\dfrac1{10}+\dfrac1{15}=\dfrac{-3+2}{30}=-\dfrac1{30}\), so u = −30 cm. \(m=\dfrac vu=\dfrac{-10}{-30}=+\dfrac13=0.33\). The positive sign shows that the image is erect and virtual; its size is one-third of the object.

Example 24

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the position, nature and size of the image. Calculate the magnification of the lens.

Answer: The image is real, inverted, 4 cm high, formed 30 cm from the lens on the right side; m = −2.

u = −15 cm, f = +10 cm, h = +2 cm. \(\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{10}-\dfrac1{15}=\dfrac1{30}\), so v = +30 cm; the image is real and on the right side of the lens. \(m=\dfrac vu=\dfrac{30}{-15}=-2\), so \(h'=-2\times2=-4\) cm. The image is inverted, twice the size of the object.

IMPORTANT POINTS TO REMEMBER

• Refraction: the bending of a ray of light as it passes from one medium to another. (a) A ray travelling from a rarer to a denser medium (say water or glass) bends towards the normal. (b) A ray travelling from a denser to a rarer medium bends away from the normal.
• Laws of refraction: (a) sin i / sin r is constant for a particular pair of media and equals the refractive index of the medium into which the light enters; (b) the incident ray, the refracted ray and the normal all lie in the same plane.
• Lateral displacement: the perpendicular distance between the emergent ray and the original path of the incident ray.
• Lens: a piece of transparent material bounded by two curved surfaces. A lens thicker in the middle and thinner at the edges is a convex (converging) lens; a lens thicker at the edges and thinner in the centre is a concave (diverging) lens.
• Optical centre: the central point of a lens; a ray of light passing through it is not deviated.
• Images formed by lenses: a convex lens forms real and inverted images for all positions of an object at and beyond the focus F. When the object is between F and O, the image is virtual and erect. A concave lens always forms a virtual, erect and diminished image, wherever the object is.
• Lens formula: \(\dfrac1f=\dfrac1v-\dfrac1u\).
• Power of a lens: the reciprocal of its focal length in metres; it is measured in dioptres (D).

Refractive indices of various substances relative to vacuum, for light of wavelength 589 nm

Solids (at 20 °C)

Liquids (at 20 °C)

Gases (at 0 °C, 1 atm)

(The source gave the wavelength as “589 mm”; it is 589 nm.)

EXERCISE 1

The source gave answers only for the numerical problems (Q29–Q42); the answers to the other questions were written for this document.

A. Very Short Answer Type Questions

What is the maximum angle of refraction when a ray of light is refracted from glass into air?

Answer: 90°. When the angle of incidence in glass equals the critical angle, the refracted ray grazes the surface at 90°.

What should be the position of an object relative to a biconvex lens so that this lens behaves like a magnifying glass?

Answer: Between the focus and the optical centre of the lens (within one focal length), so that an erect, enlarged virtual image is formed.

Can the absolute refractive index of a medium be less than unity?

Answer: No. The absolute refractive index = c/v, and the speed of light in a medium is never greater than its speed in vacuum, so the refractive index is always greater than (or equal to) 1.

To a fish under water viewing obliquely a fisherman standing on the bank of a lake, does the man look taller or shorter than what he actually is?

Answer: Taller. Rays from the man go from air (rarer) into water (denser) and bend towards the normal; the fish sees the image higher up, so the man appears taller.

Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?

Answer: Yes, it changes. When viewed obliquely, the apparent depth is less than the apparent depth seen when looking vertically, i.e. the apparent depth decreases and the bottom appears even more raised.

A substance has a critical angle of 45° for yellow light. What is its refractive index?

Answer: √2 ≈ 1.41.

\(\mu=\dfrac1{\sin C}=\dfrac1{\sin45^\circ}=\sqrt2=1.41\).

What is the critical angle for a material of refractive index √2?

Answer: 45°.

\(\sin C=\dfrac1\mu=\dfrac1{\sqrt2}\), so C = 45°.

A ray of light is incident normally on a glass slab. What is the angle of refraction?

Answer: 0°. A ray incident along the normal passes without bending.

What is the power of the combination of a convex lens and a concave lens of the same focal length?

Answer: Zero. The powers are +1/f and −1/f, so their sum is zero.

How is the power of a lens related to its focal length?

Answer: Power is the reciprocal of the focal length (in metres): P = 1/f.

Define the critical angle for total internal reflection.

Answer: The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

B. Short Answer Type Questions

Explain the shining of an air bubble in water.

Answer: Light in water that falls on the surface of the air bubble goes from a denser medium (water) to a rarer medium (air). If the angle of incidence is greater than the critical angle, the light is totally internally reflected from the bubble surface. A large amount of light is reflected towards the eye, so the bubble shines.

For the same angle of incidence, the angles of refraction in media P, Q and R are 35°, 25°, 15° respectively. In which medium will the velocity of light be minimum?

Answer: In medium R.

Since \(\mu=\dfrac{\sin i}{\sin r}\), the smaller the angle of refraction, the larger the refractive index. Medium R (r = 15°) has the largest refractive index. As μ = c/v, the speed of light is smallest in R.

Define focus and principal focus of a lens.

Answer: Principal focus of a convex lens: the point on the principal axis at which rays incident parallel to the principal axis actually meet after refraction (for a concave lens, the point from which they appear to diverge). A lens has two principal foci, F₁ and F₂, one on each side. Focus is the point at which rays of light from a point source meet (or appear to meet) after refraction; for parallel rays it is the principal focus.

A virtual image, we always say, cannot be caught on a screen. Yet when we “see” a virtual image, we are obviously bringing it onto the “screen”, i.e. the retina of our eye. Is there a contradiction?

Answer: No contradiction. A virtual image is a place where the rays only appear to meet; no light actually reaches it, so it cannot be caught on a screen. The eye lens converges the diverging rays that enter the eye and forms a real image on the retina, so what falls on the retina is a real image of the virtual image.

A convex lens is held in water. What would be the change in the focal length?

Answer: The focal length increases. The relative refractive index of the glass with respect to water is smaller than that with respect to air, so the lens bends the rays less and its focal length becomes larger.

Why do goggles (sun glasses) have zero power even though their surfaces are curved?

Answer: Both surfaces are curved by the same amount, so the lens has uniform thickness, like a thin slab with parallel faces. The converging effect of one surface is cancelled by the diverging effect of the other, so the net power is zero.

The lens shown in the figure is made of two different materials. A point object is placed on the principal axis of this lens. How many images will be obtained?

Answer: Two images. The lens is made of two different materials, i.e. two refractive indices, so the two parts have different focal lengths, and each part forms its own image.

Refer to figures (a), (b) and (c). Give the relationship between μ₁ and μ₂ in each case.

Answer: (a) μ₂ > μ₁; (b) μ₁ = μ₂; (c) μ₁ > μ₂.

In (a) the ray bends towards the normal on entering medium 2, so medium 2 is denser (μ₂ > μ₁). In (b) the ray goes straight without bending, so the two media have the same refractive index. In (c) the ray bends away from the normal, so medium 2 is rarer (μ₁ > μ₂). (This answer is read from the bending shown in the figure.)

Images formed by totally reflected light are brighter than the images formed by ordinary reflected light. Why?

Answer: In total internal reflection almost 100% of the light is reflected. In ordinary reflection from a mirror only about 85% of the light is reflected; the rest is absorbed by the glass or transmitted because of the imperfect polish. So images formed by total internal reflection are brighter.

Can light travelling from air to glass suffer total internal reflection? Justify your answer.

Answer: No. Total internal reflection occurs only when light travels from a denser medium to a rarer medium. Air is rarer than glass, so light going from air to glass bends towards the normal and the angle of refraction is always less than the angle of incidence; it can never be 90°.

What are the five general features of the image formed by a plane mirror?

Answer: The image formed by a plane mirror is:

1. virtual,
2. erect,
3. of the same size as the object,
4. formed as far behind the mirror as the object is in front of it,
5. laterally inverted.

C. Long Answer Type Questions

(a) What is total internal reflection? How is the critical angle related to the refractive index? (b) A ray of light, while travelling from a denser to a rarer medium, undergoes total internal reflection. Derive the expression for the critical angle in terms of the speed of light in the respective media.

Answer: (a) When light travels from a denser medium to a rarer medium and is incident at an angle greater than the critical angle, it is completely reflected back into the denser medium. This is called total internal reflection. The critical angle C and the refractive index μ of the denser medium (with respect to the rarer medium) are related by

\[\sin C=\dfrac1\mu\]

The greater the refractive index, the smaller the critical angle.

(b) Let the speed of light be v_d in the denser medium and v_r in the rarer medium. By Snell’s law, \(\dfrac{\sin i}{\sin r}=\dfrac{v_d}{v_r}\). At the critical angle, i = C and r = 90°, so \(\dfrac{\sin C}{\sin90^\circ}=\dfrac{v_d}{v_r}\). Hence

\[\sin C=\dfrac{v_d}{v_r}\qquad\Rightarrow\qquad C=\sin^{-1}\!\left(\dfrac{v_d}{v_r}\right)\]

Discuss in detail refraction at (i) a convex surface, (ii) a concave surface.

Answer: Refraction at a spherical surface that separates a medium of refractive index μ₁ from a medium of refractive index μ₂ (radius of curvature R, pole P, centre of curvature C) obeys the relation \(\dfrac{\mu_2}{v}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R}\).

(i) Convex surface. Let a point object O be on the principal axis in medium μ₁, with the light going into the denser medium μ₂ through a convex surface. A ray OM, where M is a point at a small height h above the axis, meets the surface at M, is refracted towards the normal CM and meets the axis at I. Let ∠MOP = α, ∠MCP = γ and ∠MIP = β. For small angles, α ≈ h/OP, γ ≈ h/PC and β ≈ h/PI. The angle of incidence is i = α + γ (an exterior angle of triangle OMC), and the angle of refraction is r = γ − β (exterior angle of triangle MCI). By Snell’s law for small angles, μ₁ i = μ₂ r, so μ₁(α + γ) = μ₂(γ − β). Substituting the values: \(\dfrac{\mu_1}{OP}+\dfrac{\mu_2}{PI}=\dfrac{\mu_2-\mu_1}{PC}\). Applying the New Cartesian sign convention (OP = −u, PI = +v, PC = +R) gives \(\dfrac{\mu_2}{v}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R}\). A real image is formed on the far side of the surface.

(ii) Concave surface. The same construction applies, but the centre of curvature C is now on the side from which the light comes, so R is negative. A ray from O is refracted towards the normal but diverges (when μ₂ > μ₁); the refracted rays do not meet, and they appear to come from a point I on the same side as the object, so a virtual image is formed. The same formula \(\dfrac{\mu_2}{v}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R}\) holds with R negative, and v comes out negative, showing a virtual image on the object’s side.

Derive the lens formula for a thin lens.

Answer: Consider a thin lens of refractive index μ₂ in a medium of refractive index μ₁ (for air, μ₁ = 1), with surfaces of radii of curvature R₁ and R₂. A point object O is at a distance u from the lens.

The first surface forms an image I₁ at distance v₁ (the image the lens would form if the second surface were absent): \(\dfrac{\mu_2}{v_1}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R_1}\) …(1)

For the second surface, I₁ acts as the object, and the final image I is formed at v (for a thin lens all distances are measured from the optical centre): \(\dfrac{\mu_1}{v}-\dfrac{\mu_2}{v_1}=\dfrac{\mu_1-\mu_2}{R_2}\) …(2)

Adding (1) and (2): \(\mu_1\left(\dfrac1v-\dfrac1u\right)=(\mu_2-\mu_1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\), i.e.

\[\dfrac1v-\dfrac1u=\left(\dfrac{\mu_2}{\mu_1}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\]

When the object is at infinity, u = ∞ and v = f, so the right side equals 1/f (the lens maker’s formula). Therefore

\[\dfrac1v-\dfrac1u=\dfrac1f\]

What is the critical angle? Give one application of total internal reflection.

Answer: The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

One application: optical fibres, which carry light signals by repeated total internal reflection inside a glass fibre (also used in endoscopes). Other applications: the sparkling of a diamond, totally reflecting prisms in binoculars and periscopes, and the formation of a mirage.

State Snell’s law of refraction.

Answer: For light passing from one medium into another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media (and for light of a given colour): \(\dfrac{\sin i}{\sin r}=\text{constant}=\mu\), where μ is the refractive index of the second medium with respect to the first. (The incident ray, the refracted ray and the normal also lie in the same plane.)

What is meant by the power of a lens? What is one dioptre?

Answer: The power of a lens is its ability to converge or diverge the rays of light passing through it; it is the reciprocal of its focal length in metres, P = 1/f. A lens that bends the rays more has a greater power.

One dioptre is the power of a lens whose focal length is 1 metre (1 D = 1 m⁻¹).

D. Numerical Problems

The speed of light in air is 3 × 10⁸ m/s. Calculate the speed of light in glass, given that the refractive index of glass is 1.5.

Answer: 2 × 10⁸ m/s.

\(v=\dfrac c\mu=\dfrac{3\times10^8}{1.5}=2\times10^8\) m/s.

The refractive index of water with respect to air is 4/3. Calculate the refractive index of air with respect to water.

Answer: 3/4.

\({}_w\mu_a=\dfrac{1}{{}_a\mu_w}=\dfrac34=0.75\).

The refractive index of glass is 3/2. What is the critical angle for the glass–water surface?

Answer: About 63° (take the refractive index of water as 4/3).

\(\sin C=\dfrac{\mu_w}{\mu_g}=\dfrac{4/3}{3/2}=\dfrac89=0.889\), so C ≈ 62.7° ≈ 63°.

(The source key printed 42°, which is the critical angle for glass to air, \(\sin^{-1}\tfrac23\). For the glass–water surface the answer is about 63°.)

A ray of light travelling in air falls on the surface of a glass slab at an angle of incidence 45°. Find the angle made by the refracted ray with the normal within the slab, where the refractive index of glass is 3/2.

Answer: About 28°.

\(\sin r=\dfrac{\sin45^\circ}{1.5}=\dfrac{0.707}{1.5}=0.471\), so r ≈ 28.1°.

A ray of light travelling in air is incident on the surface of a transparent material of refractive index √3. If the angle of refraction is 30°, calculate the angle of incidence.

Answer: 60°.

\(\sin i=\mu\sin r=\sqrt3\times\dfrac12=\dfrac{\sqrt3}2\), so i = 60°.

The focal length of a convex lens is 50 cm. Calculate its power.

Answer: +2 D.

\(P=\dfrac1f=\dfrac1{0.5}=+2\) D.

A point object is placed at a distance of 12 cm from a convex lens on its principal axis. Its image is formed 18 cm from the lens on the other side. Calculate the focal length of the lens.

Answer: 7.2 cm.

u = −12 cm, v = +18 cm. \(\dfrac1f=\dfrac1v-\dfrac1u=\dfrac1{18}+\dfrac1{12}=\dfrac5{36}\), so f = 7.2 cm.

An object is placed at a distance of 20 cm from a concave lens on its principal axis. If the focal length of the lens is 20 cm, find the position of the image.

Answer: 10 cm from the lens, on the same side as the object.

u = −20 cm, f = −20 cm. \(\dfrac1v=\dfrac1f+\dfrac1u=-\dfrac1{20}-\dfrac1{20}=-\dfrac1{10}\), so v = −10 cm.

A beam of light incident parallel to the principal axis of a concave lens appears to diverge from a point 20 cm behind the lens after refraction. Calculate the power of the lens.

Answer: −5 D.

f = −20 cm = −0.2 m, so \(P=\dfrac1f=-5\) D.

A pin 2 cm long is placed at a distance of 16 cm from a convex lens of focal length 12 cm, perpendicular to the principal axis. Find the position, nature and size of the image.

Answer: The image is formed 48 cm from the lens on the other side; it is real, inverted and 6 cm long.

u = −16 cm, f = +12 cm. \(\dfrac1v=\dfrac1{12}-\dfrac1{16}=\dfrac1{48}\), so v = +48 cm. \(m=\dfrac vu=\dfrac{48}{-16}=-3\), so \(h'=-3\times2=-6\) cm.

A convex lens of focal length 20 cm and a concave lens of focal length 12.5 cm are placed in contact having the same principal axis. Calculate the power of the combined lens.

Answer: −3 D.

\(P_1=\dfrac1{0.20}=+5\) D; \(P_2=\dfrac1{-0.125}=-8\) D. \(P=P_1+P_2=5-8=-3\) D.

(The source key printed −8 D, which is only the power of the concave lens; the combination has a power of −3 D.)

Two thin lenses of power +3.5 D and −2.5 D are placed in contact. Find the power and focal length of the lens combination.

Answer: Power = +1 D; focal length = 100 cm.

P = 3.5 + (−2.5) = +1 D; \(f=\dfrac1P=1\) m = 100 cm.

An illuminated slit is kept at a distance of 40 cm in front of a convex lens of focal length 15 cm. Find the position of the screen to obtain the image.

Answer: 24 cm from the lens on the other side.

u = −40 cm, f = +15 cm. \(\dfrac1v=\dfrac1{15}-\dfrac1{40}=\dfrac{8-3}{120}=\dfrac1{24}\), so v = +24 cm. The screen must be placed 24 cm from the lens, on the side opposite to the slit.

A ray incident on a slab at an angle of 10° is shown in the figure. Find the angle of emergence θ.

Answer: θ = 10°.

The figure shows a slab with layers of different refractive indices, with the same liquid (μ = 4.1 in the figure) on both sides of a slab made of four layers with μ = 1.2, 2, 2√3 and 4. When light passes through a parallel-sided combination of layers and emerges into the same medium it started from, the emergent ray is parallel to the incident ray (by Snell’s law, μ sin θ is the same in every layer). So the emergent angle equals the incident angle, θ = 10°.

EXERCISE 2: SINGLE CORRECT ANSWER TYPE QUESTIONS

How will the image formed by a convex lens be affected if the central portion of the lens is wrapped in black paper?

1. No image will be formed
2. A full image will be formed, but it is less bright
3. A full image will be formed, but without the central portion
4. Two images will be formed, one due to each exposed half

Answer: (B) A full image will be formed, but it is less bright.

Every part of a lens forms a complete image; covering part of the lens only reduces the amount of light, so the image is dimmer.

The critical angle for light going from medium X into medium Y is θ. The speed of light in medium X is v. The speed of light in medium Y is:

1. v(1 − cos θ)
2. v/cos θ
3. v cos θ
4. v/sin θ

Answer: (D) v/sin θ.

\(\sin\theta=\dfrac{v_X}{v_Y}\), so \(v_Y=\dfrac{v}{\sin\theta}\).

One surface of a lens is convex and the other is concave. If the radii of curvature are r₁ and r₂ respectively, the lens will be convex if:

1. r₁ > r₂
2. r₁ = r₂
3. r₁ < r₂
4. r₁ = 1/r₂

Answer: (C) r₁ < r₂.

The lens is thicker in the middle (convex) when the convex surface is more strongly curved, i.e. has the smaller radius of curvature.

An object is immersed in a fluid. In order that the object becomes invisible, it should:

1. behave as a perfect reflector
2. absorb all light falling on it
3. have refractive index one
4. have a refractive index exactly matching that of the surrounding fluid

Answer: (D) have a refractive index exactly matching that of the surrounding fluid.

Then there is no bending or reflection at its surface, so it cannot be seen.

The refractive index of glass with respect to air is 3/2. Then the refractive index of air with respect to glass is:

1. 3/4
2. 2/3
3. 1/3
4. 3

Answer: (B) 2/3.

The refractive index of glass with respect to air is 1.5 and the refractive index of water with respect to air is 4/3. What will be the refractive index of glass with respect to water?

1. 1
2. 1.5
3. 1.125
4. −10

Answer: (C) 1.125.

\({}_w\mu_g=\dfrac{1.5}{4/3}=1.125\).

The refractive index of a medium depends upon:

1. the nature of the material of the medium
2. the optical density of the medium
3. the wavelength of light
4. all of these

Answer: (D) all of these.

If the refractive index of water with respect to air is 4/3, then the refractive index of air with respect to water will be:

1. 4 × 3
2. 3/4
3. √(4/3)
4. √(3/4)

Answer: (B) 3/4.

A ray of light is incident normally on a rectangular piece of glass. The value of the angle of refraction will be:

1. 180°
2. 90°
3. 45°
4. 0°

Answer: (D) 0°.

What is the angle of deviation?

1. The angle between the reflected ray and the incident ray
2. The angle between the reflected ray and the refracted ray
3. The angle between the incident ray and the refracted ray
4. The angle between the incident ray and the emergent ray

Answer: (D) The angle between the incident ray and the emergent ray.

The speed of light in vacuum is 3.0 × 10⁸ m/s. If the refractive index of a transparent liquid is 4/3, the speed of light in the liquid is:

1. 2.25 × 10⁸ m/s
2. 3 × 10⁸ m/s
3. 4 × 10⁸ m/s
4. 4.33 × 10⁸ m/s

Answer: (A) 2.25 × 10⁸ m/s.

\(v=\dfrac c\mu=\dfrac{3\times10^8}{4/3}=2.25\times10^8\) m/s.

A swimming pool appears to be 2 m deep. Its actual depth is (μ for water = 1.33):

1. 2.66 m
2. 2 m
3. 2.34 m
4. 2.54 m

Answer: (A) 2.66 m.

Actual depth = μ × apparent depth = 1.33 × 2 = 2.66 m.

To get a real and inverted image of the same size as that of the object, the object should be placed in front of the convex lens at:

1. F
2. 2F
3. between F and 2F
4. away from 2F, where F is the focus

Answer: (B) 2F.

A spherical mirror and a spherical lens each have a focal length of −10 cm. The mirror and the lens are:

1. both convex
2. both concave
3. mirror is convex and lens is concave
4. mirror is concave and lens is convex

Answer: (B) both concave.

A negative focal length means a concave mirror and a concave lens.

The power of a lens having a focal length of 50 cm is:

1. 1/2 D
2. 2 D
3. 3 D
4. 0.2 D

Answer: (B) 2 D.

\(P=\dfrac1{0.5}=2\) D.

The focal length of a lens of power −2.0 D is:

1. −2.0 m
2. 0.2 m
3. −0.5 m
4. 0.5 m

Answer: (C) −0.5 m.

\(f=\dfrac1P=\dfrac1{-2}=-0.5\) m.

Two lenses of power +5 D and −5 D are placed in close contact. The focal length of the combination is:

1. zero
2. ∞
3. zero or ∞
4. none of these

Answer: (B) ∞.

The net power is zero, so the focal length is infinite.

A student needs a lens of power −2.0 dioptre to correct his distant vision. The focal length of the lens is:

1. +50 cm
2. −50 cm
3. 100 cm
4. −100 cm

Answer: (B) −50 cm.

The focal length of coloured goggles (without number) is:

1. zero
2. infinity
3. between zero and infinity
4. none of these

Answer: (B) infinity.

They have zero power.

Where should an object be placed so that a real and inverted image of very large size is obtained using a convex lens?

1. At the focus
2. At 2F
3. Between F and 2F
4. Beyond 2F

Answer: (A) At the focus.

(Strictly, the image of very large size is obtained for an object just beyond the focus; at the focus it is at infinity. The key answer is (A).)

A convex lens is:

1. thicker at the middle, thinner at the edges
2. diverging
3. thicker at the edges, thinner in the middle
4. of uniform thickness everywhere

Answer: (A) thicker at the middle, thinner at the edges.

A glass rod of refractive index 1.42 is immersed in kerosene. The refractive index of kerosene is 1.42. Then the rod will:

1. appear bent
2. appear raised above the liquid
3. become invisible
4. none of the above

Answer: (C) become invisible.

The refractive indices are equal, so there is no bending or reflection at the rod’s surface.

The power of a lens whose focal length is 25 cm is:

1. 4 dioptre
2. 25 dioptre
3. 0.04 dioptre
4. 2.5 dioptre

Answer: (A) 4 dioptre.

\(P=\dfrac1{0.25}=4\) D.

A thin lens is made with a material having refractive index μ = 1.5. Both sides are convex. It is dipped in water (μ = 1.33). It will behave like:

1. a convergent lens
2. a divergent lens
3. a rectangular slab
4. a prism

Answer: (A) a convergent lens.

The glass is still optically denser than water, so the biconvex lens still converges the light (with a longer focal length).

Choose the correct option:

1. If the final rays are converging, we have a real image
2. If the incident rays are converging, we have a real image
3. If the image is virtual, the corresponding object is called a virtual object
4. The image of a virtual object is called a virtual image

Answer: (A) If the final rays are converging, we have a real image.

A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black:

1. the image will be shifted backward
2. the image will not be shifted
3. the intensity of the image will decrease
4. both (B) and (C)

Answer: (D) both (B) and (C).

The lower half still forms the complete image at the same place, but with half the light, so the intensity decreases.

The minimum distance between an object and its real image formed by a convex lens of focal length f is:

1. f
2. 2f
3. 3f
4. 4f

Answer: (D) 4f.

The distance is least when the object is at 2f and the image is at 2f on the other side: 2f + 2f = 4f.

Proof for i = e

(a) Double Convex Lens : It has both the surfaces convex.

(a) Double Concave Lens : It has both the surfaces concave. (Fig.)