**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?**

**Sol.** Given, radius of curvature, R = – 36 m

$$\therefore\space\text{Focal length f =}\frac{\text{R}}{2}=\frac{36}{2}=-18\space\text{cm}$$

Height of object O = 2.5 cm

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\Rarr\space=-\frac{1}{18}=-\frac{1}{v}-\frac{1}{27}\\\frac{1}{v}=\frac{1}{18}+\frac{1}{27}\\=\frac{-3+2}{54}=-\frac{1}{54}$$

∴ Distance of screen from mirror v = – 54 cm.

Magnification for mirror

$$m =-\frac{v}{u}=\frac{\text{I}}{\text{Q}}\\=-\frac{(-54)}{-27}=\frac{1}{2.5}$$

I = – 5 cm

The negative sign shows that an inverted image is formed in front of the mirror. Thus, the screen should be placed at a distance 54 cm and image real, inverted and magnified in nature.

If we move the candle near to the mirror the screen should be moved away from mirror. As the object distance is less than focal length and no screen is required, because the image formed is virtual.

**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?**

**Sol.** Given, focal length f = + 15 cm

Distance of object u = – 12cm

Size of the object O = 4.5 cm

Using formula,

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\frac{1}{15}=\frac{1}{v}-\frac{1}{12}\\\Rarr\space\frac{1}{v}=\frac{1}{15}+\frac{1}{12}\\=\frac{4+5}{60}=\frac{9}{60}$$

Distance of image from the mirror v = 6.7 cm.

The image is formed behind the mirror.

The magnification,

$$\text{m}=-\frac{v}{u}=\frac{\text{I}}{\text{O}}\\\frac{-6.7}{-12}=\frac{1}{4.5}$$

Size of image I = 2.5 cm

As I is positive, so image is erect and virtual.

The magnification

$$\text{m}=\frac{\text{I}}{\text{O}}=\frac{2.5}{4.5}=\frac{25}{45}=\frac{5}{9}$$

As the object moves away from the mirror, the image also moves away from the mirror (as u → ∞, v → f) and the size of image will be decreased.

**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?**

**Sol. Case I:** When tank is filled with water

the apparent depth = 9.4 cm

Height of water t = 12.5 cm

So, real depth = 12.5 cm

Refractive index of water

$$\mu_{w}=\frac{\text{Real depth}}{\text{Apparent depth}}\\=\frac{12.5}{9.4}=1.33$$

**Case II:** When tank is filled with the liquid

Refractive index of liquid µ = 1.63,

$$\mu=\frac{\text{Real depth}}{\text{Apparent depth}}\\\therefore\space 1.63=\frac{12.5}{\text{Apparent depth}}$$

∴ The microscope is moved by = 9.4 – 7.67 = 1.73 cm

**4. Figure (a) and (b) show refraction of a 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 in 45° with the normal to a water-glass interface [Fig. (c)].**

**Sol.** Given, from Fig. (a)

Angle of incidence i = 60°

Angle of refraction r = 35°

Refractive index of glass w.r.t. air,

$$^{a}\mu_{g}=\frac{\text{sin}60\degree}{\text{sin}35\degree}=\frac{0.8660}{0.5736}$$

= 1.51 …(i)

From Fig. (b)

Angle of incidence i = 60°,angle of refraction r = 47°

Refractive index of water w.r.t. to air

$$^{a}\mu_{w}=\frac{\text{sin}60\degree}{\text{sin}47\degree}=\frac{0.8660}{0.7314}$$

= 1.18 …(ii)

From Fig. (c)

Angle of incidence i = 45°, let r be the angle of refraction

Refractive index of glass w.r.t. water

$$^{w}\mu_{g}=\frac{\text{sin} \space45\degree}{\text{sin}\space r\degree}\space...\text{(iii)}\\\text{As, we know that}^{w}\mu_{g}=\frac{^{a}\mu_{g}}{^{a}\mu_{w}}$$

Putting the value of ^{w}m_{g} in Eq. (iii), we get

$$\frac{^{a}\mu_{g}}{^{a}\mu_{w}}=\frac{0.7071}{\text{sin r}}\\\frac{1.51}{1.32}=\frac{0.7071}{\text{sin r}}$$

From equations (i) and (ii), w

e get

$$\text{sin r =}\frac{1.32×0.7071}{1.51}=0.6181$$

r = 38.2°

**5. 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.)**

**Sol.** Let the bulb is placed at point O

AB = AC = r

If the light incident at an angle equal to critical angle i_{c}, then a circular area is formed only.

The light source is 80 cm below the water surface i.e. AO = 80 cm, µ_{w} = 1.33

Using the formula for critical angle,

$$\text{sin}\space i_{c}=\frac{1}{\mu_{w}}\\\text{sin}_{i_{c}}=\frac{1}{1.33}=0.75\\i_c = 48.6°\\\text{In} \Delta \text{OAB tan i}_\text{c =}\frac{\text{AB}}{\text{AO}}\\\text{or}\space\text{tan}\space i_{c}=\frac{r}{l}$$

r = l tan i_{c} = 80 tan 48.6

r = 80 × 1.1345 = 90.7 cm

Area of circular surface of w

ater

A = πr^{2}

A = 3.14 × (90.7)^{2} = 25865.36 cm^{2}

A = 2.58 m^{2}

**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°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. 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.** Given, angle of minimum deviation δ_{m} = 40°

The angle of the prism A = 60°

Refractive index of glass w.r.t. air

$$^{a}\mu_{g}=\frac{\text{sin}\bigg(\frac{\text{A}+\delta_{m}}{2}\bigg)}{\text{sin}\frac{\text{A}}{2}}\\=\frac{\text{sin}\bigg(\frac{60\degree+40\degree}{2}\bigg)}{\text{sin 30\degree}}=\frac{\text{sin} 50\degree}{\text{sin} 30\degree}\\\\^{a}\mu_{g}=\frac{0.766}{0.5}=1.532$$

When prism is immersed in water

The refractive index of water w.r.t. air

^{a}m_{w} = 1.33

Refractive index of glass with respect to water

$$^{a}\mu_{g}=\frac{\text{sin}\bigg(\frac{\text{A +}\delta'_{m}}{2}\bigg)}{\text{sin}\frac{\text{A}}{2}}\\\therefore\space\frac{^{a}\mu_{g}}{^{a}\mu_{w}}=\frac{\text{sin}\bigg(\frac{\text{A+}\delta'm}{2}\bigg)}{\text{sin} 30\degree}\\\text{sin}\bigg(\frac{A+\delta'_m}{2}\bigg)=\frac{1.532×\text{sin} 30\degree}{1.33}\\\text{sin}\bigg(\frac{A + \delta'_m}{2}\bigg)=\frac{0.5×1.532}{1.33}=0.5759\\\text{sin}\bigg(\frac{\text{A + }\delta '_m}{2}\bigg)=\text{sin} 35\degree 10'$$

δ’_{m} = 2 (30°10’) – 60°

= 70°20’ – 60°

δ’_{m} = 10°20’

Thus, the angle of minimum deviation become 10°20’.

**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?**

**Sol.** For double convex lenses R_{1} = R, R_{2} = – R

Focal length f = + 20 cm

By Lens maker’s formula

$$\frac{1}{f}=(^{a}\mu_{g}-1)\bigg(\frac{1}{\text{R}}_{1}-\frac{1}{\text{R}}_{2}\bigg)\\\frac{1}{20}=(1.55-1)\bigg(\frac{1}{\text{R}}+\frac{1}{\text{R}}\bigg)\\\frac{1}{20}=0.55×\frac{2}{\text{R}}$$

R = 0.55 × 2 × 20 = 22 cm

Thus, the radius of curvature is 22 cm.

**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?**

**Sol.** Here, the point P is on the right side of lens acts as vertical object.

(a) Given, object distance

u = 12 cm

Focal length f = + 20 cm

$$\text{By Lens formula,}\space\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{v}-\frac{1}{12}=\frac{1}{20}\\\frac{1}{v}=\frac{1}{20} + \frac{1}{12}\\=\frac{3+5}{60}\\=\frac{8}{60}$$

v = 7.5 cm

Thus, the beam converges at a distance of 7.5 cm on the right side of lens.

(b) Distance of object from the lens u = 12 cm

Focal length t = – 16 cm

Using Lens formula,

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{v}-\frac{1}{12}=-\frac{1}{16}\\\frac{1}{v}=\frac{1}{12}-\frac{1}{16}=\frac{4-3}{48}$$

v = 48 cm

So, the beam converges on the right side of lens at a distance of 48 cm.

**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?**

**Sol.** Object size, O = 3 cm

Focal length f = – 21 cm

Distance of object from the concav

e lens u = – 14cm)

By Lens formula,

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{v} + \frac{1}{14}=-\frac{1}{12}\\\Rarr\space\frac{1}{v}=-\frac{1}{21}-\frac{1}{14}\\=\frac{-2-3}{42}\\=-\frac{5}{42}$$

v = – 8.4 cm

The magnification

$$\text{m}=-\frac{v}{u}=\frac{\text{I}}{\text{O}}\\\frac{(-8.4)}{-14}=\frac{\text{I}}{3}$$

I = 1.8 cm

Since, I is positive, the image formed is virtual and erect at a distance of 8.4 cm in front of lens. If the object moves further away from the lens, the image moves towards the lens. The size of image decreases gradually.

**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? Ignore thickness of the lenses.**

**Sol.** Here, focal length of convex lens f_{1} = 30 cm.

Focal length of concave lens f_{2} = – 20 cm

We know that,

$$\frac{1}{f}=\frac{1}{f}_{1} + \frac{1}{f}_{2}\\= \frac{1}{30}-\frac{1}{20}\\=\frac{2-3}{60}=-\frac{1}{60}$$

f = – 60 cm

Since, the focal length of combination is negative. Hence, combination behaves like a diverging lens i.e. as a concave lens.

**11. A compound microscope consists of an objective lens of focal length 2.0 cm and an eye-piece 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 (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.** Given, focal length of objective lens, fo = 2 cm

Focal length of eye-piece fe = 6.25 cm

Distance between both these lenses

v = 15 cm

(a) Distance of final image from eye-piece

ve = – 25 cm

Using the Lens formula for eye-piece

$$\frac{1}{V_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}\\\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\\=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$$

u_{e} = – 5 cm

As the distance between objective and eye-piece (v_{o} + u_{e}) = 15 cm

L = v_{o} + u_{e}

Distance of image formed by objective lens

v_{o} = L – |u_{e}| = 15 – 5 = 10 cm

For objective lens

$$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}\\\frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}\\=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=-\frac{4}{10}$$

u_{o} = – 2.5 cm

So, the object should at distance 2.5 cm in front of convex lens.

Magnifying power of compound microscope

$$m=\frac{v_{0}}{u_{0}}\bigg(1+\frac{d}{f_{e}}\bigg)\\=\frac{10}{2.5}\bigg(1+\frac{25}{6.25}\bigg)\space(\because d=25\text{cm})$$

m = 20

(b) The final image will be formed at infinity only if the image formed by the objective is in the focal plane of the eye-piece. Thus, here v_{e} = – ∞, u_{e} = f_{e} = 6.25 cm

Image distance of objective lens

v_{o} = L – f_{e} = 15 – 6.25 = 8.75 cm

Using Lens formula of objective lens

$$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}\\\frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}\\=\frac{1}{8.75}-\frac{1}{2}=\frac{2-8.75}{17.5}\\u_{0}=-\frac{17.5}{6.75}=-2.59\space\text{cm}$$

Magnifying power of the microscope

$$m=\frac{v_{0}}{u_{0}}\bigg(1+\frac{d}{f_{e}}\bigg)\\=\frac{8.75}{2.59}\bigg(1+\frac{25}{6.25}\bigg)$$

m = 13.51

**12. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eye-piece 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 of the microscope.**

**Sol.** Given, focal length of objective f_{0} = 8 mm = 0.8 cm

Focal length of eye-piece f_{o} = 2.5 cm

Distance of objective lens – u_{o} = – 9 mm = – 0.9 cm

Distance of image from eye-piece

v_{e} = – d = – 25 cm

By Lens equation for eye-piece

$$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}\\\text{or}\space\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\\=-\frac{1}{25}-\frac{1}{2.5}=\frac{-1-10}{25}=-\frac{11}{25}$$

u_{e} = – 2.27 cm

By Lens equation for objective

$$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}\\\text{or}\space\space\frac{1}{v_{0}}=\frac{1}{f_{0}} + \frac{1}{u_{0}}\\=\frac{1}{0.8}-\frac{1}{0.9}\\\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}$$

Distance of image for objective lens

v_{o} = 7.2 cm

Separation between two lenses

L = |u_{e}| + |v_{o}|

= 2.27 + 7.2

L = 9.47 cm

Magnifying power of compound microscope

$$m=\frac{v_{0}}{u_{0}}\bigg(1+\frac{d}{f_{e}}\bigg)=\frac{7.2}{0.9}\bigg(1+\frac{25}{2.5}\bigg)$$

m = 88

**13. A small telescope has an objective lens of focal length 144 cm and an eye piece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?**

**Sol.** Given, focal length of objective lens f_{o} = 144 cm

Focal length of eye-piece f_{e }= 6 cm

Magnifying power of the telescope when the final image is formed at ∞.

$$m=\frac{f_{0}}{f_{e}}=-\frac{144}{6}=-24$$

∴ Separation between lenses L = f_{o} + f_{e }= 144 + 6

= 150 cm.

**14. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye-piece 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 × 10 ^{6} m, and the radius of lunar orbit is 3.8 × 10^{8}m.**

**Sol.** Given, focal length of objective lens f_{o} = 15 m

Focal length of eye-piece f_{e} = 1 cm = 0.01 m

(a) Angular magnification

$$m=\frac{f_{0}}{f_{e}}=\frac{15}{0.01}=1500$$

Let d be the diameter of the image of the moon formed by the objective lens.

$$\therefore\space\text{Angle subtended by the image =}\frac{d}{f_{0}}=\frac{d_{i}}{15}$$

(b) Diameter of moon d_{o} = 3.48 × 10^{6} m

Radius of lunar orbit r = 3.8 × 10^{8} m

The angle subtended by the moon

$$=\frac{\text{Diameter of moon}}{\text{Radius of lunar orbit}}\\=\frac{3.48×10^{6}}{3.8×10^{8}}$$

But we know that the angle subtended by the image is equal to the angle subtended by the object.

$$\therefore\space\frac{d_{i}}{15}=\frac{3.48×10^{6}}{3.8×10^{8}}\\\text{or}\space d_{i}=\frac{3.48×15×10^{\normalsize-2}}{3.8}$$

= 13.73 × 10^{–2}m

or d_{i} = 13.73 cm

Thus, the diameter of the image is 13.73 cm.

**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.**

**(b) ****a conv****ex 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.**

**[Note: This ****exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]**

$$\textbf{Sol.}\space\text{(a) The mirror formula}\space\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\\\text{According to question,}\\\text{f < u < 2f}\\\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}\text{or}-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}\\\text{Add}\space\frac{1}{f}\space\text{on both sides, we get}\\\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0\space\text{...(i)}\\\frac{1}{f}-\frac{1}{2f}<\frac{1}{v}\space\bigg[\because\frac{1}{f}-\frac{1}{u}=\frac{1}{v}\bigg]\\\frac{1}{2f}<\frac{1}{v}$$

v = 2f

As, f is negative, hence v will also be negative. So, the image formed is real and image lies beyond 2f.

(b) For convex mirror f > 0. As object always placed on left side of mirror i.e, object distance u > 0.

$$\text{Lens formula,}\space\frac{1}{v}=\frac{1}{f}-\frac{1}{v}\text{as f > 0, and u < 0}$$

Thus, v is always positive. The image formed is virtual. It is independent from the location of object.

$$\text{(c) For convex mirror f} > 0, u < 0\space\text{and}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\text{so}\frac{1}{u}>\frac{1}{f}\text{or v} < f $$

Thus, image always located between pole and focus, as v < |u|. So the image is always diminished.

(d) For concave mirror f < 0.

As object is placed between pole and focus.

i.e. f < u < 0

$$\therefore\space\frac{1}{f}-\frac{1}{u}>0\\\text{By Lens formula,}\\\frac{1}{f}-\frac{1}{u}=\frac{1}{v}>0\Rarr v>0$$

It means that v is positive, image formed up right and virtual.

$$\frac{1}{v}<\frac{1}{u}$$

v > |u| so image is enlarged.

**16. 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.** Given, thickness of glass slab (real depth) = 15 cm.

Refractive index of glass ^{o}m_{g} with respect to air = 1.5

$$^o\mu_g=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{15}{^a\mu_g}$$

Apparent depth of pin y = 15/1.5 = 10 cm

Distance by which the pin appears to be raised

= Real depth – Apparent depth

= 15 – 10 = 5 cm

The answer does not depend on the location of the slab.

**17. (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 angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.**

**(b) ****What is the answ****er if there is no outer covering of the pipe?****Sol.** (a) Given, refractive index of the glass fibre with respect to air

µ_{2} = ^{a}µ_{g} = 1.68

Refractive index of the outer material with respect to air

µ_{1} = ^{a}µ_{outer} = 1.44

Let the critical angle be i_{c}.

$$\mu=\frac{\mu_{2}}{\mu_{1}}=\frac{1}{\text{sin} i_{c}}\\\text{sin}\space i_{c}=\frac{\mu_{1}}{\mu_{2}}=\frac{1.44}{1.68}=0.8571$$

i_{c} = 59°

The total internal reflection will take place for the angle of incidence i will be greater than the critical angle i_{c}.

i.e., i > 59° or when angle of refraction, r < r_{max}

where r_{max} = 90° – i_{c} = 90° – 59° = 31°

$$\therefore\space^{a}\mu_{g}=\frac{\text{sin i}_{max}}{\text{sin r}_{max}}=1.68$$

or sin i_{max} = 1.68 sin 31° = 1.68 × 0.5156

i_{max} = sin^{–1 }(0.8662) = 60°

Thus, all the rays 0 < i < 60°, will suffer total internal reflection in the pipe.

(b) If there is no outer covering of the pipe then reflection inside the pipe will take place from the glass to air

$$\text{sin}\space i'_{c}=\frac{\mu_{1}}{\mu_{2}}=\frac{1}{1.68}=0.5952\\\text{Critical angle i’c = 36.5°}\\\text{Now, i = 90, we have r = 36.5°, r = 1.68 =}\frac{\text{sin 90\degree}}{\text{sin r}}$$

So i’ f = 90 – 36.5 = 53.5°

Here, i’ is greater than the critical angle. Thus, all the rays incident at an angles in the range zero to 90° will suffer total internal reflection.

**18. Answer the following questions:**

**(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain. **

**(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?**

**(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?**

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

**(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?**

**Sol.** (a) Yes, plane mirror and convex mirror both produces the real image if the rays incident on the plane or convex mirror are converging to a point behind the mirror. Because they are reflected to a point in front of the mirror. In other words we can say a plane or convex mirror can produce a real image if the object is virtual.

(b) No, there is no contradiction because virtual image formed by mirror acts as virtual object for eye lens. Our eye lens is convergent and it forms a real image of virtual object on retina.

(c) As the rays of light travels from rarer to denser medium, they bends towards the normal. So, the fisherman appears taller.

(d) Yes, the apparent depth will be decrease, further, when water tank is viewed obliquely as compared to the real depth.

$$\text{(e) Refractive index}\spaceµ _{diamond}> µ_{glass} \text{Refractive index µ =}\frac{1}{\text{sin}i_{c}}$$

where, i_{c} is the critical angle.

As the refractive index of diamond is more than that of water, so the critical angle for glass is more than diamond. A diamond cutter, cuts the diamond at large range of angle of incidence to ensure that light entering in the diamond suffers multiple total external reflections and gives the sparkling in diamond.

**19. The image of a small electric bulb fixed on the wall of a rooms 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.** Suppose the object is placed at u in front of the lens and the distance of image from the lens is (3 – u)

i.e., v = (3 – u)n

$$\text{From Lens formula,}\space\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{(3-u)}-\frac{1}{(-u)}=\frac{1}{f}\\\text{or}\space\frac{1}{(3-u)} + \frac{1}{u}=\frac{1}{f}\\\text{or}\space\frac{u+3-u}{u(3-u)}=\frac{1}{f}$$

3f = 3u – u^{2}

u^{2} – 3u + 3f = 0

$$\text{Now,}\space u=\frac{-(-3)\pm\sqrt{9-4×(3f)}}{2}\\u=\frac{+3\sqrt{9-2f}}{2}$$

Condition for image to be formed real

9 – 12f ≥ 0

or

9 ≥ 12f or f ≤ 0.75m

Thus, the maximum possible focal length of the lens required for his purpose is 0.75m.

**20. 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.** Given, distance between screen and object

a = 90 cm

Distance between two positions of the lens

d = 20 cm

By displacement formula

$$f=\frac{a^{2}-d^{2}}{4a}=\frac{(90)^{2}-(20)^{2}}{4×90}\\=\frac{7700}{360}=21.4\space\text{cm}$$

**21. (a) Determine the ‘effective focal length’ of the combination of the two lenses in questions 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination 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 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.**

**Sol.** (a) Given, the focal length of convex lens

f_{1} = 30 cm

The focal length of concave lens

f_{2} = –20 cm, d = 8 cm

Now,

$$\frac{1}{f}=\frac{1}{f_{1}} + \frac{1}{f_{2}}-\frac{d}{f_{1}f_{2}}\\\frac{1}{f}=\frac{1}{30}-\frac{1}{20}-\frac{8}{30×(-20)}\\=\frac{20-30+8}{30×20}\\\frac{1}{f}=-\frac{2}{600}$$

f = – 300 cm

(i) Let the incident beam falls on convex lens and assume that concave lens is not there.

u_{1} = ∞, f_{1} = 30 cm

Using the Lens formula

$$\frac{1}{f_{1}}=\frac{1}{v_{1}}-\frac{1}{u_{1}}\\=\frac{1}{30}=\frac{1}{v_{1}}-\frac{1}{\infty}$$

Position of the image formed by convex lens

v_{1} = 30 cm

This image acts as an object for concave lens.

Now, distance of object.

u_{2} = + (30 – 8) = 22 cm

f_{2} = – 20 cm

Using Lens formula

$$\frac{1}{f_{1}}=\frac{1}{v_{1}} - \frac{1}{u_{1}}\\\frac{1}{30}=\frac{1}{v_{1}}-\frac{1}{\infty}$$

Position of the image formed by convex lens

v_{1} = 30 cm

This image acts as an object for concave lens.

Now, distance of object.

u_{2} = + (30 – 8) = 22 cm

f_{2} = – 20 cm

Using Lens formula

$$\frac{1}{f_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}\\\frac{1}{v_{2}}=-\frac{1}{20}+\frac{1}{22}=-\frac{1}{220}$$

Distance of final image from lens v_{2} = – 220 cm.

Thus, the parallel beam would appear to diverge from a point at a distance 220 – 4 = 216 cm from the centre of two lens system.

(ii) Let us take that the parallel beam first falls on concave lens.

u_{1} = – ∞, f_{1} = –20 cm, v_{1} = ?

$$\frac{1}{v_{1}}+\frac{1}{\infty}=\frac{1}{-20}$$

v_{1} = – 20 cm

This acts as an object for convex lens.

u_{2} = – (20 + 8) = – 28 cm, f_{2} = 30 cm, v_{2} = ?

$$\frac{1}{v_{2}}+\frac{1}{28}=\frac{1}{30}\\\frac{1}{v_{2}}=\frac{1}{30}-\frac{1}{28}\\=\frac{14-15}{420}=-\frac{1}{420}$$

v_{2} = – 420 cm

The beam appears to diverge from a point 420 – 4 (= 416) cm on the left of the centre of the two lens system. From these two cases it concludes that the answer depends on which side of the lens system the parallel beam is incident. So, the notion of effective focal length does not useful here.

(b) Given, size of object O_{1} = 1.5 cm Distance from the convex lens u_{1} = – 40 cm,

$$\text{For the convex lens,}\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\\=\frac{1}{v_{1}}=\frac{1}{30} + \frac{1}{(\normalsize-40)}\\=\frac{4-3}{120}=\frac{1}{120}$$

v_{1} = 120cm

Magnification produced by convex lens

Now, this image acts as an object for concave lens.

∴ For concave lens u_{2} = 120 – 8 = 112 cm, f_{2} = – 20 cm

$$\frac{1}{v_{2}}=\frac{1}{-20} + \frac{1}{112}\\=\frac{112+20}{112×20}=\frac{-92}{112×20}\\\text{v}_{2}=\frac{-112×20}{92}\text{cm}$$

Magnification produced by concave lens

$$\text{m}_{2}=\frac{v_{2}}{u_{2}}=\frac{(-112×20)}{92×112}=\frac{20}{92}\\\text{Magnification produced by combination}\\m = m_1 × m_2\\\text{m = (–3)X}\\=\bigg(-\frac{20}{90}\bigg)=\frac{60}{92}$$

Magnification produced by combination

m = m_{1} × m_{2}

m = (–3)X

$$=\bigg(-\frac{20}{90}\bigg)=\frac{60}{92}$$

m = 0.652

Size of image = m × size of objects

= 0.652 × 1.5 = 0.98 cm

Thus, the magnification produced by the two lens system is 0.652 and size of image is 0.98 cm.

**22. 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 material of the prism is 1.524.**

**Sol.** Angle of prism, A = 60°

Refractive index µ = 1.524

The critical angle is i_{c} because it just suffers total internal refraction.

$$\text{sin}\space i_{c}=\frac{1}{\mu}\\=\frac{1}{1.524}$$

= 0.6561

i_{c} = 41°

For a prism r_{1} + r_{2} = A here r_{2} = i_{c}

∴ r_{1} + i_{c} = A

r_{1} + 41° = 60°

⇒ r_{1} = 19°

$$\text{Now}\space \mu=\frac{\text{sin}i_{1}}{\text{sin}r_{1}}$$

or sin i_{1} = 1.524 sin 19° = 1.524 × 0.3256

or i_{1} = sin^{–1} (0.4962)

i_{1} = 29°75’

So, the angle should be 29°75’.

**23. 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 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) Given, focal length f = 10 cm, u = – 9 cm

Size of object = 1 mm

We know that

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\\frac{1}{10}=\frac{1}{v}+\frac{1}{9}\\\Rarr\space\frac{1}{v}=\frac{1}{10}-\frac{1}{9}=-\frac{1}{90}\\\text{or v = – 90 cm}\\\text{The magnification by the lens}\\m=+\frac{v}{u}=-\frac{(+90)}{-9}=10\\m=\frac{v}{u}=\frac{\text{I}}{\text{O}}$$

I = O × m = 1 × 10 = 10 mm

Area of each square in virtual image = (10)^{2} = 100 mm^{2}

Thus, the magnification is 10 and area of each square in the virtual image is 10 mm^{2}.

$$\text{(b) Angular magnification, m =}\frac{d}{u}=\frac{25}{9}=2.8$$

(c) No, they are equal if v = d.

(where d is the least distance of distinct vision).

**24. (a) At what distance should the lens he held from the figure in question 29 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) Given, v = – 25 cm and f = 10 cm.

For maximum magnifying power the image should be formed at least distant of distinct vision.

By Lens formula

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\\Rarr\space\frac{1}{10}=\frac{1}{-25}-\frac{1}{u}\\=\frac{1}{u}=\frac{1}{-25}-\frac{1}{10}=\frac{-2-5}{50}=-\frac{7}{50}$$

= – 7.14 cm

$$\text{(b) Magnification m =}\frac{v}{u}=\frac{-25}{-7.14}=3.5\\\text{(c) Magnifying power =}\frac{d}{u}=\frac{25}{7.14}=3.5$$

Yes, the magnification equal to the magnifying power because, the image formed at least distance of distinct vision, i.e., 25 cm.

**25. What should be the distance between the object in question 24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?**

**Sol.** Given, area of image, A_{I} = 6.25 mm^{2}

Area of object,

A_{o} = 1 mm^{2}

Focal length of lens, f = 10 cm

$$\text{Linear magnification , m =}\sqrt{\frac{\text{A}_{\text{I}}}{\text{A}_{0}}}=\sqrt{\frac{6.25}{\text{I}}}=2.5\\\text{Again, magnification, m =}\frac{v}{u}$$

v = m × u= 2·5 × u …(i)

From Lens formula

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

From Eq. (i),

$$\text{or}\space\frac{1}{10}=\frac{1}{u}\bigg(\frac{1-2.5}{2.5}\bigg)\\\text{or}\space u=\frac{-1.5×10}{2.5}=-6\text{cm}$$

or v = 2.5u = 2.5 (–6) = – 15cm

Thus, the virtual image is formed at a distance of 15 cm. But it is less than least distance of distinct vission. So, it cannot be seen by the eyes distinctly.

**26. 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 than does a magnifying glass provide angular magnification?**

**(b) ****In ****viewing through a magnifying glass, one usually positions one’s eyes 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 objectiv****e and the eye-piece of a compound microscope have short focal lengths?**

**(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) Since, the size of image is much bigger than the size of object and, angular size of image is equal to angular size of object. Due to help of a magnifying glass one can see the objects placed closer than the least distance of distant vision. As closer the object, larger be the angular size.

(b) Yes, the angular magnification changes. As the distance between eye and magnifying glass is increased, the angular magnification decreases.

(c) We cannot make the lenses with very small focal length easily.

(d) The angular magnification produced by the eye-piece of a compound microscope

$$=\bigg(\frac{25}{f_{e}}+1\bigg).$$

If f_{e} is small, the angular magnification will be large.

Further, magnification of objective lens is $$\frac{v}{u}$$ As object lies close to focus of objective lens u ≈ f_{0}. To increase this magnification

$$\bigg(\frac{v}{f_{0}}\bigg).$$

f_{o} should be smaller.

Thus f_{o} and f_{e} both are small.

(e) When we place our eyes very close to the eye-piece of a compound microscope, we are not able to collect the refracted light in large amount because the field of view will be reduced. So, the image is blurred.

The best position of the eye for viewing through a compound microscope is at the eye ring attached to the eye-piece.

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

**Sol.** Given, focal length of objective, f_{o} = 1.25 cm

Focal length of eye-piece, f_{e} = 5 cm

Least distance of distinct vision, d = 25 cm

Angular magnification, m_{e} = 30

The magnification produced by eye-piece

$$m_{e}=1+\frac{d}{f_{e}}=1+\frac{25}{5}=6$$

The magnification produced by microscope

m = m_{o} × m_{e}

30 = m_{o} × 6

where, m_{o} is the magnification produced by objective lens.

m_{o} = 5

Again, we know that magnification of objective lens

$$m_{0}=\frac{v_{0}}{u_{0}}\\5=\frac{-v_{0}}{u_{0}}\\\text{or}\space\space\text{v}_{0}=-5u_{0}\qquad\text{...(i)}\\\frac{1}{f_{0}}=\frac{1}{v_{0}}-\frac{1}{u_{0}}\\\frac{1}{1.25}=\frac{1}{-5u_{0}}-\frac{1}{u_{0}}=\frac{6}{5\mu_{0}}$$

From Eq. (i), we get

$$\mu_{0}=-\frac{6}{5}×1.25=-1.5\space\text{cm}$$

v_{o} = –5u_{o} = – 5 (– 1.5) = 7.5 cm

Thus, the object is placed at a distance of 1.5 cm from the objective lens to get the desired magnification.

Now, applying the Lens formula for eye-piece

$$\frac{1}{f_{e}}=\frac{1}{v_{e}}-\frac{1}{u_{e}}\\\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\\=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}\\(\because v_{e}=-25\space\text{cm})$$

u_{e} = – 4.17 cm

The separation between objective and eye-piece.

|v_{o}|+ |u_{e}| = 4.17 + 7.5

= 11.67 cm

Thus, the microscope is adjusted such as the distance between eye-piece and objective is 11.67 cm.

**28. A small telescope has an objective lens of focal length 140 cm and an eye-piece 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) For normal adjustment of telescope the magnification

$$m=\frac{f_{o}}{f_{e}}=-\frac{140}{5}=-28$$

(b) When final image is formed at least distance of distinct vision, the magnification

$$m=\frac{f_{0}}{f_{e}}\bigg(1+\frac{f_{e}}{d}\bigg)=\frac{140}{5}\bigg(1+\frac{5}{25}\bigg)$$

m = 28 (1 + 0.2) = 33.6

**29. (a) For the telescope described in Q. 28 (a), what is the separation between the objective lens and the eye-piece?**

**(b) ****If this telescope ****is used to view a 100m 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 m?**

**Sol.** Give, f_{o} = 140 cm, f_{e} = 5 cm and d = 145 cm

(a) In normal adjustment, the separation between eye-piece and objective lens

= f_{o} + f_{e} = 140 + 5 = 145 cm

(b) Height of tower, O_{t} = 100m

Distance of tower, u = 3 km = 3000 m

The angle subtended

$$\theta_{0}=\frac{\text{O}_t}{u}=\frac{100}{3000}=\frac{1}{30}\space\text{rad}\space\text{...(i)}\\\text{The angle subtended by the image.}\\\theta_{i}=\frac{\text{I}_{t}}{f_{0}}=\frac{\text{I}_{t}}{140}\\(∵\space\text{I}_{t}= height of image tower)\\\text{As}\space\theta_o = \theta_i\\\frac{1}{30}=\frac{\text{I}_{t}}{140}\\\text{I}_{t}=\frac{14}{3}=4.7\text{cm}$$

Thus, the height of image is 4.7 cm.

(c) When image formed at distance, d = 25 cm, then magnification produced by eye-piece

$$m=1+\frac{d}{f_{e}}=1+\frac{25}{5}=6$$

Let I be the height of the final image of the tower and size of the image formed by objective

$$\text{O}=\frac{14}{3}\text{cm}\\= 4.7 \text{cm}\\\text{or}\space m=\frac{\text{I}}{\text{O}}$$

∴ Height of final image = m × O = 6 × 4.7 = 28.2 cm

Thus, the height of final image of tower is 28.2 cm.

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

**Sol.** Given, distance between objective mirror and secondary mirror

d = 20 mm

Radius of curvature of objective mirror

= R_{1} = 220 mm

$$\therefore\space\text{Focal length of objective mirror}, f_1 =\frac{220}{2}=110\space \text{mm}$$

Radius of curvature of secondary mirror = R_{2 }= 140 mm

$$\therefore\space\text{Focal length of secondary mirror}, f_2 =\frac{140}{2}=70\space \text{mm}$$

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for secondary mirror.

So, the object distance for small mirror u = f_{1} – d

i.e., u = 110 – 20 = 90 mm

$$\text{By mirror formula,}\space\frac{1}{v} + \frac{1}{u}=\frac{1}{f_{2}}\\\frac{1}{v}=\frac{1}{f_{2}}-\frac{1}{u}=\frac{1}{70}-\frac{1}{90}\\\frac{9-7}{630}=\frac{2}{630}$$

v = 315 mm or v = 31.5 cm

Thus, the final image is formed at 315 mm away from secondary mirror.

**31. Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in figure. 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 1.5m away?**

**Sol.** Given, deflection, θ = 3.5°

Distance between screen and mirror, x = 1.5m

When mirror turns by angle θ, the reflected ray turned by 2 θ.

$$\therefore\space 2\theta=3×3.5=7\degree=\frac{7\pi}{180}\text{rad}\\\text{Again, in Δ AOS}\\\text{tan}\space 2\theta=\frac{\text{AS}}{\text{sin}\space\theta}\\\text{tan}\bigg(\frac{7\pi}{180}\bigg)=\frac{\text{AS}}{1.5}=\frac{\text{d}}{1.5}\\\text{or}\space d=1.5\space\text{tan}\bigg(\frac{7\pi}{180}\bigg)\\\text{For small angle,}\\\text{tan}\frac{7\pi}{180}≈\frac{7\pi}{180}\text{rad}$$

Again, in Δ AOS

$$\text{tan 2}\theta =\frac{\text{AS}}{\text{sin}\theta}\\\text{tan}\bigg(\frac{7\pi}{180}\bigg)=\frac{\text{AS}}{\text{1.5}}=\frac{d}{1.5}\\\text{or}\space d=1.5\space\text{tan}\bigg(\frac{7\pi}{180}\bigg)\\\text{For small angle,}\\\text{tan}\frac{7\pi}{180}≈\frac{7\pi}{180}\\d=1.5×\frac{7\pi}{180}=0.18\space\text{m}$$

**32. 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?**

**Sol.** Given, focal length of convex lens of glass, f_{1} = 30 cm.

Focal length

of combination of convex lens and plano-concave liquid lens f = 45 cm.

Refractive index of lens, µ_{g} = 1.5

Let f_{2} be the focal length of the plano-concave lens made of liquid between the convex lens and mirror.

For combined lens

$$\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{f}\\\frac{1}{30}+\frac{1}{f_{2}}=\frac{1}{45}\\\Rarr\space\frac{1}{f_{2}}=\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$$

f_{2} = – 90 cm

We know that radii of curvature of two surfaces of plano-concave lens made of liquid formed between foci convex lens and plane mirror are –R and ∞.

For the convex lens of glass, R_{1} = R, R_{2} = – R

Using Lens maker’s formula

$$\frac{1}{f_{2}}=(\mu_{1}-1)\bigg(\frac{1}{R_{1}}-\frac{1}{R_{2}}\bigg)\\=\bigg(\frac{3}{2}-1\bigg)\bigg(\frac{1}{\text{R}}+\frac{1}{\text{R}}\bigg)\\\frac{1}{30}=\frac{1}{2}×\frac{2}{\text{R}}$$

∴ R = 30 cm

Again, R_{1} = – R = – 30 cm, R_{2} = ∞

Using the Lens maker’s formula

$$\frac{1}{f_{2}}=(\mu-1)\bigg(\frac{1}{\text{R}_{1}}-\frac{1}{\text{R}_{2}}\bigg)\\-\frac{1}{90}=(\mu_{1}-1)\bigg(\frac{1}{30}-\frac{1}{\infty}\bigg)\\\frac{1}{90}=\frac{1}{30}(\mu_{1}-1)\\\mu_{1}=1+\frac{1}{3}=\frac{4}{3}=1.33\\\text{Thus, the refractive index of liquid is}\space\frac{4}{3}.$$

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