Class 12 Physics Chapter 9 Important Questions (2026-27)

Class 12 Physics Chapter 9 Ray Optics and Optical Instruments: important questions with answers on lenses, mirrors and prisms for CBSE 2026-27.

Important Questions for Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments

Looking for Class 12 Physics Chapter 9 important questions? This page covers Ray Optics and Optical Instruments with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on mirrors, lenses, the prism, the microscope and the telescope. Attempt them after finishing the NCERT chapter.

Very Short Answer Questions (1 Mark)

Refraction and Refractive Index

Question 1

A glass lens of refractive index 1.5 is placed in a trough of liquid. What must be the refractive index of the liquid in order to make the lens disappear?

Answer: The refractive index of the liquid must be 1.5 (equal to that of the glass lens).

Question 2

A converging lens of refractive index 1.5 is kept in a liquid medium having the same refractive index. What would be the focal length of the lens in this medium?

Answer: The lens will act like a plane sheet of glass, so its focal length will be infinite (∞).

Question 3

A glass lens of refractive index 1.45 disappears when immersed in a liquid. What is the value of the refractive index of the liquid?

Answer: The refractive index of the liquid must be 1.45.

Question 4

When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer.

Answer: No, the energy carried by the light wave remains the same. Energy \(E = h\nu\), and frequency \(\nu\) does not change when light moves between media.

Question 5

When monochromatic light travels from one medium to another, its wavelength changes but frequency remains the same. Explain.

Answer: If \(v_1, v_2\) are velocities and \(\lambda_1, \lambda_2\) are wavelengths in the two media, then \(v_1/\lambda_1 = v_2/\lambda_2 = \nu\). When light enters a denser medium (\(v_1 > v_2\)), the wavelength and speed decrease, but the frequency remains unchanged.

Question 6

For the same value of angle of incidence, the angles of refraction in three media A, B, C are 15°, 25°, 35° respectively. In which medium would the velocity of light be minimum?

Answer: In medium A — smallest angle of refraction means largest refractive index, hence smallest velocity.

Question 7

State the conditions for the phenomenon of total internal reflection to occur.

Answer: (i) Light must travel from an optically denser to an optically rarer medium. (ii) The angle of incidence in the denser medium must be greater than the critical angle for the two media.

Question 8

Calculate the speed of light in a medium whose critical angle is 30°.

Answer: \(\sin i_c = \dfrac{1}{\mu} \Rightarrow \mu = \dfrac{1}{\sin 30^\circ} = 2\).

Speed \(v = \dfrac{c}{\mu} = \dfrac{3 \times 10^8}{2} =\) \(1.5 \times 10^8\ \text{m/s}\)

Question 9

When light travels from a rarer medium to a denser medium, the speed of light decreases. Does the reduction in speed imply a reduction in the energy?

Answer: No. Energy depends on frequency (\(E = h\nu\)), which does not change.

Lenses and Their Combinations

Question 10

How does the power of a convex lens vary if the incident red light is replaced by violet light?

Answer: Power increases — from the Lens Maker's formula, \(\mu_{\text{violet}} > \mu_{\text{red}}\), so \(P_{\text{violet}} > P_{\text{red}}\).

Question 11

When red light passing through a convex lens is replaced by blue light, how will the focal length change?

Answer: Focal length will decrease (since \(\mu_{\text{blue}} > \mu_{\text{red}}\)).

Question 12

If the wavelength of light incident on a convex lens is increased, how will its focal length change?

Answer: Focal length will increase (as refractive index decreases with increasing wavelength).

Question 13

Two thin lenses of power +4 D and −2 D are in contact. What is the focal length of the combination?

Answer: \(P_{\text{net}} = 4 + (-2) = 2\ \text{D}\). \(f = \dfrac{1}{P} =\) 0.5 m = 50 cm

Question 14

Two thin lenses of power +6 D and −2 D are in contact. What is the focal length of the combination?

Answer: \(P_{\text{net}} = 6 - 2 = 4\ \text{D}\). \(f =\) 0.25 m = 25 cm

Question 15

A converging lens is kept coaxially in contact with a diverging lens — both being of equal focal lengths. What is the focal length of the combination?

Answer: Infinity — powers cancel.

Question 16

Two thin lenses of power −4 D and +2 D are placed in contact coaxially. Find the focal length of the combination.

Answer: \(P_{\text{net}} = -4 + 2 = -2\ \text{D}\). \(f = -0.5\ \text{m} =\) −50 cm

Question 17

Two thin lenses of power −2 D and +2 D are placed in contact coaxially. What is the focal length of the combination?

Answer: \(P_{\text{net}} = 0\) → focal length \(f =\) infinity.

Question 18

Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?

Answer: When \(\mu_{\text{lens}} = \mu_{\text{liquid}}\).

Question 19

How would a biconvex lens appear when placed in a trough of liquid having the same refractive index as that of the lens?

Answer: It appears as a plane sheet of glass (invisible).

Question 20

A biconvex lens made of transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.

Answer: It will behave as a diverging lens, because \(\mu_{\text{lens}} < \mu_{\text{water}}\), so the factor \((\mu_r - 1) < 0\), making \(f\) negative.

Question 21

A biconvex lens made of transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.

Answer: It will behave as a converging lens, since \(\mu_{\text{lens}} > \mu_{\text{water}}\), so \((\mu_r - 1) > 0\).

Question 22

A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens?

Answer: The lens will behave as a converging lens (since \(\mu_{\text{lens}} < \mu_{\text{medium}}\)).

Question 23

A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens?

Answer: Object must be at the focus of the convex lens (rays emerge parallel and reflect back along same path): \(f =\) 20 cm.

Question 24

An object is kept in front of a concave lens. What is the nature of the image formed?

Answer: The image is virtual, erect and diminished.

Prisms and Dispersion

Question 25

How does the angle of minimum deviation of a glass prism vary if incident violet light is replaced with red light?

Answer: It decreases — since \(\mu_{\text{red}} < \mu_{\text{violet}}\), \(\delta_{\text{red}} < \delta_{\text{violet}}\).

Question 26

How does the angle of minimum deviation of a glass prism of refractive index 1.5 change if it is immersed in a liquid of refractive index 1.3?

Answer: The angle of deviation decreases (relative refractive index decreases).

Question 27

Write the relationship between angle of incidence i, angle of prism A, and angle of minimum deviation for a triangular prism.

Answer: \(A + \delta_m = 2i\), so \(i = \dfrac{A + \delta_m}{2}\).

Scattering of Light

Question 28

Why does the bluish colour predominate in a clear sky?

Answer: According to Rayleigh's law, intensity of scattered light \(I \propto \dfrac{1}{\lambda^4}\). Blue light (shorter wavelength) is scattered much more than red, giving the sky its blue colour.

Question 29

Why does bluish colour predominate in a clear sky?

Answer: Same as Question 28 — Rayleigh scattering.

Mirrors and Image Formation

Question 30

When an object is placed between f and 2f of a concave mirror, would the image formed be (i) real or virtual and (ii) diminished or magnified?

Answer: (i) Real, (ii) Magnified.

Optical Instruments

Question 31

You are given three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope?

(Table: L₁: P = 3 D, aperture 8 cm; L₂: P = 6 D, aperture 1 cm; L₃: P = 10 D, aperture 1 cm)

Answer: Objective = L₁ (less power, more aperture); Eyepiece = L₃ (more power, less aperture).

Question 32

The objective lenses of two telescopes have the same apertures but their focal lengths are in the ratio 1 : 2. Compare the resolving powers of the two telescopes.

Answer: Ratio of resolving power = 1 : 1. Resolving power is independent of the focal length of the objective.

Question 33

Why must both the objective and the eyepiece of a compound microscope have short focal lengths?

Answer: For high magnification, since \(m \propto \dfrac{L}{f_o} \cdot \dfrac{D}{f_e}\), smaller focal lengths give larger magnification.

Short Answer Questions (2–3 Marks)

Prism, Refraction and Refractive Index

Question 34

A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when the angle of incidence is 3/4 of the angle of prism. Calculate the speed of light in the prism.

Answer: A = 60°, i = 3/4 × 60 = 45°.

\(\delta_m = 2i - A = 90^\circ - 60^\circ = 30^\circ\).

\(\mu = \dfrac{\sin[(A+\delta_m)/2]}{\sin(A/2)} = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx 1.414\)

\(v = \dfrac{c}{\mu} = \dfrac{3 \times 10^8}{\sqrt{2}} \approx\) \(2.12 \times 10^8\ \text{m/s}\)

Question 35

Define refractive index of a transparent medium. A ray of light passes through a triangular prism. Plot a graph showing the variation of angle of deviation with angle of incidence.

Answer: Refractive index is the ratio of the speed of light in vacuum to its speed in the medium: \(\mu = \dfrac{c}{v}\).

Graph: \(\delta\) vs i — decreases to a minimum (at i = e) and then increases. The minimum value is \(\delta_m\).

Question 36

(i) What is the relation between critical angle and refractive index of a material? (ii) Does critical angle depend on the colour of light? Explain.

Answer: (i) \(\sin i_c = \dfrac{1}{\mu}\).

(ii) Yes. Critical angle depends on the refractive index, which varies with wavelength — so different colours have different critical angles.

Question 37

Red light is deviated less than blue light by a prism. Give reason. (ii) Give the formula to determine the refractive index of a prism in minimum deviation condition.

Answer: (i) Since \(\lambda_{\text{red}} > \lambda_{\text{blue}}\), \(\mu_{\text{red}} < \mu_{\text{blue}}\), hence red light deviates less.

(ii) \(\mu = \dfrac{\sin[(A+\delta_m)/2]}{\sin(A/2)}\)

Question 38

A ray of light incident on an equilateral glass prism (\(\mu_g = \sqrt{3}\)) moves parallel to the base of the prism inside it. Find the angle of incidence for this ray.

Answer: Ray parallel to base → minimum deviation → i = e. Inside, angle of refraction \(r = A/2 = 30^\circ\).

\(\mu = \dfrac{\sin i}{\sin r} \Rightarrow \sqrt{3} = \dfrac{\sin i}{\sin 30^\circ} \Rightarrow \sin i = \dfrac{\sqrt{3}}{2} \Rightarrow i =\) 60°

Question 39

A ray of light passes through an equilateral prism in such a way that the angle of incidence equals the angle of emergence, each being 3/4 times the angle of prism. Determine (i) the angle of deviation and (ii) the refractive index of the prism.

Answer: A = 60°, i = e = 45°.

(i) \(\delta = i + e - A = 45 + 45 - 60 =\) 30°

(ii) \(\mu = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx\) 1.414

Question 40

Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism. Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation.

Answer: Standard derivation. Using geometry of prism:

\(A = r_1 + r_2\), \(\delta = (i_1 - r_1) + (i_2 - r_2) = i_1 + i_2 - A\).

At minimum deviation: \(i_1 = i_2 = i\), \(r_1 = r_2 = r = A/2\), \(\delta = 2i - A\).

\(\mu = \dfrac{\sin i}{\sin r} =\) \(\dfrac{\sin[(A+\delta_m)/2]}{\sin(A/2)}\)

Question 41

Draw a ray diagram showing the path of a ray of light entering through a triangular glass prism. Deduce the expression for the refractive index of glass prism in terms of the angle of minimum deviation and angle of the prism.

Answer: Same derivation as Question 40: \(\mu = \dfrac{\sin[(A+\delta_m)/2]}{\sin(A/2)}\).

Question 42

A ray PQ is incident normally on the face AB of a triangular prism of refracting angle 60°, made of material of refractive index \(2/\sqrt{3}\). Trace the path of the ray as it passes through the prism. Also calculate the angle of emergence and angle of deviation.

Answer: Ray enters face AB normally → goes straight → hits face AC at angle 60°.

Critical angle: \(\sin i_c = \dfrac{1}{\mu} = \dfrac{\sqrt{3}}{2} \Rightarrow i_c = 60^\circ\).

Since angle of incidence on AC equals critical angle, the ray grazes along AC. Angle of emergence = 90°, angle of deviation = 30°.

Total Internal Reflection

Question 43

Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices for the two rays '1' and '2' are 1.35 and 1.45. Trace the path of these rays after entering through the prism.

Answer: Angle of incidence on second face = 45°.

Critical angle for ray 1: \(\sin i_c = \dfrac{1}{1.35} \Rightarrow i_c \approx 47.8^\circ\). Since 45° < 47.8°, ray 1 refracts out.

Critical angle for ray 2: \(\sin i_c = \dfrac{1}{1.45} \Rightarrow i_c \approx 43.6^\circ\). Since 45° > 43.6°, ray 2 undergoes total internal reflection.

Question 44

Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices for the two rays '1' and '2' are 1.3 and 1.5. Trace the path of these rays after entering through the prism.

Answer: Angle on second face = 45°.

Ray 1: critical angle \(\sin i_c = 1/1.3 \Rightarrow i_c \approx 50.3^\circ\). 45° < 50.3° → ray refracts out.

Ray 2: critical angle \(\sin i_c = 1/1.5 \Rightarrow i_c \approx 41.8^\circ\). 45° > 41.8° → ray undergoes total internal reflection.

Question 45

Draw a ray diagram to show how a right-angled isosceles prism may be used to bend the path of light rays by 90°. Write the necessary condition in terms of the refractive index of the material of this prism.

Answer: Ray enters normally on one face, strikes the hypotenuse at 45°. For total internal reflection to occur, critical angle < 45°, so \(\sin 45^\circ > \dfrac{1}{\mu} \Rightarrow \mu > \sqrt{2} \approx 1.414\).

Question 46

Three light rays red (R), green (G) and blue (B) are incident on a right angled prism 'abc' at face 'ab'. Refractive indices for red, green and blue are 1.39, 1.44, 1.47 respectively. Out of the three, which colour ray will emerge out of face 'ac'? Justify. Trace the path of these rays after passing through face 'ab'.

Answer: Critical angles:

Red: \(i_c = \sin^{-1}(1/1.39) \approx 46^\circ\)
Green: \(i_c \approx 44^\circ\)
Blue: \(i_c \approx 42.8^\circ\)

Incident angle on ac is 45° for all three. Since 45° > 46° is FALSE for red, red refracts out. Green and blue have critical angle < 45°, so they undergo total internal reflection. Only red emerges out.

Question 47

A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of the ray through the prism. From which face will the ray emerge?

Answer: The critical angle for \(\mu = 1.5\) is \(i_c \approx 41.8^\circ\).

On face AC, angle of incidence = 30° (for typical 60° prism geometry). Since 30° < 41.8°, the ray will emerge from face AC.

Question 48

An illuminated bulb is at the bottom of a tank containing a liquid of refractive index \(\mu\) up to height H. Find the expression for the diameter of an opaque disc, floating symmetrically on the liquid surface in order to cut off the light from the bulb.

Answer: Light escapes within a cone of half-angle equal to critical angle \(i_c\).

\(\tan i_c = \dfrac{r}{H} \Rightarrow r = H\tan i_c = \dfrac{H}{\sqrt{\mu^2 - 1}}\).

Diameter of disc = \(\dfrac{2H}{\sqrt{\mu^2 - 1}}\)

Question 49

A ray of light is incident on a glass prism of refractive index \(\mu\) and refracting angle A. If it just suffers total internal reflection at the other face, obtain an expression relating the angle of incidence, angle of prism and critical angle.

Answer: At second face, angle of refraction = critical angle \(i_c\), so \(r_2 = i_c\).

At first face: \(r_1 = A - i_c\) (from \(r_1 + r_2 = A\)).

By Snell's law: \(\sin i = \mu\sin r_1 = \mu\sin(A - i_c)\).

So \(\sin i = \mu\sin(A - i_c)\).

Question 50

Define refractive index of a medium. In the given ray diagram, calculate the speed of light in the liquid of unknown refractive index. (Width of liquid surface = 30 cm, depth = 40 cm; light just undergoes total internal reflection)

Answer: \(\mu = \dfrac{c}{v}\).

Critical angle: \(\sin i_c = \dfrac{30}{\sqrt{30^2+40^2}} = \dfrac{30}{50} = 0.6\).

\(\mu = \dfrac{1}{\sin i_c} = \dfrac{1}{0.6} = \dfrac{5}{3}\).

\(v = \dfrac{3 \times 10^8}{5/3} =\) \(1.8 \times 10^8\ \text{m/s}\)

Question 51

A small bulb (assumed point source) is placed at the bottom of a tank containing water to a depth of 80 cm. Find the area of the surface of water through which light from the bulb can emerge. (Refractive index of water = 4/3)

Answer: Critical angle: \(\sin i_c = \dfrac{3}{4}\), so \(\tan i_c = \dfrac{3}{\sqrt{7}}\).

Radius of illuminated circle: \(r = h\tan i_c = 0.8 \times \dfrac{3}{\sqrt{7}} \approx 0.907\ \text{m}\).

Area = \(\pi r^2 \approx \pi \times 0.823 \approx\) \(2.59\ \text{m}^2\)

Question 52

Define the term 'critical angle' for a pair of media. A point source of monochromatic light S is kept at the centre of the bottom of a cylinder of radius 15 cm. The cylinder contains water (refractive index 4/3) to a height of 7 cm. Draw the ray diagram and calculate the area of the water surface through which the light emerges in air.

Answer: Critical angle: \(\sin i_c = \dfrac{3}{4}\), \(\tan i_c = \dfrac{3}{\sqrt{7}}\).

Radius of illuminated region: \(r = h\tan i_c = 7 \times \dfrac{3}{\sqrt{7}} \approx 7.94\ \text{cm}\).

But cylinder has radius 15 cm, so all of the water surface (both luminous and outside) exists. Total area of water surface receiving light = \(\pi \times 15^2 = 225\pi \approx 707\ \text{cm}^2\)

Question 53

A fish in a water tank sees the outside world as if it (the fish) is at the vertex of a cone whose circular base coincides with the water surface. Given the depth of water 'h' and critical angle 'ic', find by drawing a suitable ray diagram the relationship between the radius of the cone and 'h'.

Answer: \(\tan i_c = \dfrac{r}{h}\), so radius \(r =\) \(h\tan i_c\). Diameter of the base = \(2h\tan i_c\).

Lens Formula and Lens Maker's Formula

Question 54

The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of the lens is 12 cm, find the refractive index of the material of the lens.

Answer: R₁ = 10 cm, R₂ = −15 cm, f = 12 cm.

\(\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) \Rightarrow \dfrac{1}{12} = (\mu - 1)\left(\dfrac{1}{10} + \dfrac{1}{15}\right) = (\mu - 1)\cdot\dfrac{1}{6}\)

\(\mu - 1 = \dfrac{6}{12} = 0.5 \Rightarrow \mu =\) 1.5

Question 55

A biconvex lens has focal length 2/3 times the radius of curvature of either surface. Calculate the refractive index of lens material.

Answer: For biconvex lens with both surfaces of same radius R: \(f = \dfrac{R}{2(\mu - 1)}\).

Given \(f = \dfrac{2R}{3}\): \(\dfrac{2R}{3} = \dfrac{R}{2(\mu - 1)} \Rightarrow \mu - 1 = \dfrac{3}{4} \Rightarrow \mu =\) 1.75

Question 56

Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is 0.3 m and the refractive index of the material is 1.5.

Answer: Plano-convex: R₁ = ∞, R₂ = −R.

\(\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{\infty} - \dfrac{1}{-R}\right) = \dfrac{\mu - 1}{R}\)

\(R = f(\mu - 1) = 0.3 \times 0.5 =\) 0.15 m = 15 cm

Question 57

A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?

Answer: \(P_1 = 1/0.25 = 4\ \text{D}\), \(P_2 = -1/0.2 = -5\ \text{D}\). \(P_{\text{net}} = \) −1 D.

System is diverging (negative power).

Question 58

A convex lens of focal length 30 cm is placed coaxially in contact with a concave lens of focal length 40 cm. Determine the power of the combination. Will the system be converging or diverging?

Answer: \(P_1 = 1/0.3 = 3.33\ \text{D}\), \(P_2 = -1/0.4 = -2.5\ \text{D}\). \(P_{\text{net}} =\) +0.83 D.

System is converging.

Question 59

A convex lens of focal length f₁ is kept in contact with a concave lens of focal length f₂. Find the focal length of the combination.

Answer: \(\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}\) (both signs as per convention).

So \(F = \dfrac{f_1 f_2}{f_1 + f_2}\).

Question 60

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power for viewing distant objects when (i) in normal adjustment, (ii) final image at least distance of distinct vision. (b) Also find the separation in each case.

Answer: (i) Normal: \(m = \dfrac{f_o}{f_e} = \dfrac{140}{5} =\) 28. Separation = 140 + 5 = 145 cm.

(ii) Near point: \(m = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right) = 28 \times \left(1 + \dfrac{5}{25}\right) = 28 \times 1.2 =\) 33.6. Separation = \(f_o + \dfrac{f_e D}{f_e + D} = 140 + \dfrac{5 \times 25}{30} = 140 + 4.17 \approx\) 144.17 cm

Question 61

(a) Define refractive index of a medium. (b) In a given ray diagram, calculate the speed of light in the liquid of unknown refractive index. (Width of liquid surface = 30 cm, depth = 40 cm)

Answer: Same as Question 50. \(\mu = 5/3\), \(v = 1.8 \times 10^8\ \text{m/s}\).

Question 62

A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power for viewing distant objects in normal adjustment? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

Answer: Magnifying power = \(\dfrac{f_o}{f_e} = \dfrac{150}{5} =\) 30.

Object angular size = \(\dfrac{100}{3000} = \dfrac{1}{30}\) rad. Image height at focus of objective = \(f_o \times \text{angular size} = 150 \times \dfrac{1}{30} =\) 5 cm

Question 63

A convex lens is used to obtain a magnified image of an object on a screen 10 m from the lens. If the magnification is 19, find the focal length of the lens.

Answer: Real image, \(v = 10\ \text{m}\), \(|m| = 19\).

\(\dfrac{v}{u} = -19\) (magnified real image, u is distance to object). Actually \(|m| = \dfrac{v}{u} = 19 \Rightarrow u = \dfrac{10}{19}\).

\(\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{10} + \dfrac{19}{10} = \dfrac{20}{10} = 2 \Rightarrow f =\) 0.5 m = 50 cm

Question 64

An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object.

Answer: \(u + v = 90\), \(\dfrac{v}{u} = 2 \Rightarrow v = 2u \Rightarrow 3u = 90 \Rightarrow u = 30\ \text{cm}, v = 60\ \text{cm}\).

\(\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{60} + \dfrac{1}{30} = \dfrac{1}{20} \Rightarrow f =\) 20 cm

Nature: Convex lens of focal length 20 cm.

Question 65

The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length is 20 cm, calculate the object and image distances.

Answer: Erect image → virtual → \(m = +4 = \dfrac{v}{u}\) (with sign convention \(u\) negative, \(v\) negative).

So \(v = -4u\) relative magnitudes: \(|v| = 4|u|\). Using lens formula: \(\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\), with sign convention \(u = -x, v = -4x\), \(f = 20\):

\(\dfrac{1}{20} = \dfrac{1}{-4x} - \dfrac{1}{-x} = \dfrac{1}{-4x} + \dfrac{1}{x} = \dfrac{3}{4x} \Rightarrow x = 15\ \text{cm}\).

Object distance = 15 cm; image distance = 60 cm (virtual, same side).

Lens in Different Media

Question 66

A convex lens made of glass (μ = 1.5) is dipped in turn in (i) a medium of μ = 1.6, (ii) a medium of μ = 1.3. (a) Will it behave as a converging or a diverging lens in the two cases? (b) How will its focal length change in the two media?

Answer: (i) Medium μ = 1.6: lens μ (1.5) < medium μ (1.6), so \(\mu_r = \dfrac{1.5}{1.6} < 1\) → the factor \((\mu_r - 1)\) is negative → diverging. Focal length becomes negative (magnitude changes).

(ii) Medium μ = 1.3: lens μ (1.5) > medium μ (1.3), so \(\mu_r > 1\) → converging. Focal length becomes longer (magnitude increases, still positive).

Question 67

A converging lens has focal length 20 cm in air. It is made of material of refractive index 1.6. It is immersed in a liquid of refractive index 1.3. Calculate its new focal length.

Answer: In air: \(\dfrac{1}{f_a} = (\mu_g - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) = (0.6)K = \dfrac{1}{20} \Rightarrow K = \dfrac{1}{12}\).

In liquid: \(\dfrac{1}{f_w} = \left(\dfrac{\mu_g}{\mu_l} - 1\right)K = \left(\dfrac{1.6}{1.3} - 1\right) \times \dfrac{1}{12} = \left(\dfrac{0.3}{1.3}\right) \times \dfrac{1}{12} = \dfrac{0.3}{15.6}\)

\(f_w = \dfrac{15.6}{0.3} =\) 52 cm

Question 68

A convex lens made of glass (μ = 1.5) is dipped in turn in (i) a medium of μ = 1.65, (ii) a medium of μ = 1.33. (a) Converging or diverging in the two cases? (b) How does the focal length change?

Answer: (i) μ = 1.65 > 1.5 → diverging, f becomes negative.

(ii) μ = 1.33 < 1.5 → converging, f becomes longer.

Question 69

An equiconvex lens of refractive index μ₁, focal length f and radius of curvature R is immersed in a liquid of refractive index μ₂. For (i) μ₂ > μ₁, (ii) μ₂ < μ₁, draw the ray diagrams and find the focal length in terms of original focal length and refractive indices.

Answer: Lens in air: \(\dfrac{1}{f} = (\mu_1 - 1)\dfrac{2}{R}\).

Lens in medium: \(\dfrac{1}{f'} = \left(\dfrac{\mu_1}{\mu_2} - 1\right)\dfrac{2}{R}\).

So \(f' = f \cdot \dfrac{\mu_1 - 1}{\mu_1/\mu_2 - 1}\)

(i) μ₂ > μ₁ → diverging, f' negative. (ii) μ₂ < μ₁ → converging, f' positive (larger than f).

Mirror Formula and Mirrors

Question 70

With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.

Answer: Consider an object AB between F and C of a concave mirror. Image is A'B'. Using similar triangles:

\(\triangle A'B'P \sim \triangle ABP\) → \(\dfrac{A'B'}{AB} = \dfrac{v}{u}\) (magnitudes).

Also \(\triangle A'B'F \sim \triangle MPF\) giving \(\dfrac{v - f}{f}\) relation. Combining and applying sign convention:

\(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\)

Question 71

Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

Answer: Mirror formula: \(\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}\). Both f and u are negative. Let \(u = -x\) where \(2f < x < f\) (or \(|u|\) between |f| and 2|f|).

Solving gives \(v < -2f\), i.e., image is real (v negative) and beyond 2f.

Question 72

Use the mirror equation to show that (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f, (b) a convex mirror always produces a virtual image, (c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Answer: (a) As in Question 71 — real image beyond 2f.

(b) Convex mirror: \(f > 0\), \(u < 0\). Then \(\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} > 0\) always, so \(v > 0\) → virtual image regardless of u.

(c) Concave mirror with \(f < u < 0\): \(\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} > 0\), \(v > 0\) → virtual; and since |v| > |u| → enlarged.

Question 73

Calculate the distance of an object of height h from a concave mirror of focal length 10 cm so as to obtain a real image of magnification 2.

Answer: Real image, m = −2 = −v/u → v = 2u. Using \(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\):

\(\dfrac{1}{2u} + \dfrac{1}{u} = \dfrac{1}{-10}\) → \(\dfrac{3}{2u} = -\dfrac{1}{10}\) → u = −15 cm (object 15 cm in front of mirror).

Question 74

A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis has its image coinciding with itself. What is the focal length of the lens?

Answer: Rays from the object must be parallel after passing through the lens, so they reflect back on themselves. Object is at the focus: f = 20 cm.

Question 75

A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power for viewing distant objects? If the telescope views a 100 m tall tower 3 km away, find the height of the image formed by the objective.

Answer: Same as Question 62 — m = 30, image height = 5 cm.

Optical Instruments

Question 76

Draw a ray diagram of a reflecting type telescope. State two advantages of this telescope over a refracting telescope.

Answer: Standard ray diagram (concave mirror objective + eyepiece).

Advantages: (i) No chromatic aberration (mirror objective); (ii) No spherical aberration (paraboloidal mirror); (iii) Larger aperture possible, giving higher resolving power; (iv) Image is brighter.

Question 77

Draw a ray diagram of an astronomical telescope in normal adjustment position. State two drawbacks of this type of telescope.

Answer: Ray diagram: parallel rays → objective → image at f_o → eyepiece → parallel rays. M = f_o/f_e.

Drawbacks: (i) Chromatic aberration; (ii) Low resolving power (large aperture difficult to manufacture); (iii) Image is inverted and faint.

Question 78

Draw a ray diagram of a compound microscope. Write the expression for its magnifying power.

Answer: Standard ray diagram (objective + eyepiece).

When final image is formed at least distance of distinct vision: \(m = -\dfrac{v_o}{u_o}\left(1 + \dfrac{D}{f_e}\right)\).

When at infinity: \(m = -\dfrac{v_o}{u_o}\cdot\dfrac{D}{f_e}\) or simplified form.

Question 79

Draw a labelled ray diagram of an astronomical telescope in near point position. Write the expression for its magnifying power.

Answer: Ray diagram: telescope focused such that final image is at near point.

Magnifying power: \(m = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right)\)

Question 80

Draw a labelled ray diagram showing the image formation of an astronomical telescope in normal adjustment position. Write the expression for magnifying power.

Answer: Ray diagram: normal adjustment.

Magnifying power: \(m = -\dfrac{f_o}{f_e}\)

Question 81

Draw a ray diagram for the formation of image in a compound microscope. Write the expression for its magnifying power.

Answer: Same as Question 78.

Question 82

Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over a refracting telescope.

Answer: Same as Question 76. Advantages: (i) No chromatic aberration, (ii) No spherical aberration with paraboloid mirror, (iii) Higher brightness and resolving power.

Question 83

A ray of light incident on a glass prism and undergoes minimum deviation when the angle of incidence is 3/4 of the prism angle. Calculate the speed of light in the prism (assume equilateral prism).

Answer: Same as Question 34. v ≈ 2.12 × 10⁸ m/s.

Long Answer Questions (5 Marks)

Mirror Equation Derivations

Question 84

With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.

Answer: Same as Question 70. Final expression: \(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\).

Question 85

Answer: Same as Question 72.

Refraction at Spherical Surface

Question 86

Draw a ray diagram to show the formation of the image of an object placed on the axis of a convex refracting surface, of radius of curvature R, separating two media of refractive indices n₁ and n₂ (n₂ > n₁). Use this diagram to deduce the relation \(\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}\).

Answer: For paraxial rays, using small-angle approximation and sign convention, we get:

From geometry: \(\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}\)

This is the refraction formula for a single spherical surface.

Question 87

Define power of a lens. Write its SI unit. Two thin convex lenses of focal lengths f₁ and f₂ are placed in contact coaxially. Derive the expression for the effective focal length of the combination.

Answer: Power of a lens: Ability to converge/diverge light, defined as \(P = \dfrac{1}{f}\). SI unit: dioptre (D).

For two thin lenses in contact: \(\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2}\).

So \(P = P_1 + P_2\), and the effective focal length is \(F = \dfrac{f_1 f_2}{f_1 + f_2}\).

Compound Microscope

Question 88

Draw a ray diagram showing the image formation by a compound microscope. Hence obtain the expression for total magnification when the image is formed at infinity.

Answer: Standard ray diagram with objective and eyepiece.

Magnification due to objective: \(m_o = \dfrac{v_o}{u_o} = \dfrac{L}{f_o}\) (when image is at infinity, L is tube length ≈ distance from second focal point of objective to first focal point of eyepiece).

Magnification due to eyepiece: \(m_e = \dfrac{D}{f_e}\).

Total: \(m = m_o \times m_e = \dfrac{L}{f_o} \cdot \dfrac{D}{f_e}\)

Question 89

(i) Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working. (ii) Why must both the objective and the eye-piece of a compound microscope have short focal lengths?

Answer: (i) Standard diagram. Working: objective forms a real, inverted, magnified image at the eyepiece's focus. This image acts as an object for the eyepiece, which forms a final virtual, magnified image viewed by the eye.

(ii) \(m \propto \dfrac{1}{f_o}\) and \(m \propto \dfrac{1}{f_e}\) — smaller focal lengths give larger magnification. Also smaller f means angular magnification is large for close viewing.

Question 90

Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain.

Answer: Magnifying power (final image at infinity) is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object when both are at the least distance of distinct vision from the eye.

For short f_o and f_e, the magnification is higher.

Question 91

Draw a labelled ray diagram of a compound microscope. Why are the objective and the eye-piece chosen to have small focal length?

Answer: Same as Question 89 (ii): (1) Larger magnification with short focal lengths; (2) Avoids aberrations for larger bending of rays.

Question 92

(a) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision. (b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The distance between the objective and eyepiece is 14 cm. If the least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eyepiece.

Answer: (a) Same ray diagram. Final image at D = 25 cm.

(b) Total m = 20, m_e = 5 ⇒ m_o = 4. Tube length = 14 cm, D = 20 cm.

For eyepiece: m_e = 1 + D/f_e = 5 ⇒ D/f_e = 4 ⇒ f_e = 5 cm. Then distance from objective image to eyepiece = 5 cm (since m_e = D/f_e when image at near point), and v_o = 14 − 5 = 9 cm.

For objective: m_o = v_o/u_o = 4 ⇒ u_o = 2.25 cm. \(\dfrac{1}{f_o} = \dfrac{1}{9} - \dfrac{1}{-2.25} = \dfrac{1}{9} + \dfrac{4}{9} = \dfrac{5}{9}\) ⇒ f_o = 1.8 cm

Astronomical Telescope

Question 93

Draw a labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly its working. (ii) An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment?

Answer: (i) Standard ray diagram. Working: parallel rays from distant object form image at focus of objective. This image (at focus of eyepiece in normal adjustment) is viewed through eyepiece, giving magnified angular view.

(ii) P_o = 1 D → f_o = 1 m; P_e = 10 D → f_e = 0.1 m. Magnifying power = f_o/f_e = 10.

Question 94

Draw a labelled ray diagram of a refracting telescope. Define its magnifying power and write the expression for it. Write two important limitations of a refracting telescope over a reflecting type telescope.

Answer: Ray diagram (standard).

Magnifying power: the ratio of the angle subtended by the final image on the eye to the angle subtended by the object directly: \(M = \dfrac{f_o}{f_e}\).

Limitations: (i) Chromatic aberration (blurred multicoloured image); (ii) Low light-gathering power due to difficulty of making large-aperture lenses.

Question 95

Draw a labelled ray diagram of a refracting telescope in normal adjustment. Write two main considerations required of an astronomical telescope.

Answer: Same diagram. Considerations: (i) objective should have large focal length and large aperture (high magnification, high light gathering, high resolving power); (ii) eyepiece should have small focal length.

Question 96

A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power for viewing distant objects? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

Answer: Same as Question 62 — M = 30, h' = 5 cm.

Question 97

(i) A giant refracting telescope has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification? (ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m and the radius of the lunar orbit is 3.8 × 10⁸ m.

Answer: (i) Angular magnification = \(\dfrac{f_o}{f_e} = \dfrac{15}{0.01} =\) 1500.

(ii) Angular diameter of the moon = \(\dfrac{3.48 \times 10^6}{3.8 \times 10^8} = 9.16 \times 10^{-3}\) rad.

Diameter of image at focus of objective: \(d' = f_o \times \theta = 15 \times 9.16 \times 10^{-3} \approx\) 0.137 m = 13.7 cm

Question 98

(i) A giant refracting telescope has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification? (ii) If the telescope views the moon, find the diameter of the image of the moon formed by the objective. (Moon diameter 3.42 × 10⁶ m, orbital radius 3.8 × 10⁶ m)

Answer: (i) \(m = \dfrac{15}{0.01} =\) 1500.

(ii) Angular diameter of the moon = \(\dfrac{3.42 \times 10^6}{3.8 \times 10^6} = 0.9\) rad.

Diameter of image = \(15 \times 0.9 =\) 13.5 m

Key Formulas to Remember

Concept	Formula
Refractive index	\(\mu = \dfrac{c}{v} = \dfrac{\sin i}{\sin r}\)
Critical angle	\(\sin i_c = \dfrac{1}{\mu}\)
Prism (min deviation)	\(\mu = \dfrac{\sin[(A+\delta_m)/2]}{\sin(A/2)}\)
Mirror formula	\(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\)
Lens formula	\(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\)
Lens Maker's formula	\(\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\)
Power of lens	\(P = \dfrac{1}{f}\) (in metres)
Lenses in contact	\(\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}\)
Magnification (mirror/lens)	\(m = \dfrac{v}{u}\)
Telescope magnification	\(M = \dfrac{f_o}{f_e}\) (normal adjustment)
Microscope magnification	\(M = -\dfrac{v_o}{u_o} \cdot \dfrac{D}{f_e}\) (image at infinity)
Scattering (Rayleigh)	\(I \propto \dfrac{1}{\lambda^4}\)
Refraction at spherical surface	\(\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}\)

FAQs

Are these questions enough for CBSE Class 12 Physics Chapter 9?

Yes. These questions cover every major topic — refraction, total internal reflection, lenses, prisms, mirrors, dispersion and scattering, optical instruments (microscope and telescope) — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.

Which topics are most important for board exams?

Lens Maker's formula and lens combinations, refraction through prisms (minimum deviation derivation), total internal reflection (conditions, critical angle calculations), mirror and lens formulas with sign conventions, magnifying power of microscope and telescope (ray diagrams and derivations), and scattering of light are consistently high-weightage topics.

How should I prepare for numericals in this chapter?

Practise lens and mirror formula problems, lens combination power calculations, critical angle → speed of light problems, prism minimum deviation → refractive index, image formation with sign conventions, and magnification/height of image in telescopes. Focus on applying the sign convention carefully — this is where most mistakes are made.