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

Question 9.1

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?

1. Given Data

Object height, $h = 2.5 \text{ cm}$
Object distance, $u = -27 \text{ cm}$ (sign convention)
Radius of curvature, $R = -36 \text{ cm}$
Focal length, $f = R/2 = -18 \text{ cm}$

2. Image Distance Calculation

Using Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$$\frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{-18} – \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27}$$ $$\frac{1}{v} = \frac{-3 + 2}{54} = \frac{-1}{54} \implies v = -54 \text{ cm}$$

The screen should be placed **54 cm** from the mirror on the same side as the object.

3. Nature and Size

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

$$m = -\frac{-54}{-27} = -2$$ $$h’ = m \times h = -2 \times 2.5 = -5.0 \text{ cm}$$

The image is **Real, Inverted, and Magnified (5 cm high)**.

4. Moving Object

If the candle is moved closer to the mirror (between f and 2f), the image moves away (beyond 2f). If moved closer than the focus ($<18$ cm), the image becomes virtual and cannot be obtained on a screen.

Question 9.2

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.

1. Given Data

$h = 4.5 \text{ cm}$, $u = -12 \text{ cm}$, $f = +15 \text{ cm}$ (Convex mirror).

2. Image Location

$$\frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{15} – \frac{1}{-12} = \frac{1}{15} + \frac{1}{12}$$ $$\frac{1}{v} = \frac{4 + 5}{60} = \frac{9}{60} \implies v = \frac{60}{9} \approx 6.7 \text{ cm}$$

Image forms **6.7 cm behind the mirror**.

3. Magnification

$$m = -\frac{v}{u} = -\frac{6.7}{-12} = +0.56$$ $$h’ = m \times h = 0.56 \times 4.5 = 2.5 \text{ cm}$$

Image: Virtual, Erect, Diminished (2.5 cm). Moving needle away moves image towards focus (F).

Question 9.3

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?

1. Refractive Index Calculation

Real Depth $d_{real} = 12.5 \text{ cm}$, Apparent Depth $d_{app} = 9.4 \text{ cm}$.

$$\mu = \frac{d_{real}}{d_{app}} = \frac{12.5}{9.4} \approx 1.33$$

2. New Shift Calculation

New Refractive Index $\mu’ = 1.63$.

$$\text{New } d’_{app} = \frac{12.5}{1.63} \approx 7.67 \text{ cm}$$ $$\text{Shift Required} = d_{app} – d’_{app} = 9.4 – 7.67 = 1.73 \text{ cm}$$

Microscope must be moved UP by 1.73 cm.

Question 9.4

Figures 9.27(a) and (b) show refraction of a ray in air incident at $60^\circ$ 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 $45^\circ$ with the normal to a water-glass interface.

1. Calculate Indices from Air

From (a): $i=60^\circ, r=35^\circ \implies \mu_g = \frac{\sin 60}{\sin 35} \approx 1.51$.
From (b): $i=60^\circ, r=47^\circ \implies \mu_w = \frac{\sin 60}{\sin 47} \approx 1.32$.

2. Water to Glass Refraction

$i = 45^\circ$. Applying Snell’s Law: $\mu_w \sin i = \mu_g \sin r$.

$$\sin r = \frac{\mu_w}{\mu_g} \sin 45^\circ = \frac{1.32}{1.51} \times 0.707 \approx 0.618$$ $$r = \sin^{-1}(0.618) \approx 38.2^\circ$$

Angle of Refraction $\approx 38.2^\circ$.

Question 9.5

A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. 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.

1. Concept

Light emerges only within a cone of semi-vertical angle equal to the critical angle $C$, where $\sin C = 1/\mu$.

2. Radius of Emergence

$$R = \frac{h}{\sqrt{\mu^2 – 1}} = \frac{0.80}{\sqrt{(1.33)^2 – 1}} = \frac{0.80}{\sqrt{1.7689 – 1}} \approx 0.91 \text{ m}$$

3. Area Calculation

$$\text{Area} = \pi R^2 = 3.14 \times (0.91)^2 \approx 2.6 \text{ m}^2$$

Emergent Area $\approx 2.6 \text{ m}^2$.

Question 9.6

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 $40^\circ$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^\circ$. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

1. Refractive Index of Prism

$$\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)} = \frac{\sin((60+40)/2)}{\sin(60/2)} = \frac{\sin 50^\circ}{\sin 30^\circ} = \frac{0.766}{0.5} = 1.532$$

2. In Water

Relative index $\mu’ = \mu_g / \mu_w = 1.532 / 1.33 \approx 1.15$.

$$\mu’ = \frac{\sin((A+\delta’_m)/2)}{\sin(A/2)} \implies 1.15 = \frac{\sin((60+\delta’_m)/2)}{0.5}$$ $$\sin(30 + \delta’_m/2) = 0.575 \implies 30 + \delta’_m/2 \approx 35.1^\circ$$ $$\delta’_m = 2(35.1 – 30) = 10.2^\circ$$

New deviation $\approx 10.2^\circ$.

Question 9.7

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?

Lens Maker’s Formula

For double convex lens, $R_1 = +R, R_2 = -R$.

$$\frac{1}{f} = (\mu – 1) \left( \frac{1}{R_1} – \frac{1}{R_2} \right) = (\mu – 1) \frac{2}{R}$$ $$\frac{1}{20} = (1.55 – 1) \frac{2}{R} = \frac{1.1}{R}$$ $$R = 20 \times 1.1 = 22 \text{ cm}$$

Radius of Curvature = 22 cm.

Question 9.8

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?

Concept

Point P acts as a **Virtual Object**. So, $u = +12 \text{ cm}$.

(a) Convex Lens (f = +20)

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{12} = \frac{3+5}{60} = \frac{8}{60}$$ $$v = 7.5 \text{ cm}$$

(b) Concave Lens (f = -16)

$$\frac{1}{v} = \frac{1}{-16} + \frac{1}{12} = \frac{-3+4}{48} = \frac{1}{48}$$ $$v = 48 \text{ cm}$$

(a) 7.5 cm, (b) 48 cm.

Question 9.9

An object of size 3.0 cm is placed 14 cm 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?

1. Calculation

$u = -14 \text{ cm}, f = -21 \text{ cm}$.

$$\frac{1}{v} = \frac{1}{-21} + \frac{1}{-14} = \frac{-2-3}{42} = \frac{-5}{42}$$ $$v = -8.4 \text{ cm}$$ $$m = \frac{v}{u} = \frac{-8.4}{-14} = 0.6$$ $$h’ = 0.6 \times 3.0 = 1.8 \text{ cm}$$

2. Description

Image is **Virtual, Erect, Diminished**, located 8.4 cm from lens. As object moves away, image moves towards focus and becomes smaller.

Question 9.10

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?

Calculation

$f_1 = +30 \text{ cm}, f_2 = -20 \text{ cm}$.

$$\frac{1}{F} = \frac{1}{30} + \frac{1}{-20} = \frac{2-3}{60} = -\frac{1}{60}$$ $$F = -60 \text{ cm}$$

System is **Diverging** with f = -60 cm.

Question 9.11

A compound microscope consists of an objective lens of focal length 2.0 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?

(a) Final Image at D (25 cm)

Eyepiece: $v_e = -25 \text{ cm}$. Using lens formula, find $u_e = -5 \text{ cm}$.
Objective image distance $v_o = L – |u_e| = 15 – 5 = 10 \text{ cm}$.
Using lens formula for objective ($f_o=2$):

$$\frac{1}{u_o} = \frac{1}{10} – \frac{1}{2} = \frac{1-5}{10} = -0.4 \implies u_o = -2.5 \text{ cm}$$

(b) Final Image at Infinity

Eyepiece: $u_e = f_e = 6.25 \text{ cm}$.
Objective image distance $v_o = 15 – 6.25 = 8.75 \text{ cm}$.
Using lens formula for objective:

$$\frac{1}{u_o} = \frac{1}{8.75} – \frac{1}{2} \implies u_o \approx -2.59 \text{ cm}$$

Object Distances: (a) 2.5 cm, (b) 2.59 cm.

Question 9.12

A person with a normal near point (25 cm) using a compound microscope with 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? Calculate the magnifying power.

1. Objective Analysis

$u_o = -0.9 \text{ cm}, f_o = 0.8 \text{ cm}$.

$$\frac{1}{v_o} = \frac{1}{0.8} – \frac{1}{0.9} = \frac{9-8}{7.2} \implies v_o = 7.2 \text{ cm}$$

2. Eyepiece Analysis

Final image at $v_e = -25 \text{ cm}$, $f_e = 2.5 \text{ cm}$.

$$\frac{1}{u_e} = \frac{1}{-25} – \frac{1}{2.5} = \frac{-1-10}{25} \implies u_e \approx -2.27 \text{ cm}$$

3. Results

Separation $L = |v_o| + |u_e| = 7.2 + 2.27 = 9.47 \text{ cm}$.
Magnification $M = \frac{v_o}{u_o} (1 + \frac{D}{f_e}) = \frac{7.2}{0.9} (1 + \frac{25}{2.5}) = 8 \times 11 = 88$.

Question 9.13

A small telescope has an objective lens of focal length 144 cm 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?

Formulas (Normal Adjustment)

$$M = \frac{f_o}{f_e} = \frac{144}{6} = 24$$ $$L = f_o + f_e = 144 + 6 = 150 \text{ cm}$$

Magnification = 24, Separation = 150 cm.

Question 9.14

(a) Giant telescope: $f_o=15 \text{ m}, f_e=1.0 \text{ cm}$. Angular magnification? (b) Diameter of moon image if moon diameter $3.48 \times 10^6 \text{ m}$, orbit $3.8 \times 10^8 \text{ m}$?

(a) Magnification

$$M = \frac{f_o}{f_e} = \frac{1500 \text{ cm}}{1 \text{ cm}} = 1500$$

(b) Image Diameter

Angle subtended by moon $\alpha = \frac{\text{Dia}_{moon}}{\text{Orbit}} = \frac{3.48 \times 10^6}{3.8 \times 10^8} \approx 0.0091 \text{ rad}$.
Image diameter $d = f_o \alpha = 15 \text{ m} \times 0.0091 \approx 0.137 \text{ m} = 13.7 \text{ cm}$.

Question 9.15

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… (and other cases).

Deduction (a)

Mirror formula: $1/v = 1/f – 1/u$. For concave mirror, $f<0$, $u<0$.
If $f < u < 2f$ (numerically), then $1/f < 1/u < 1/2f$.
Substituting signs: $-1/f < -1/u < -1/2f$.
Solving shows $v$ is negative and $|v| > 2f$. Thus, image is Real and Beyond 2f.

Question 9.16

A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5.

Shift Formula

$$\text{Normal Shift} = t \left( 1 – \frac{1}{\mu} \right) = 15 \left( 1 – \frac{1}{1.5} \right)$$ $$\text{Shift} = 15 \left( 1 – \frac{2}{3} \right) = 15 \times \frac{1}{3} = 5 \text{ cm}$$

Shift = 5 cm. Independent of slab location.

Question 9.17

Figure 9.28 shows 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 angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place?

1. Critical Angle

$$\sin C = \frac{\mu_2}{\mu_1} = \frac{1.44}{1.68} \approx 0.857 \implies C \approx 59^\circ$$

For TIR, angle of incidence at interface $i > C$, i.e., $i > 59^\circ$.

2. Input Angle Range

Using geometry and Snell’s law at input face, maximum acceptance angle $i_{max} \approx 60^\circ$.

Question 9.18

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Condition for Real Image

For a real image to be formed on a screen, distance between object and screen $D \ge 4f$.

$$f_{max} = \frac{D}{4} = \frac{3}{4} = 0.75 \text{ m}$$

Max Focal Length = 0.75 m.

Question 9.19

A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens.

Displacement Method Formula

$D = 90 \text{ cm}$, $d = 20 \text{ cm}$.

$$f = \frac{D^2 – d^2}{4D} = \frac{90^2 – 20^2}{4 \times 90} = \frac{8100 – 400}{360}$$ $$f = \frac{7700}{360} \approx 21.4 \text{ cm}$$

Focal Length $\approx 21.4 \text{ cm}$.

Question 9.20

(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. 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.

(a) Effective Focal Length

Incident parallel beam ($u_1=\infty$) on Convex Lens ($f_1=30$). Image at $v_1 = 30$.
This acts as object for Concave Lens ($f_2=-20$) at distance 8 cm away. $u_2 = 30 – 8 = +22$.
Using lens formula: $1/v_2 = 1/-20 + 1/22 \implies v_2 \approx -220 \text{ cm}$.
Effective focal length $\approx 220 \text{ cm}$.

(b) Magnification

$u_1 = -40$. $1/v_1 = 1/30 – 1/40 \implies v_1 = 120$. Mag $m_1 = 120/-40 = -3$.
Object for lens 2: $u_2 = 120 – 8 = 112$.
$1/v_2 = 1/-20 + 1/112 \implies v_2 \approx -24.3$. Mag $m_2 = -24.3/112 \approx -0.217$.
Total Mag $M = m_1 \times m_2 = -3 \times -0.217 \approx 0.65$.
Image Size $h’ = 0.65 \times 1.5 \approx 0.98 \text{ cm}$.

Question 9.21

At what angle should a ray of light be incident on the face of a prism of refracting angle $60^\circ$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

1. Critical Angle

$$\sin C = \frac{1}{1.524} \approx 0.656 \implies C \approx 41^\circ$$

2. Refraction Angles

For grazing emergence, $r_2 = C = 41^\circ$.
$r_1 = A – r_2 = 60 – 41 = 19^\circ$.

3. Incident Angle

$$\sin i = \mu \sin r_1 = 1.524 \times \sin 19^\circ \approx 1.524 \times 0.325 \approx 0.496$$ $$i = \sin^{-1}(0.496) \approx 29.75^\circ$$

Incidence Angle $\approx 30^\circ$.

Question 9.22

A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. (a) Magnification produced? Area of virtual image? (b) Angular magnification? (c) Are they equal?

(a) Linear Magnification

$u = -9 \text{ cm}, f = 10 \text{ cm}$ (Assuming f=10 for standard calculation, or if f=9 then u=-9 means image at infinity). Text says f=9, u=9. This places object at focus. Image at infinity. Linear magnification is technically infinite/undefined.

Correction: Let’s assume standard Near Point viewing. If $v=-25$, then $1/u = 1/-25 – 1/9 \implies u \approx -6.6$. Then $m = v/u \approx 3.8$.

If viewed at $u=-9$ (at focus), image is at $\infty$. Angular magnification is relevant here.

(b) Angular Magnification

$$M = \frac{D}{f} = \frac{25}{9} \approx 2.8$$

(c) Equality

They are not equal unless image is at near point (where $M = 1 + D/f$).

Question 9.23

(a) At what distance should the lens be held from the card sheet in Exercise 9.22 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?

(a) Distance for Max Power

Image at near point $v = -25 \text{ cm}$. $f = 10 \text{ cm}$ (or 9 from previous? let’s stick to 9).
$\frac{1}{u} = \frac{1}{-25} – \frac{1}{9} = \frac{-9-25}{225} = \frac{-34}{225} \implies u \approx -6.6 \text{ cm}$.

(b) Magnification

$m = \frac{v}{u} = \frac{-25}{-6.6} \approx 3.8$.

(c) Equality

Yes, for image at near point, Linear Magnification = Angular Magnification.

Question 9.24

What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25 \text{ mm}^2$?

Calculation

Area magnification $A’ = 6.25 \implies$ Linear mag $m = \sqrt{6.25} = 2.5$.
$v = 2.5 u$ (Virtual image, v is negative, u is negative, ratio positive).
Lens formula: $1/v – 1/u = 1/f \implies 1/2.5u – 1/u = 1/9$.
$1/u (0.4 – 1) = 1/9 \implies -0.6/u = 1/9 \implies u = -5.4 \text{ cm}$.

Distance = 5.4 cm.

Question 9.25

Answer the following questions: (a) In what sense does a magnifying glass provide angular magnification? (b) Does angular magnification change if the eye is moved back?

(a) It allows the object to be placed closer than the near point (25 cm), thus increasing the angle subtended at the eye.

(b) Yes, it changes slightly because the angle subtended at the eye depends on the distance from the lens.

Question 9.26

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?

Setup Calculation

Assume normal adjustment (Image at D=25cm).
$M_e = 1 + D/f_e = 1 + 25/5 = 6$.
Total $M = M_o \times M_e \implies 30 = M_o \times 6 \implies M_o = 5$.
Since $M_o \approx v_o/u_o$ (ignoring sign), let $v_o = 5u_o$.
Using lens formula for objective: $1/5u_o – 1/-u_o = 1/1.25 \implies u_o = 1.5 \text{ cm}$, $v_o = 7.5 \text{ cm}$.
For eyepiece, $u_e$ for image at D is $u_e = -4.17 \text{ cm}$.
Separation $L = |v_o| + |u_e| = 7.5 + 4.17 = 11.67 \text{ cm}$.

Object at 1.5 cm. Separation 11.67 cm.

Question 9.27

A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when (a) normal adjustment, (b) image at 25cm?

Formulas

(a) $M = f_o/f_e = 140/5 = 28$.

(b) $M = \frac{f_o}{f_e}(1 + f_e/D) = 28(1 + 5/25) = 28 \times 1.2 = 33.6$.

Question 9.28

(a) For telescope in Q9.27, separation? (b) Height of tower image?

(a) Separation $L = f_o + f_e = 145 \text{ cm}$.

(b) $\alpha = 100/3000 = 1/30$ rad. Image height $h = f_o \alpha = 140 \times 1/30 \approx 4.7 \text{ cm}$.

Question 9.29

A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?

Calculation

Primary (Concave, $R_1=220$): Image at $v_1 = -110 \text{ mm}$.
Secondary (Convex, $R_2=140, f_2=70$) is 20mm away.
Object distance for secondary $u_2 = 110 – 20 = +90 \text{ mm}$ (Virtual Object).
Mirror Formula: $1/v_2 + 1/90 = 1/70 \implies 1/v_2 = 1/70 – 1/90 = 2/630$.
$v_2 = 315 \text{ mm} = 31.5 \text{ cm}$.

Final image: 31.5 cm from secondary mirror.

Question 9.30

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

Concept

If mirror rotates by $\theta$, reflected ray rotates by $2\theta$.

$$d = D \tan(2\theta) = 1.5 \times \tan(7^\circ) \approx 1.5 \times 0.1228 = 0.184 \text{ m} = 18.4 \text{ cm}$$

Displacement = 18.4 cm.

Question 9.31

Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror… Distance changes from 45cm to 30cm. Refractive index of liquid?

1. Focal Lengths

Without liquid: $f_g = 30 \text{ cm}$. Since $\mu=1.5$, $R = f_g = 30 \text{ cm}$.
With liquid: Combined $F = 45 \text{ cm}$.

2. Liquid Lens

$1/F = 1/f_g + 1/f_l \implies 1/45 = 1/30 + 1/f_l \implies 1/f_l = -1/90$.
Liquid lens is plano-concave ($R_1=-30, R_2=\infty$).
$1/f_l = (\mu_l – 1)(-1/30) \implies -1/90 = -(\mu_l – 1)/30$.
$1/3 = \mu_l – 1 \implies \mu_l = 1.33$.

Refractive Index of Liquid = 1.33