Chapter 9 Ray Optics and Optical Instruments

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NCERT Solutions Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments

Chapter 9 of Physics for Class 12 is Ray Optics and Optical Instruments. In this chapter, light is studied as a straight line. This chapter provides the foundation for the field of optics and introduces the basics of reflection and refraction. Ray optics investigates the geometry of light (i.e., how it changes direction/angle when it passes from one medium to another) as well as how to create images with mirrors and lenses. Understanding ray optics is crucial to understanding how humans see (the eye) and developing advanced types of technology such as fiber optics, telescopes, and microscopes. As a result, ray optics is one of the most important chapters making up the total score for the CBSE Board Exam, JEE (Joint Entrance Examination), and NEET (National Eligibility Entrance Test).

Solutions to questions in NCERT Class XII Physics Chapter 9 (Ray Optics and Optical Instruments) are a comprehensive collection of solutions. It provides an easily understood way to apply the laws of reflection and trace rays from simple reflections through complex optical devices such as lenses and lens systems.

1.0Class 12 Physics Chapter 9 Ray Optics and Optical Instruments: Key Concepts

Class 12 Physics Chapter 9, Ray Optics and Optical Instruments, explains the behaviour of light during reflection and refraction and the working of optical instruments. The main concepts covered in this chapter include:

Reflection from Spherical Mirrors: Learn the mirror formula, magnification, and image formation by concave and convex mirrors using the Cartesian sign convention.
Refraction and Total Internal Reflection: Understand Snell’s law, refractive index, critical angle, and the conditions required for total internal reflection. The chapter also explains applications of TIR, such as optical fibres.
Refraction Through Curved Surfaces and Lenses: Study refraction at spherical surfaces, the lens maker’s formula, and the combination of two or more thin lenses.
Refraction Through a Prism: Understand the path of light through a prism and the relationship between the angle of incidence, angle of emergence, and angle of minimum deviation.
Dispersion and Scattering of Light: Learn how white light separates into its component colours and understand the formation of a rainbow and the blue colour of the sky.
Optical Instruments: Study the working of the human eye, microscope, and telescope, including the basic principles of their use for viewing objects clearly.

2.0NCERT Solutions Class 12 Physics Chapter 9 Ray Optics and Optical Instruments : Detailed Solutions

SOLVED EXAMPLES

Suppose that the lower half of the concave mirror's reflecting surface is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror? Sol. You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).
A mobile phone lies along the principal axis of a concave mirror, as shown in figure. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror?

Concave mirror with a mobile phone placed along the principal axis, showing the pole, focus F, and centre of curvature C.

Sol.

Ray diagram showing image formation of a mobile phone by a concave mirror, with object and image positions along the principal axis.

The ray diagram for the formation of the image of the phone is shown in figure. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., $B^{'} C = BC$ .Magnification is not uniform because different parts of the object are at different distances (u) from the mirror. Yes, the distortion depends entirely on where the phone is placed relative to the mirror.

An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Sol. The focal length $f = - 15/2 cm = - 7.5 cm$

(i) The object distance $u = - 10 cm$ . Then Eq, gives

$\frac{1}{v} + \frac{1}{- 10} = \frac{1}{- 7.5}$

$v = \frac{10 \times 7.5}{- 2.5} = - 30 cm$

The image is 30 cm from the mirror on the same side as the object. Also, magnification $m = - \frac{v}{u} = - \frac{( - 30 )}{( - 10 )} = - 3$ The image is magnified, real and inverted. (ii) The object distance $u = - 5 cm$ . Then from Eq.

$\frac{1}{v} + \frac{1}{- 5} = \frac{1}{- 7.5} or v = \frac{5 \times 7.5}{( 7.5 - 5 )} = 15 cm$

This image is formed at 15 cm behind the mirror. It is a virtual image. Magnification $= m = - \frac{v}{u} = - \frac{15}{( - 5 )} = 3$ The image is magnified, virtual and erect.

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of $R = 2 m$ . If the jogger is running at a speed of $5 ms^{- 1}$ , how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m and (d) 9 m away.

Sol. From the mirror equation, Eq., $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ we get,

$v = \frac{f u}{u - f}$

For convex mirror, since $R = 2 m, f = 1 m$ . Then for $u = - 39 m$ ,

$v = \frac{( - 39 ) \times 1}{- 39 - 1} = \frac{39}{40} m$

Since the jogger moves at a constant speed of $5 ms^{- 1}$ , after 1s the position of the image v (for $u = - 39 + 5 = - 34$ ) is (34/35)m. The shift in the position of image in 1s is

$\frac{39}{40} - \frac{34}{35} = \frac{1365 - 1360}{1400} = \frac{5}{1400} = \frac{1}{280} m$

Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is $(1/280) ms^{- 1}$ . Similarly, it can be seen that for $u = - 29 m, - 19 m$ and -9 m , the speed with which the image appears to move is $\frac{1}{150} ms^{- 1}, \frac{1}{60} ms^{- 1}$ and $\frac{1}{10} ms^{- 1}$ , respectively. Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she move closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.

Light from a point source in air falls on a spherical glass surface ( $n = 1.5$ and radius of curvature $= 20 cm$ ). The distance of the light source from the glass surface is 100 cm. At what position the image is formed? Sol. We use the relation given by Equation,

$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$

Here,

$u = - 100 cm, v = ?, R = + 20 cm, n_{1} = 1 and n_{2} = 1.5.$

We then have

$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} or v = + 100 cm$

The image is formed at a distance of 100 cm from the glass surface, in the direction of incident light.

A magician during a show makes a glass lens with $n = 1.47$ disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water? Sol. The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means $n_{1} = n_{2}$ . This gives $1/ f = 0$ or $f \to \infty$ . The lens in the liquid will act like a plane sheet of glass. No, the liquid is not water. It could be glycerine.
(i) If $f = 0.5 m$ for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm . What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass $= 1.5$ )

Sol.

(i) $Power (P) = \frac{1}{f} = \frac{1}{0.5} = + 2$ dioptre. (ii) Here, we have $f = + 12 cm$ ,

$R_{1} = + 10 cm, R_{2} = - 15 cm$

Refractive index of air is taken as unity.

We use the lens formula of equation,

$\frac{1}{f} = (\frac{n _{2}}{n _{1}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

The sign convention has to be applied for f, $R_{1}$ and $R_{2} . n_{1} = 1& n_{2} = n$ Substituting the values, we have

$\frac{1}{12} = (n - 1) (\frac{1}{10} - \frac{1}{- 15})$

This gives $n = 1.5$

(iii) For a glass lens in air,

$n_{2} = 1.5, n_{1} = 1, f = + 20 cm .$

Hence, the lens formula gives

$\frac{1}{20} = 0.5 [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$

For the same glass lens in water,

$n_{2} = 1.5 n_{1} = 1.33. Therefore, \frac{1.33}{f} = (1.5 - 1.33) [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$

Combining these two equations, we find $f = + 78.2 cm$ .

Find the position of the image formed by the lens combination given in the figure.

Lens combination with two lenses of different focal lengths, showing their separation and object position for image formation.

Sol. Image formed by the first lens

$\frac{1}{v _{1}} - \frac{1}{u _{1}} = \frac{1}{f _{1}} \frac{1}{v _{1}} - \frac{1}{- 30} = \frac{1}{10}$

or $v_{1} = 15 cm$

The image formed by the first lens serves as the object for the second. This is at a distance of ( $15 - 5$ ) $cm = 10 cm$ to the right of the second lens. Though the image is real, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens.

$\frac{1}{v _{2}} - \frac{1}{10} = \frac{1}{- 10} or v_{2} = \infty$

The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.

$\frac{1}{v _{3}} - \frac{1}{u _{3}} = \frac{1}{f _{3}} or \frac{1}{v _{3}} = \frac{1}{\infty} + \frac{1}{30}$

or $v_{3} = 30 cm$ The final image is formed 30 cm to the right of the third lens.

EXERCISE QUESTION WITH SOLUTIONS

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm . At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved ? Sol. $h_{1} = 2.5 cm, u = - 27 cm, R = - 36 cm$

$f = \frac{R}{2} = - \frac{- 36}{2} = - 18 cm \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \Rightarrow \frac{1}{- 27} + \frac{1}{v} = \frac{1}{- 18}$

$v = - 54 cm$

$\frac{h _{2}}{h _{1}} = - \frac{v}{u}$

$\frac{h _{2}}{2.5} = - [\frac{- 54}{- 27}]$

$\Rightarrow h_{2} = - 5 cm$ If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. Sol. $f = + 15 cm, h_{1} = 4.5 cm, u = - 12 cm, v =$ ? Using mirror equation,

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f} \Rightarrow \frac{1}{- 12} + \frac{1}{v} = \frac{1}{15} v = 6.7 cm \frac{h _{2}}{h _{1}} = - \frac{v}{u} \Rightarrow \frac{h _{2}}{4.5} = - (\frac{6.7}{- 12}) h_{2} = + 2.5 cm m = \frac{h _{2}}{h _{1}} = \frac{2.5}{4.5} = 0.56$

If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again? Sol. Case-I Actual depth, $d_{AC} = 12.5 cm$ Apparent depth, $d_{AP} = 9.4 cm$

$n = \frac{d _{AC}}{d _{AP}} = \frac{12.5}{9.4} = 1.33$

Case-II $d_{AP} = \frac{d _{AC}}{n} = \frac{12.5}{1.63} = 7.67 cm$ So, We should move the microscope by distance $9.4 - 7.67 = 1.73 cm$

Figure (a) and (b) show refraction of an incident ray in air incident at 60° with the normal to a glass-air and water air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is $4 5^{\circ}$ with the normal to a water-glass interface [figure (c)].

Refraction diagrams showing light rays passing between air, glass, and water at different interfaces and angles of incidence.

Sol. From fig. (a)

$_{a} n_{g} = \frac{s i n i}{s i n r} = \frac{s i n 6 0 ^{\circ}}{s i n 3 5 ^{\circ}} = \frac{0.8660}{0.5736} = 1.51$

From fig. (b),

$_{a} n_{w} = \frac{s i n i}{s i n r} = \frac{s i n 6 0 ^{\circ}}{s i n 4 7 ^{\circ}} = \frac{0.8660}{0.7314} = 1.184$

From fig (c),

$_{w} n_{g} = \frac{_{a} n _{g}}{_{a} n _{w}} = \frac{s i n i}{s i n r}$

or $\frac{1.51}{1.184} = \frac{s i n 4 5 ^{\circ}}{s i n r} = \frac{0.7071}{s i n r}$ or $sin r = \frac{1.184 \times 0.7071}{1.51} = 0.5546$

$∴ r ≃ 33.6 8^{\circ} .$

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33 . (Consider the bulb to be a point source.)

Point light source at the bottom of a water tank, showing the critical angle and circular region through which light emerges from the water surface.

Sol.

$sin θ_{C} = \frac{1}{n} = \frac{1}{1.33} θ_{C} ≃ 48. 6^{\circ} r = h tan θ_{C} = 0.80 tan 48. 6^{\circ} = 90.7 cm$

So, area of the surface of water $= π r^{2}$

$= 3.14 \times (0.907)^{2} = 2.61 m^{2}$

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is $6 0^{\circ}$ . If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. Sol. When the prism is placed in air:

$δ_{m} = 4 0^{\circ}, A = 6 0^{\circ}$

$∴$ Refractive index of the prism material is

$n_{g} = \frac{sin \frac{A + δ _{m}}{2}}{sin \frac{A}{2}} = \frac{sin \frac{6 0 ^{\circ} + 40}{2}}{sin \frac{6 0 ^{\circ}}{2}} = \frac{sin 5 0 ^{\circ}}{sin 3 0 ^{\circ}} = \frac{0.7660}{0.5000} = 1.532$

When the prism is placed in water :

$_{w} n_{g} = \frac{s i n \frac{A + δ _{m}^{'}}{2}}{s i n \frac{A}{2}} or \frac{_{a} n _{g}}{_{a} n _{w}} = \frac{s i n \frac{6 0 ^{\circ} + δ _{m}^{'}}{2}}{s i n \frac{6 0 ^{\circ}}{2}}$

or $\frac{1.532}{1.33} = \frac{s i n \frac{6 0 ^{\circ} + δ _{m}^{'}}{2}}{s i n 3 0 ^{\circ}}$ or

$sin \frac{6 0 ^{\circ} + δ _{m}^{'}}{2} = \frac{1.532}{1.33} \times 0.5 = 0.5759$

$∴ 3 0^{\circ} + \frac{δ _{m}^{'}}{2} = sin^{- 1} (0.5759) = 35.1 6^{\circ}$ or

$δ^{'}_{m} = 10.3 2^{\circ} .$

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm? Sol. $\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{20} = (1.55 - 1) (\frac{1}{R} - (- \frac{1}{R}))$

$R = 22 cm$

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P . At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Sol:

Converging light beam passing through convex and concave lenses placed before point P, showing the resulting convergence points.

9. An object of size 3 cm is placed 14cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? Sol. $h_{1} = 3 cm, u = - 14 cm, f = - 21 cm$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \frac{1}{v} - \frac{1}{( - 14 )} = \frac{1}{- 21} v = - 8.4 cm$

$\frac{h _{2}}{h _{1}} = \frac{v}{u} \frac{h _{2}}{3} = \frac{- 8.4}{- 14} \Rightarrow h_{2} = 1.8 cm .$

So, the image is diminished in size.

What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses. Sol. $\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{30} + \frac{1}{( - 20 )}$ $f = - 60 cm$ (Diverging lens)
A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case? Sol.

$f_{0} = 2 cm f_{e} = 6.25 cm L = 15 cm$

(a) When final image is obtained at least distance of distinct vision.

$v_{e} = - 25 cm \frac{1}{v _{e}} - \frac{1}{u _{e}} = \frac{1}{f _{e}} \frac{1}{- 25} - \frac{1}{u _{e}} = \frac{1}{6.25} u_{e} = - 5 cm, L = 15 cm L = v_{0} + ∣ u_{e} ∣ = 15 v_{0} = 10 cm \frac{1}{v _{0}} - \frac{1}{u _{0}} = \frac{1}{f _{0}}$

Using lens formula for objective lens,

$\frac{1}{10} - \frac{1}{u _{0}} = \frac{1}{2} u_{0} = - 2.5 cm M = \frac{v _{0}}{∣ u _{0} ∣} (1 + \frac{D}{f _{e}}) = \frac{10}{2.5} (1 + \frac{25}{6.25}) = 20$

(b) When final image is formed at infinity :

$v_{e} = \infty, u_{e} = f_{e} = 6.25 v_{0} = L - ∣ u_{e} ∣ = 15 - 6.25 = 8.75 cm \frac{1}{v _{0}} - \frac{1}{u _{0}} = \frac{1}{6.25} u_{0} = - 2.59 cm M = \frac{v _{0}}{u _{0}} \frac{D}{f _{e}} = \frac{8.75}{2.59} \times \frac{25}{6.25} = 13.51$

A person with a normal near point (25 cm) using a compound microscope with a objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses. How much is the magnifying power of the microscope? Sol. Here,

$f_{0} = 0.8 cm, u_{0} = - 0.9 cm, v_{0} = ?$

For the objective, As,

$\frac{1}{v _{0}} - \frac{1}{u _{0}} = \frac{1}{f _{0}}$

∴

$\frac{1}{v _{0}} = \frac{1}{f _{0}} + \frac{1}{u _{0}} = \frac{1}{0.8} - \frac{1}{0.9} = \frac{0.9 - 0.8}{0.9 \times 0.8} = \frac{0.1}{0.8 \times 0.9}$

$v_{0} = \frac{0.8 \times 0.9}{0.1} = 7.2 cm$

For the eyepiece,

$= \frac{1}{v _{e}} - \frac{1}{u _{e}} = \frac{1}{f _{e}}$

$= \frac{1}{- 25} - \frac{1}{u _{e}} = \frac{1}{2.5} = \frac{- 1}{25} - \frac{1}{2.5} = \frac{1}{u _{e}} = \frac{- 25}{11} = - 2.27 cm$

Hence the separation between the two lenses

$= v_{0} + ∣ u_{e} ∣ = 7.2 + 2.27 = 9.47 cm$

Magnifying power,

$m = m_{0} \times m_{e} = \frac{v _{0}}{∣ u _{0} ∣} (1 + \frac{D}{f _{e}}) = \frac{7.2}{0.9} (1 + \frac{25}{2.5}) = 88$

A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0 cm . What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? Sol. $M = - \frac{f _{0}}{f _{e}} = \frac{- 144}{6} = - 24$

$L = f_{0} + f_{e} = 144 + 6 = 150 cm$

(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) 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 $3.48 \times 1 0^{6} m$ , and the radius of lunar orbit is $3.8 \times 1 0^{8} m$ . Sol.

Refracting telescope diagram showing the objective lens forming an image of the Moon, with the Moon’s diameter and orbital radius indicated.

(a) $M = \frac{f _{0}}{f _{e}} = \frac{- 15}{1 0 ^{- 2}} = - 1500$ (b) $α = \frac{Dia.of moon}{Radius}$

$α = \frac{3.48 \times 1 0 ^{6}}{3.8 \times 1 0 ^{8}} = \frac{d}{15} d = 13.73 cm .$

Use the mirror equation to deduce 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 independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Sol. (a) From mirror formula, $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ Now for a concave mirror, $f < 0$ and for an object on the left, $u < 0$ .

$∴ 2 f < u < f or \frac{1}{2 f} > \frac{1}{u} > \frac{1}{f}$

or $- \frac{1}{2 f} < - \frac{1}{u} < - \frac{1}{f}$ or $\frac{1}{f} - \frac{1}{2 f} < \frac{1}{f} - \frac{1}{u} < \frac{1}{f} - \frac{1}{f}$ or $\frac{1}{2 f} < \frac{1}{v} < 0$ This implies that $v < 0$ so that image is formed on left. Also the above inequality implies.

$2 f > v or ∣2 f ∣ < ∣ v ∣ [∵ 2 f and v are negative]$

i.e., The real image is formed beyond 2f.

(b) For a convex mirror, $f > 0$ and for an object on left, $u < 0$ . From mirror formula,

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

This implies that $\frac{1}{v} > 0$ or $v > 0$ This shows that whatever be the value of u a convex mirror forms a virtual image on the right.

(c) For convex mirror, $f > 0$ and for an object on the left $u < 0$ , so mirror formula, $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ implies that $\frac{1}{v} > \frac{1}{f} [∵ - \frac{1}{u}$ is a +ve quantity ] or This shows that the image is located between the pole and the focus of the mirror. (d) From mirror formula, $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ For a concave mirror, $f < 0$ and for an object located between the pole and focus of a concave mirror, $f < u < 0$

$∴ \frac{1}{f} > \frac{1}{u} or \frac{1}{f} - \frac{1}{u} > 0 or \frac{1}{v} > 0$

i.e., a virtual image is formed on the right side of mirror.

$Also \frac{1}{v} < \frac{1}{∣ u ∣} or v > ∣ u ∣ ∴ ∣ m ∣ = \frac{v}{∣ u ∣} > 1$

i.e., image is enlarged.

A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass $= 1.5$ . Does the answer depend on the location of the slab ? Sol. $x = t (1 - \frac{1}{μ}) = 15 (1 - \frac{1}{1.5}) = 5 cm$ The answer does not depend on location of glass slab.
(a) Figure shows a cross-section of a 'light-pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angle of the incident rays with the axis of the pipe for which total reflections inside the pipe take place as shown in the figure.

Cross-section of a glass fibre light pipe showing incident rays entering the core and undergoing total internal reflection at the inner surface.

(b) What is the answer if there is no outer covering of the pipe ? Sol. (a) Given $n_{1} = 1.68, n_{2} = 1.44$ ,

$n_{1} sin i_{c}^{'} = n_{2} sin 9 0^{\circ}$

∴ Critical angle $i^{'}_{c}$ is given by

$sin i_{c}^{'} = \frac{n _{2}}{n _{1}} = \frac{1.44}{1.68} = 0.8571$

So $i_{c}^{'} = 5 9^{o}$ Total internal reflection will occur if the angle $i^{'} > i_{^{'}}$ , i.e., if $i^{'} > 5 9^{\circ}$ or when $r < r_{max}$ , where $r_{max} = 9 0^{\circ} - 5 9^{\circ} = 3 1^{\circ}$ . Using snell's law,

$\frac{s i n i _{m a x}}{s i n r _{m a x}} = 1.68$

or $sin i_{max} = 1.68 \times sin r_{max}$

$= 1.68 \times sin 3 1^{\circ}$

$= 1.68 \times 0.5150$

$= 0.8652$

$∴ i_{max} ≃ 6 0^{\circ}$ Thus all incident rays which make angles in the range $0 < i < 6 0^{\circ}$ with the axis of the pipe will suffer total internal reflections in the pipe.

(b) If the outer covering of pipe is not present, then refractive index of outer pipe $= 1$ ,

$i_{c}^{'} = sin^{- 1} (\frac{1}{1.68}) = 36. 5^{\circ},$

Now, $i = 9 0^{\circ}$ will have $r = 36. 5^{\circ}$ and $i^{'} = 53. 5^{\circ}$ which is greater than $i^{'}_{c}$ . Thus all incident rays (in the range $53. 5^{\circ} < i < 9 0^{\circ}$ ) will suffer TIR.

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose? Sol. The minimum distance between an object and its real image is 4f.

$∴ 4 f_{m a x} = D$

or $f_{max} = \frac{D}{4} = \frac{3 m}{4} = 0.75 m$ [Derive $4 f_{max} = D$ ]. [Newton's displacement formula]

A screen is placed 90 cm from an object, the image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens. Sol. As shown in figure let O and I be the positions of object and image respectively and $L_{1}$ and $L_{2}$ be the two conjugate positions of the lens.

Convex lens with two conjugate positions L₁ and L₂ forming an image on a screen placed at a fixed distance from the object.

Obviously, $x + 20 + x = 90 cm$ or $x = 35 cm$ when the lens is in position $L_{1}$ we have

$u = - x = - 35 cm, v = 20 + x = 20 + 35 = 55 cm ∴ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{55} + \frac{1}{35} = \frac{7 + 11}{385} = \frac{18}{385}$

or $f = \frac{385}{18} = 21.4 cm$ .

(a) Determine the 'effective focal length' of the combination of the two lenses in Question 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangements. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image. Sol. (a) (i) Let a parallel beam of light be incident from the left on the convex lens first. Then

$f_{1} = 30 cm u_{1} = - \infty$

As $\frac{1}{v _{1}} - \frac{1}{u _{1}} = \frac{1}{f _{1}}$

$∴ \frac{1}{v _{1}} = \frac{1}{f _{1}} + \frac{1}{u _{1}} = \frac{1}{30} - \frac{1}{\infty} = \frac{1}{30}$

or $v_{1} = + 30 cm$ This image becomes a virtual object for the second lens so that

$f_{2} = - 20 cm u_{2} = + (30 - 8) = + 22 cm$

Now, $\frac{1}{v _{2}} = \frac{1}{f _{2}} + \frac{1}{u _{2}}$

$= \frac{1}{v _{2}} = - \frac{1}{20} + \frac{1}{22} = \frac{- 11 + 10}{220} = \frac{- 1}{220}$

Or $v_{2} = - 220 cm$ The parallel incident beam appears to diverge from a point $220 - 4 = 216 cm$ from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first. Then

$f_{1} = - 20 cm, u_{1} = - \infty$

As $\frac{1}{v _{1}} = \frac{1}{f _{1}} + \frac{1}{u _{1}} = \frac{1}{- 20} + \frac{1}{- \infty} = - \frac{1}{20}$ or $v_{1} = - 20 cm$

This image becomes a real object for the second lens so that

$f_{2} = + 30 cm, u_{2} = - (20 + 8) = - 28 cm$

Now, $\frac{1}{v _{2}} = \frac{1}{f _{2}} + \frac{1}{u _{2}} = \frac{1}{30} - \frac{1}{28} = \frac{14 - 15}{420} = \frac{- 1}{420}$ or $v_{2} = - 420 cm$ Thus the parallel incident beam appear to diverge from a point 420-4 = 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. The notion of effective focal length, therefore, does not seem to be meaningful for the system.

(b) here $u_{1} = - 40 cm, f_{1} = 30 cm$

$As \frac{1}{v _{1}} - \frac{1}{u _{1}} = \frac{1}{f _{1}} ∴ \frac{1}{v _{1}} + \frac{1}{40} = \frac{1}{30} or \frac{1}{v _{1}} = \frac{1}{30} - \frac{1}{40} = \frac{4 - 3}{120} = \frac{1}{120} or v_{1} = 120 cm$

Magnitude of magnification due to the first (convex) lens is $m_{1} = \frac{v _{1}}{∣ u _{1} ∣} = \frac{120}{40} = 3$ This image becomes a virtual object for the second lens so that

$u_{2} = + (120 - 8) = + 112 cm f_{2} = - 20 cm$

now

$\frac{1}{v _{2}} = \frac{1}{f _{2}} + \frac{1}{u _{2}} = - \frac{1}{20} + \frac{1}{112} = \frac{- 112 + 20}{112 \times 20} = \frac{- 92}{112 \times 20}$

Or $v_{2} = - \frac{112 \times 20}{92} cm = - 24.35 cm$ Magnitude of magnification due to the second (concave) lens is

$m_{2} = \frac{∣ v _{2} ∣}{u _{2}} = \frac{112 \times 20}{92 \times 112} = \frac{20}{92}$

Net magnitude of magnification due to the two-lens system is

$m = m_{1} \times m_{2} = \frac{3 \times 20}{92} = 0.652$

Size of image,

$h_{2} = mh_{1} = 0.652 \times 1.5 = 0.98 cm$

At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? the refractive index of the prism is 1.524. Sol. The refracted ray QR will just suffer total internal reflection if it is incident at the critical angle $i_{c}$ .

Prism ray diagram showing a light ray entering a 60° prism and reaching the second face at the critical angle for total internal reflection.

Thus $r_{2} = i_{c}$ Now $sin i_{c} = \frac{1}{n} = \frac{1}{1.524} = 0.6562$

$∴ i_{c} = sin^{- 1} (0.6562) ≃ 4 1^{\circ}$

but, $r_{1} + r_{2} = A$

$∴ r_{1} = A - r_{2} = A - i_{c} = 6 0^{\circ} - 4 1^{\circ} = 1 9^{\circ}$

From Snell's law, $n = \frac{s i n i _{1}}{s i n r _{1}}$

$∴ sin i_{1} sin i_{1} sin i_{1} = n sin 1 9^{\circ} = 1.524 \times 0.3256 = 0.4962$

Hence, $i_{1} = sin^{- 1} (0.4962) ≃ 3 0^{\circ}$ .

A card sheet divided into squares each of size $1 mm^{2}$ is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens ? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

Sol. (a) $h_{i} = 1 mm, u = - 9 cm, f = + 10 cm$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{( - 9 )} = \frac{1}{10} v = - 90 cm m = \frac{v}{u} = \frac{- 90}{- 9} = 10 A = 100 mm^{2} = 1 cm^{2}$

(b) $M = \frac{D}{u} = \frac{- 25}{- 9} = 2.8$ (c) The magnification in part (a) is the not the same as magnifying power in part (b). The magnification magnitude is $(\frac{v}{u})$ & the magnifying power is $(\frac{D}{∣ u ∣})$ . The two quantities will be equal when the image is formed at the near point (25 cm).

(a) At what distance should the lens be held from the card sheet in Question 22 in order to view the squares distinctly with the maximum possible magnifying power ? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case ? Explain.

Sol. (a) Maximum magnifying power is obtained when the image is at the near point (25 cm). Thus,

$v = - 25 cm, f = + 10 cm, u = ? \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$∴ \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = - \frac{1}{25} - \frac{1}{10} = \frac{- 2 - 5}{50} = \frac{- 7}{50}$ Or $u = - \frac{50}{7} = - 7.14 cm$ So lens should be held 7.14 cm away from the card sheet. (b) Magnitude of magnification is

$m = \frac{v}{u} = \frac{- 25}{- 50/7} = 3.5$

Yes, the magnifying power is equal to the magnitude of magnification because image is formed at the least distance of distinct vision.

What should be the distance between the object in Question 23 and the magnifying glass if the virtual image of each square have an area of $6.25 mm^{2}$ ? Would you be able to see the squares distinctly with your eyes very close to the magnifier? Sol. Here, the magnification in area

$= \frac{6.25 mm ^{2}}{1 mm ^{2}} = 6.25$

∴ Linear magnification,

$m = 6.25 = 2.5$

$m = \frac{v}{u}$

∴

$v = mu = 2.5 u$

Now

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{2.5 u} - \frac{1}{u} = \frac{1}{10}$

or or

$\frac{1 - 2.5}{2.5 u} = \frac{1}{10} u = \frac{( - 1.5 ) \times 10}{2.5} = - 6 cm$

Hence $v = 2.5 u = 2.5 \times (- 6) = - 15 cm$ As the virtual image is closer than the normal near point (25 cm), it cannot be seen by the eye distinctly.

Answer the following questions: (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass, one usually position one's eye very close to the lens. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. what then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal length? (e) When viewing through a compound microscope, our eyes should be positioned not on the eye-piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eye-piece? Sol. (a) It is true that the angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. When a magnifying glass is not used, an object has to be placed at a distance of 25 cm. But the use of magnifying glass allows us to place the object much closer to the eye than at 25 cm. The closer object has larger angular size than the same object at 25 cm. It is in this sense that a magnifying lens produces angular magnification.

(b) Yes, the angular magnification decreases slightly if the eye is moved back. This is because angle subtended at the eye would be slightly less than the angle subtended at the lens. The effect is negligible when image is at much larger distance. (c) First, grinding lenses of very small focal lengths is not easy. More important, if we decrease focal length, both spherical and chromatic aberrations become large. So in practice we cannot get a magnifying power of more than 3 or so with a simple convex lens. However using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) The magnifying power of a compound microscope is given by

$m = m_{o} \times m_{e} = \frac{v _{o}}{u _{o}} \times (1 + \frac{D}{f _{e}}) = \frac{f _{o}}{u _{o} - f _{o}} \times (1 + \frac{D}{f _{e}})$

Angular magnification $(m_{0})$ of objective will be large when $u_{o}$ is slightly greater than $f_{o}$ . Since microscope is used for viewing very close objects, so $u_{o}$ is small. consequently $f_{o}$ has to be small. Moreover, the angular magnification $(m_{e})$ of the eyepiece will be large so $f_{e}$ has to be small. (e) When we place our eyes too close to the eyepiece of compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially.

Hence, the clarity of the image get blurred. The best position of the eye for viewing through a compound microscope is at the eye ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope? Sol. We assume the microscope in common usage, i.e., the final image is formed at the least distance of distinct vision,

$D = 25 cm, f_{e} = 5 cm$

∴ Angular magnification of the eyepiece is

$m_{e} = 1 + \frac{D}{f _{e}} = 1 + \frac{25}{5} = 6$

As total magnification, $m = m_{e} \times m_{0}$ ∴ Angular magnification of the eye piece is

$m_{0} = \frac{m}{m _{e}} = \frac{30}{6} = 5$

As real image is formed by the objective, therefore,

$m_{0} = \frac{v _{o}}{u _{0}} = - 5 or v_{0} = - 5 u_{0} f_{0} = 1.25 cm$

Now, $\frac{1}{v _{0}} - \frac{1}{u _{0}} = \frac{1}{f _{0}}$ or

$\frac{1}{- 5 u _{0}} - \frac{1}{u _{0}} = \frac{1}{1.25}$

$\frac{- 6}{5 u _{0}} = \frac{1}{1.25}$

$u_{0} = - \frac{6 \times 1.25}{5} = - 1.5 cm$

Thus the object should be held at 1.5 cm in front of the objective lens. Also

$v_{0} = - 5 u_{0} = - 5 \times (- 1.5) = 7.5 cm$

As $\frac{1}{v _{e}} - \frac{1}{u _{e}} = \frac{1}{f _{e}}$ ∴

$\frac{1}{u _{e}} = \frac{1}{v _{e}} - \frac{1}{f _{e}} = \frac{1}{- 25} - \frac{1}{5} [v_{e} = - D = - 25 cm] = \frac{- 1 - 5}{25} = - \frac{6}{25} or u_{e} = \frac{- 25}{6} = - 4.17 cm$

∴ Separation between the objective and the eyepiece

$= ∣ u_{e} ∣ + ∣ v_{0} ∣ = 4.17 + 7.5 = 11.67 cm$

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity) ? (b) the final image is formed at the least distance of distinct vision $(25 cm)$ ?

Sol. (a) $M = \frac{- f _{0}}{f _{e}} = \frac{- 140}{5} = - 28$

(b)

$M = \frac{- f _{0}}{f _{e}} (1 + \frac{f _{e}}{D}) = \frac{- 140}{5} (1 + \frac{5}{25}) M = - 33.6$

(a) For the telescope described in Question 27 (a), what is the separation between the objective lens and the eyepiece?

(b) 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? (c) What is the height of the final image of the tower if it is formed at 25 cm?

Sol. (a)

$f_{0} = 140 cm f_{e} = 5 cm L = f_{0} + f_{e} = 140 + 5 = 145 cm$

(b) $tan θ = \frac{h}{b}$ , If $θ$ is very small then $tan θ ≃ θ$

Refracting telescope diagram showing the objective lens and eyepiece positions, with image formation of a distant tower.

$θ = \frac{100}{3000} = \frac{1}{30} radian$

angle subtended by image

$= \frac{y}{f _{0}} = \frac{y}{140}$

from (i) and (ii)

$\frac{1}{30} = \frac{y}{140}$

$\Rightarrow y = \frac{14}{3} cm = 4.7 cm$

$h_{1} = \frac{14}{3} cm m = \frac{h _{2}}{h _{1}} h_{2} = mh_{1} = 6 \times \frac{14}{3} = 28 cm$

A cassegrain telescope uses two mirrors as shown in fig. such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm , where will the final image of an object at infinity be ?

Cassegrain telescope diagram showing the objective mirror, secondary mirror, and eyepiece arranged for image formation.

Sol. The image formed by the larger (concave) mirror acts as a virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at 110 mm from the larger mirror. The distance of the virtual object for the smaller mirror $= 110 - 20 = 90 mm$ . For the small convex mirror, we have

$u = 90 mm, f = 70 mm, v = ?$

Using mirror formula,

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{70} - \frac{1}{90} = \frac{1}{315}$

$∴ v = 315 mm$ Thus the image is formed at 315 mm from the smaller mirror.

Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Galvanometer mirror diagram showing a normally incident light ray and the reflected spot displaced on a distant screen.

Sol.

Diagram showing points A, B, and S with a displacement d marked between the relevant positions.

$tan 2 θ = \frac{d}{AO}; tan 7^{\circ} = \frac{d}{1.5} d = 1.5 tan 7^{\circ} = 1.5 \times 0.1228 = 0.1842 m d = 18.42 cm$

Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid ?

Equiconvex lens placed over a liquid layer on a plane mirror, showing a needle on the principal axis and its inverted image.

Sol. $f = 45 cm, n = 1.5$ For lens $L \to \frac{1}{f _{1}} = (n - 1) (\frac{1}{R} - \frac{1}{- R})$

$∵ \Rightarrow \Rightarrow \Rightarrow \frac{1}{f _{1}} = (n - 1) \times \frac{2}{R} \frac{1}{f _{1}} = (1.5 - 1) \times \frac{2}{R} R \Rightarrow f_{1} = 30 cm \frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} \frac{1}{45} = \frac{1}{f _{1}} + \frac{1}{f _{2}} \frac{1}{45} = \frac{1}{30} + \frac{1}{f _{2}} \frac{1}{45} - \frac{1}{30} = \frac{1}{f _{2}} \frac{2 - 3}{90} = \frac{1}{f _{2}} \Rightarrow \frac{- 1}{90} = \frac{1}{f _{2}} f_{2} = - 90 cm$

for plano convex liquid lens

$\frac{1}{f _{2}} = (n_{2} - 1) (\frac{1}{R} - \frac{1}{\infty}) \frac{1}{90} = (n_{2} - 1) \times \frac{1}{30} n_{2} = 1.33$

Hence R.I. of liquid is 1.33.

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Get NCERT Solutions for Class 12 Physics chapter-wise, covering exercise answers, formulas, and detailed solutions to help students understand topics and practise questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 9

Clear Ray Diagrams: The solutions use clear ray diagrams for mirrors, lenses, prisms, and optical instruments to help students understand image formation and the path of light.
Correct Use of Cartesian Sign Convention: The solutions explain the New Cartesian Sign Convention clearly, helping students use the correct signs while solving mirror and lens formula questions.
Step-by-Step Derivations: Important derivations, including the lens maker’s formula and prism-related formulas, are explained in simple steps so that students can follow each calculation.
Clear Understanding of Optical Instruments: The solutions explain normal adjustment and the least distance of distinct vision for microscopes and telescopes, making it easier to understand their magnifying power.
Real-Life Applications of Ray Optics: The solutions connect concepts with familiar examples such as mirages, myopia, and hypermetropia to help students understand how ray optics is used in everyday life.

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NCERT Solutions Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments

1.0Class 12 Physics Chapter 9 Ray Optics and Optical Instruments: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 9 Ray Optics and Optical Instruments : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTION WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 9

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