Ray Optics And Optical Instruments Class 12 Important Questions: CBSE Physics Chapter 9

1 Mark Questions

1.  A person standing before a concave mirror cannot see his image unless he is beyond the centre of curvature. Why?

Ans: Let a man stand beyond focus i.e., between focus and centre of curvature, then the image formed will be real and inverted and is formed beyond C (beyond him). Thus, he cannot see the image. 

But if he stands beyond C, the image will be formed between focus and centre of curvature which is in front of him and thus he will be able to see his reflected image.

2. For what angle of incidence, the lateral shift produced by a parallel-sided glass plate is maximum?

Ans: We know that lateral shift $d$ is given as,

$$d={\displaystyle \frac{t}{\cos r}}\sin (i-r)$$

For lateral shift $d$ to be maximum, $sin(i-r)$ must be maximum i.e., $i-r$ must be minimum. This happens when $i={90}^{{}^{\circ }}$.

$$\Rightarrow d={\displaystyle \frac{t}{\cos r}}\sin ({90}^{\circ }-r)$$ 

$\Rightarrow D=t$

Then, we can say that lateral shift is maximum.

3. You read a newspaper, because of the light it reflects. Then why do you not see even a faint image of yourself in the newspaper?

Ans: We know that image is formed due to the regular reflection of light. 

However, when we read a newspaper, there is diffused (irregular) reflection of light, thus we are not able to see even a faint image of ourselves on the newspaper.

4. A substance has a critical angle of $$\mathbf{4}{\mathbf{5}}^{{}^{\circ }}$$ for a yellow light, then what is its refractive index?

Ans: We know that the refractive index is given as below:

$$\mu ={\displaystyle \frac{1}{\sin C}}$$
Substituting the values, we have

$\mu ={\displaystyle \frac{1}{\sin {45}^{{}^{\circ }}}}={\displaystyle \frac{1}{\left(1/\sqrt{2}\right)}}$

$\Rightarrow \mu =\sqrt{2}$ 

5. An object is placed between the pole and the focus of a concave mirror produces a virtual and enlarged image. Justify using mirror formula.

Ans: We know that the mirror formula is as given below,

$${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$$

$$\Rightarrow v={\displaystyle \frac{uf}{u-f}}$$

Now magnification, $m={\displaystyle \frac{v}{u}}$.

$\Rightarrow m={\displaystyle \frac{{\displaystyle \frac{uf}{u-f}}}{u}}$

$\Rightarrow m={\displaystyle \frac{f}{u-f}}$

For a concave mirror, 

$$f=-ve$$

$$u=-ve$$ 

Given $$u<f$$; 

$$\therefore m=+ve$$, hence enlarged image is produced.

6. A converging and diverging lens of equal focal lengths are placed coaxially in contact. Find the focal length and power of the combination.

Ans: We know that the lens formula is as given below,

$${\displaystyle \frac{1}{F}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}$$

For converging lens $$f_1=+f$$

For diverging lens $$f_2=-f$$

$\Rightarrow {\displaystyle \frac{1}{F}}={\displaystyle \frac{1}{f}}-{\displaystyle \frac{1}{f}}$

$\Rightarrow F={\displaystyle \frac{1}{0}}\Rightarrow F=\mathrm{\infty }$

Since  $$P={\displaystyle \frac{1}{F}}={\displaystyle \frac{1}{\mathrm{\infty }}}=0$$

Hence, the power of the combination of the lens is $$P=0$$.

7. The refractive index of a material of a convex lens is $$n_1$$. It is immersed in a medium of refractive index $$n_2$$. A parallel beam of light is incident on the lens. Trace the path of the emergent rays when $$n_2>n_1$$.

Ans: We know that for  $$n_2>n_1$$ then the convex lens behaves as a concave lens as shown below.

8. In a telescope the focal length of the objective and the eye piece are $\mathbf{60}\mathbf{cm}$ and $\mathbf{5}\mathbf{cm}$ respectively. What is (1) its magnification power? (2) the tube length?   

Ans: 

1. We know that magnification $$M=-{\displaystyle \frac{f_o}{f_e}}={\displaystyle \frac{-60}{5}}=-12$$

2. The tube length is $$L=f_o+f_e=60+5=65cm$$.

9. Show the variation of $u$ and $v$ in the case of a convex mirror.

Ans: The variation of $u$ and $v$ in a convex mirror is as shown below.

10. Two lenses having focal length $$f_1$$ and $$f_2$$ are placed coaxially at a distance $$x$$ from each other. What is the focal length of the combination?

Ans: The formula for the focal length of the combination of two lenses when placed coaxially at a distance $x$ from each other is as given below:

$${\displaystyle \frac{1}{F}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}-{\displaystyle \frac{x}{f_1{f}_2}}$$

11. Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

Ans: Myopia and hypermetropia are common eye defects.

A myopic or hypermetropic person need not necessarily suffer a partial loss in their eyes’ ability of accommodation.

Myopia occurs when the eye-balls engage in elongation from the front to the back whereas hypermetropia occurs when the eye-balls shorten themselves. 

On the other hand, when the eye-lens completely loses its ability of adjusting itself, then the defect is called presbyopia.

2 Mark Questions

1. What are optical fibres? Give their one use.

Ans: Optical fibres are thin and long strands of fine quality glass or quartz coated with a thin layer of material with refractive index less than that of the strands. 

They work on the principle of total internal reflection and thus, they avoid any loss in transfer of information.  

Uses 

Optical fibres are often used in medical investigations i.e., one can examine the inside view of stomach and intestine by a method called endoscopy.

2. How do the focal lengths of a lens change with increase in the wavelength of the light?

Ans: We know that $${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$$.

Also, $$\mu \propto {\displaystyle \frac{1}{\lambda }}$$.

Clearly, when wavelength $(\lambda )$ increases, refractive index $$(\mu )$$  decreases. 

Similarly, $\mu \propto f$.

Clearly, as refractive index $$(\mu )$$ decreases, focal length $(f)$ increases.

3. Show with a ray diagram, how an image is produced in a total reflecting prism?

Ans: Consider the two rays from the object PQ, as shown below.

They undergo total internal reflection firstly at the face AB and then at BC forming the final image $$P^{\prime }{Q}^{\prime }$$ (real and inverted image).

4. The radii of the curvature of the two spherical surfaces which is a lens of required focal length are not the same. It forms an image of an object. The surfaces of the lens facing the object and the image are interchanged. Will the position of the image change? Ans: As we know that, 

$${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$$

When the radii of curvature $$R_1$$ and $$R_2$$ are interchanged, the focal length of the lens also changes. Hence, the position of the image will reduce gradually.

5. A thin converging lens has focal length (f) when illuminated by violet light. State with reason how the focal length of the lens will change if violet light is replaced by red light.

Ans: We know that, 

$${\displaystyle \frac{1}{f}}=(n-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$$

Since, $n$ for violet is more that $n$ for red colour, and since $m\propto f$, we can say that the focal length of the lens will decrease when violet light is replaced by red light.

6. Thin prism of angle $${60}^{\circ }$$ gives a deviation of $${30}^{\circ }$$. What is the refractive index of the material of the prism? Ans: We know that the refractive index of a thin prism is as follows

$$n={\displaystyle \frac{\sin \left({\displaystyle \frac{A+{\delta }_m}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}}$$

Substituting the given values, we have

$n={\displaystyle \frac{\sin \left({\displaystyle \frac{60+30}{2}}\right)}{\sin {\displaystyle \frac{60}{2}}}}={\displaystyle \frac{\sin {45}^{{}^{\circ }}}{\sin {30}^{{}^{\circ }}}}$

$\Rightarrow n=1.41$, which is the refractive index of the given thin prism.

7. Although the surfaces of a goggle lens are curved it does not have any power. Why? Ans: Since the two surfaces of a goggle lens are parallel i.e., one surface convex and the other concave, the resultant power of the two surfaces is zero as powers on both surfaces are equal but opposite in sign. 

$$p=p_1+p_2=p+(-p)=0$$

8. A ray of light is incident normally on one face of the prism of apex angle $${30}^{\circ }$$ and refractive index $$\sqrt{2}$$. Find the angle of deviation for the ray of light.

Ans: Let us assume that the ray PQ falls normally on AB. 

Then, it goes straight to AC without any refraction (QR) as shown in the figure.

Also given that

$n=\sqrt{2}$;

$A={30}^{{}^{\circ }}$;

$i={30}^{{}^{\circ }}$

Applying Snell’s law for face AC,

$n={\displaystyle \frac{\sin r}{\sin {30}^{{}^{\circ }}}}$

$\Rightarrow \sin r={\displaystyle \frac{1}{\sqrt{2}}}$

$\Rightarrow r={45}^{{}^{\circ }}$

Now angle of deviation,

$\delta =r-i$

$\Rightarrow \delta ={45}^{{}^{\circ }}-{30}^{{}^{\circ }}$

$\Rightarrow \delta ={15}^{{}^{\circ }}$, is the angle of deviation.

9. Following data was recorded for values of object distance and corresponding values of image distance in the experiment on study of real image formation by a convex lens of power $$+5D$$. One of the three observations is incorrect. Identify and give reason:

Ans: Given that $P=5D$

 We know that $P={\displaystyle \frac{1}{f}}$

$\Rightarrow f={\displaystyle \frac{1}{5}}=20cm$

Then,

$2f=40cm$

To find, which of the following observations are incorrect, consider the ray diagram as shown below, where it can be seen that when the object is placed between focus and twice the focus, the image is obtained after the measure if twice the focus.

1. $u=25$ and $v=97$ is correct.

When the object is placed between $f=20cm$ and $2f=40cm$, the image is obtained after $2f=40cm$.

2. $u=30$ and $v=60$ is also correct.

When the object is placed between $f=20cm$ and $2f=40cm$, the image is obtained after $2f=40cm$.

3. $u=30$ and $v=37$ is incorrect. 

Here, when the object is placed between $f=20cm$ and $2f=40cm$, the image is obtained before $2f=40cm$.

Clearly, we can conclude that observation (3) is incorrect because both object and the image here lie between $f$ and $2f$.

10. Birds flying high in the air appear to be higher than in reality. Explain why?

Ans: Birds fly in air, which is a rarer medium when compared to the ground, which is denser.

The light from the birds when viewed will undergo refraction towards the normal. Thus, the birds appear to fly at a higher point. i.e., 

Apparent height > Real height

11. What is the focal length of a convex lens of focal length $30cm$ in contact with a concave lens of focal length $20cm$? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Ans: Given that,

Focal length of the convex lens, $$f_1=30cm$$

Focal length of the concave lens, $$f_2=-20cm$$

Focal length of the system of lenses $$=f$$

Then the equivalent focal length of a system of two lenses in contact is given as:

$${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}$$

$$\Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{30}}-{\displaystyle \frac{1}{20}}={\displaystyle \frac{2-3}{60}}=-{\displaystyle \frac{1}{60}}$$

$$\therefore f=-60cm$$

Hence, the focal length of the combination of lenses is $60cm$. The negative sign indicates that the system of lenses acts as a diverging lens.

12. 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?    

Ans: Given that,

Distance between the object and the image, $$d=3m$$

Maximum focal length of the convex lens $$=f_{max}$$

For real images, the maximum focal length is given as:

$$f_{max}={\displaystyle \frac{d}{4}}={\displaystyle \frac{3}{4}}=0.75m$$

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

13. 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.

Ans: Given that,  

Distance between the image (screen) and the object, $$D=90cm$$

Distance between two locations of the convex lens, $$d=20cm$$

Focal length of the lens $$=f$$

Focal length is related to $$d$$ and D as:

$$f={\displaystyle \frac{D^2-d^2}{4D}}={\displaystyle \frac{{\left(90\right)}^2-{\left(20\right)}^2}{4\times 90}}={\displaystyle \frac{770}{36}}=21.39cm$$

Therefore, the focal length of the convex lens is $21.39cm$.

14. You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will 

1. Deviate a pencil of white light without much dispersion,

Ans: Place the two given prisms next to each other. 

Make sure that their bases are on the opposite sides of the incident white light, with   their faces touching each other.

When the white light is incident on the first prism, it will get dispersed. This acts as the incident light for the second prism and the dispersed light this time will recombine to give white light as a result of the combination of the two prisms.

2. Disperse (and displace) a pencil of white light without much deviation.

Ans: Take the system of the two prisms as suggested in answer (a).

Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. 

This combination will now disperse the pencil of white light without much deviation.

15. A myopic person has been using spectacles of power $-1.0$ dioptre for distant vision. During old age he also needs to use separate reading glass of power $+2.0$ dioptres. Explain what may have happened.

Ans: Given that the power of the spectacles used by the myopic person is $$P=-1.0D$$.

Focal length of the spectacles,

$$f={\displaystyle \frac{1}{P}}={\displaystyle \frac{1}{-1\times {10}^{-2}}}=-100cm$$

Hence, the far point of the person is $100cm$. He might have a normal near point of $25cm$. 

When he uses the spectacles, the objects placed at infinity produce virtual images at $100cm$.

He uses the ability of accommodation of the eye-lens to see the objects placed between $100cm$ and $25cm$.

During old age, the person uses reading glasses of power, $$p^{\prime }=+2D$$.

The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at $25cm$.

16. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Ans: In the given case, the person is able to see vertical lines more distinctly than horizontal lines. 

This means that the refracting system (cornea and eye-lens) of the eye is not working effectively in different planes. This defect is called astigmatism. 

The person's eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. 

Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

17. 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

1. The telescope is in normal adjustment (i.e., when the final image is at infinity)?

Ans: Given that, 

Focal length of the objective lens, $$f_o=140cm$$

Focal length of the eyepiece, $$f_e=5cm$$

Least distance of distinct vision, $$d=25cm$$

When the telescope is in normal adjustment, its magnifying power is given as:

$m={\displaystyle \frac{f_o}{f_e}}$

$\Rightarrow m={\displaystyle \frac{140}{5}}=28$

Thus, the magnifying power is $28$. 

2. The final image is formed at the least distance of distinct vision 25cm?

Ans: When the final image is formed at $$d$$, the magnifying power of the telescope is

given as:

$m={\displaystyle \frac{f_o}{f_e}}[1+{\displaystyle \frac{f_e}{d}}]$

$\Rightarrow m={\displaystyle \frac{140}{5}}[1+{\displaystyle \frac{5}{25}}]$

$\Rightarrow m=28[1+0.2]$

$\Rightarrow m=28\times 1.2=33.6$

Thus, the magnifying power is $33.6$.

18. Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in figure below. A current in the coil produces a deflection of $${3.5}^{\circ }$$ in the mirror. What is the displacement of the reflected spot of light on a screen placed $\mathbf{1}.\mathbf{5}\mathbf{m}$ away?

Ans: Given that, 

Angle of deflection, $$\theta ={3.5}^{\circ }$$

Distance of the screen from the mirror, $$D=1.5m$$

The reflected rays get deflected by an amount twice the angle of deflection i.e., $$2\theta ={7.0}^{\circ }$$

The displacement ($$d$$) of the reflected spot of light on the screen is given as:

$$\tan 2\theta ={\displaystyle \frac{d}{1.5}}$$

$$\therefore d=1.5\times \tan 7^{\circ }=0.184m=18.4cm$$

Hence, the displacement of the reflected spot of light is $18.4cm$.

3 Marks Questions

1. Find the radius of curvature of the convex surface of a plane convex lens, whose focal length is $\mathbf{0}.\mathbf{3}\mathbf{m}$ and the refractive index of the material of the lens is $\mathbf{1}.\mathbf{5}$.

Ans: Given that,

$$\mu =1.5$$ 

$$f=0.3m$$

For plane convex lens,

$R_2=-\mathrm{\infty }$ and let $$R_1=R$$.

Substituting these values in the formula for focal length,

${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

$\Rightarrow {\displaystyle \frac{1}{0.3}}=(1.5-1)({\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{\mathrm{\infty }}})$

$\Rightarrow \left({\displaystyle \frac{1}{R}}\right)0.5={\displaystyle \frac{1}{0.3}}$

$\Rightarrow R=0.15m$

Thus, the radius of curvature is $R=0.15m$.

2. Show that the limiting value of the angle of prism is twice its critical angle. Hence define critical angle.

Ans: We know that, 

Angle of the prism is given by $$A=r_1+r_2$$.

In a case of a triangular prism where $i_1=i_2={90}^{{}^{\circ }}$, angle of refraction is given by $r_1=r_2=C$.
where, $C$ is the critical angle.
Clearly, 

$$A=r_1+r_2$$

$\Rightarrow A=C+C$

$\Rightarrow A=2C$

Therefore, the angle of incidence for which angle of refraction is $${90}^{\circ }$$, is called the critical angle.

3. Draw a labelled diagram of telescope when the image is formed at the least distance of distinct vision? Hence derive the expression for its magnifying power.

Ans:  We know that,

magnifying power $$={\displaystyle \frac{anglesubtendedbytheimageattheeye}{anglesubtendedbytheobjectattheeye}}$$

$$MP={\displaystyle \frac{\tan \beta }{\tan \alpha }}={\displaystyle \frac{\beta }{\alpha }}$$ (Since angles are very small)

$$\tan \beta ={\displaystyle \frac{A^{\prime }{B}^{\prime }}{B^{\prime }E}}$$ and $$\tan \alpha ={\displaystyle \frac{A^{\prime }{B}^{\prime }}{B^{\prime }O}}$$

$MP={\displaystyle \frac{A^{\prime }{B}^{\prime }}{B^{\prime }E}}\times {\displaystyle \frac{A^{\prime }{B}^{\prime }}{B^{\prime }O}}$

$MP={\displaystyle \frac{B^{\prime }O}{B^{\prime }E}}={\displaystyle \frac{f_o}{-v_e}}$

$$MP={\displaystyle \frac{-f_o}{-v_e}}$$......(i)

For eye piece,

${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f_e}}$

$-{\displaystyle \frac{1}{D}}-{\displaystyle \frac{1}{-v_e}}={\displaystyle \frac{1}{f_e}}$

Multiply by D,

$$-1+{\displaystyle \frac{D}{v_e}}={\displaystyle \frac{D}{f_e}}$$

$${\displaystyle \frac{D}{v_e}}={\displaystyle \frac{D}{f_e}}+1$$

$${\displaystyle \frac{1}{v_e}}={\displaystyle \frac{1}{f_e}}+{\displaystyle \frac{1}{D}}={\displaystyle \frac{1}{f_e}}(1+{\displaystyle \frac{f_e}{D}})$$

Substituting in (i),

$$MP={\displaystyle \frac{-f_o}{f_e}}(1+{\displaystyle \frac{f_e}{D}})$$

4. Drive the expression for the angle of deviation for a ray of light passing through an equilateral prism of refracting angle A.

Ans: Consider the given diagram:

At the surface AB, 

$${\delta }_1=i-r_1$$

At the surface AC,

$${\delta }_2=e-r_2$$

Thus,

$$\delta ={\delta }_1+{\delta }_2$$

$$\delta =i+e-(r_1+r_2)$$ ......(1) 

Now, in quadrilateral AQOR,

$$\mathrm{\angle }A+\mathrm{\angle }Q+\mathrm{\angle }O+\mathrm{\angle }R={360}^{\circ }$$

$$\Rightarrow \mathrm{\angle }A+\mathrm{\angle }O={180}^{\circ }$$ ......(2)

Also, in $$\mathrm{\Delta }QOR$$,

$$\mathrm{\angle }Q+\mathrm{\angle }O+\mathrm{\angle }R={180}^{\circ }$$

Or

$$r_1+r_2+\mathrm{\angle }O={180}^{\circ }$$ ......(3)

From (2) and (3),

$$r_1+r_2=A$$ ......(4)

Substituting equation (4) in equation (1)

$$\delta =i+e-A$$

Or

$$A+\delta =i+e$$

5. Draw a ray diagram to illustrate image formation by a Newtonian type reflecting telescope. Hence state two advantages of it over refracting type telescopes.

Ans: The ray diagram of the Newtonian type reflecting telescope is as shown below.

Advantages 

The image formed in a reflecting type telescope is free from chromatic aberrations. 

The image formed is very bright due to its large light-gathering power. 

6. The magnifying power of an astronomical telescope in the normal adjustment position is $100$. The distance between the objective and the eye piece is  $101cm$. Calculate the focal length of the objective and the eye piece.

Ans: Given that, 

$$f_o+f_e=101cm$$ ......(1)

$$M=\left|{\displaystyle \frac{f_o}{f_e}}\right|=100$$

$$f_o=100f_e$$  ......(2)

Substituting equation (2) in equation (1),

$f_e+100f_e=101$

$\Rightarrow 101f_e=101$ 

$\Rightarrow f_e=1cm$ 

Substituting $$f_e$$ in equation (2),

$f_o=100\times 1$ 

$f_o=100cm$ 

Thus, the focal length of the eye-piece is 1cm whereas the focal length of the objective is 100cm.

7. A convex lens made up of refractive index $${\mathbf{n}}_{\mathbf{1}}$$ is kept in a medium of refractive index $${\mathbf{n}}_{\mathbf{2}}$$. Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if 

1. $$n_1>n_2$$

Ans: When $$n_1>n_2$$, the lens behaves as a convex lens.

2. $$n_1=n_2$$

Ans: When $$n_1=n_2$$, the lens behaves as a plane plate and thus, no refraction takes place.

3. $$n_1<n_2$$

Ans: When $$n_1<n_2$$, the lens behaves as a convex lens.

8. Derive the relation  $${\displaystyle \frac{\mathbf{1}}{\mathbf{f}}}={\displaystyle \frac{\mathbf{1}}{{\mathbf{f}}_{\mathbf{1}}}}+{\displaystyle \frac{\mathbf{1}}{{\mathbf{f}}_{\mathbf{2}}}}$$, where ${\mathbf{f}}_{\mathbf{1}}$ and $${\mathbf{f}}_{\mathbf{2}}$$ are focal lengths of two thin lenses and $\mathbf{F}$ is the focal length of the combination in contact. 

Ans: Consider two thin lenses in contact having focal length $$f_1$$ and $$f_2$$.

For the first lens, 

$${\displaystyle \frac{1}{v_1}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f_1}}$$ ......(1)

For the second lens, $$I_1$$ acts as an object which forms the final image $$I$$.

Clearly, $${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{v_1}}={\displaystyle \frac{1}{f_2}}$$ ......(2)

Adding equations (1) & (2) 

$${\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}+{\displaystyle \frac{1}{v_1}}-{\displaystyle \frac{1}{v_1}}$$

$$\Rightarrow {\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$$

Using lens formula $$({\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{F}})$$, 

$${\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}={\displaystyle \frac{1}{F}}$$

For $n$ number of thin lenses in contact,  

$${\displaystyle \frac{1}{F}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}+{\displaystyle \frac{1}{f_3}}+......+{\displaystyle \frac{1}{f_n}}$$

Hence the derivation.

9. A convex lens has a focal length $\mathbf{0}.\mathbf{2}\mathbf{m}$ and made of glass $$(\mu =\mathbf{1}.50)$$is immersed in water $$(\mu =\mathbf{1}.\mathbf{33})$$. Find the change in focal length of the lens.

Ans: Given,

$f_a=0.2m$; $${}^a{\mu }_g=1.50$$

It is known that

$${\displaystyle \frac{1}{f_a}}=({}^a{\mu }_g-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$$

$\Rightarrow {\displaystyle \frac{1}{0.2}}=(1.50-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

$$\Rightarrow {\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}}=10$$ ......(1)

Now, 

$${}^w{\mu }_g={\displaystyle \frac{{}^a{\mu }_g}{{}^a{\mu }_w}}={\displaystyle \frac{1.50}{1.33}}=1.128$$

Also, the focal length of the lens when immersed in water would be

$${\displaystyle \frac{1}{f_w}}=({}^w{\mu }_g-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$$

$\Rightarrow {\displaystyle \frac{1}{f_w}}=(1.128-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

$\Rightarrow {\displaystyle \frac{1}{f_w}}=\left(0.128\right)\times 10=1.28$

$\Rightarrow f_w={\displaystyle \frac{1}{1.28}}$

$\Rightarrow f_w=0.78m$

Thus, the change in focal length is given by

$f_w-f_a=0.78-0.20=0.58m$

10. A reflecting type telescope has a concave reflector of radius of curvature $\mathbf{120}\mathbf{cm}$. Calculate the focal length of eye piece to achieve a magnification of $\mathbf{20}$.

Ans: Given that 

$$M=20$$

$$R=120cm$$ (Concave reflector)

$$f_o={\displaystyle \frac{R}{2}}={\displaystyle \frac{-120}{2}}=-60cm$$

$$M={\displaystyle \frac{f_o}{f_e}}\Rightarrow {\displaystyle \frac{-60}{f_e}}=-20$$

$$\Rightarrow f_e=3cm$$

Thus, the focal length of the eye-piece is 3cm.

11. Show that a convex lens produces an N time magnified image, when the object distances from the lens have magnitude $$(\mathbf{f}\pm {\displaystyle \frac{\mathbf{f}}{\mathbf{N}}})$$. Here f is the magnitude of the focal length of the lens. Hence find two values of object distance u, for which convex lens of power $\mathbf{2}.\mathbf{5}\mathbf{D}$ will produce an image that is four times as large as the object? 

Ans: We know that the magnifying power is given by

$$M={\displaystyle \frac{f}{v+f}}$$

For real image $$m=-N$$;

$$\Rightarrow -N={\displaystyle \frac{f}{u+f}}\Rightarrow u+f={\displaystyle \frac{-f}{N}}$$ 

$$\Rightarrow u=-(f+{\displaystyle \frac{f}{N}})$$ ......(1)

For virtual image $$m=N$$;

$$\Rightarrow N={\displaystyle \frac{f}{u+f}}$$

$$\Rightarrow u+f=-f+{\displaystyle \frac{f}{N}}$$

$$\Rightarrow u=-(f-{\displaystyle \frac{f}{N}})$$ ......(2)

From equation (1) & (2) we can say that magnification produced by a lens can be N if 

$$u=-(f\pm {\displaystyle \frac{f}{N}})$$.

Now, power of the lens $$P=2.5D$$.

$$f={\displaystyle \frac{1}{P}}={\displaystyle \frac{1}{2.5}}m$$

$$\Rightarrow f={\displaystyle \frac{1}{2.5}}\times 100=40cm$$

It is given that

$$M=\pm 4$$

$$\Rightarrow \pm 4={\displaystyle \frac{40}{u+40}}$$

$$\Rightarrow u+40=\pm 10$$

$$\Rightarrow u=-40\pm 10$$

$$\Rightarrow u=30cm$$ or $$-50cm$$

These are the two values of object distances.

12. Define total internal reflection of light. Hence write two advantages of total reflecting prisms over a plane mirror. 

Ans: The phenomenon of reflection of light when a ray of light traveling from a denser medium getting reflected back into the same denser medium provided the angle of incidence is greater than the angle called critical angle is called total internal reflection.

Advantages 

1. It does require silvering.

2. Multiple reflections do not take place in a reflecting prism. Due to this, only one image is formed, which is very bright.

13. An equiconvex lens of radius of curvature R is cut into two equal parts by a vertical plane, so it becomes a plano-convex lens. If f is the focal length of equiconvex lens, then what will be focal length of the planoconvex lens?

Ans: We know that

$${\displaystyle \frac{1}{f}}=(n-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$$

For equiconvex lens $$R_1=R_2=R$$

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

$$\Rightarrow {\displaystyle \frac{1}{f}}=(n-1)({\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}})$$

$$\Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{2(n-1)}{R}}$$   ......(1)

For plano-convex lens,

$$R_1=R$$ and $$R_2=\mathrm{\infty }$$

Clearly,

$${\displaystyle \frac{1}{f^{\prime }}}=(n-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$$

$$\Rightarrow {\displaystyle \frac{1}{f^{\prime }}}=(n-1)({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{\mathrm{\infty }}})$$

$$\Rightarrow {\displaystyle \frac{1}{f^{\prime }}}={\displaystyle \frac{(n-1)}{R}}$$ ......(2)

From (1) & (2)

$${\displaystyle \frac{f^{\prime }}{f}}=2\Rightarrow f^{\prime }=2f$$

14. A converging lens of focal length $\mathbf{6}.\mathbf{25}\mathbf{cm}$ is used as a magnifying glass if near point of the observer is $\mathbf{25}\mathbf{cm}$ from the eye and the lens is held close to the eye. Calculate

1. Distance of Object From the Lens.

Ans: It is known that,

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

Given,

$$f=6.25cm$$

$$v=-25cm$$

Substituting these values in the first expression,

$${\displaystyle \frac{1}{u}}={\displaystyle \frac{-1}{25}}-{\displaystyle \frac{1}{6.25}}={\displaystyle \frac{-1}{5}}$$

$$\Rightarrow u=-5cm$$

2. Angular Magnification

Ans: Angular magnification is given by

$$m=1+{\displaystyle \frac{D}{F}}=1+{\displaystyle \frac{25}{6.25}}$$

$\Rightarrow m=1+4=5$

3. Angular Magnification When Final Image is Formed at Infinity.

Ans: When the image is formed at infinity,

$$m={\displaystyle \frac{D}{F}}={\displaystyle \frac{25}{6.25}}=4$$

15. Draw a graph to show that variation of angle of deviation ${\delta }_{\mathbf{m}}$with that of angle of incidence i for a monochromatic ray of light passing through a glass prism of refracting angle A. Hence deduce the relation.

$$\mu ={\displaystyle \frac{\sin \left({\displaystyle \frac{\mathbf{A}+{\delta }_m}{\mathbf{2}}}\right)}{\sin \left({\displaystyle \frac{\mathbf{A}}{\mathbf{2}}}\right)}}$$

Ans: For the minimum deviation position, $$i={\displaystyle \frac{{\delta }_m+A}{2}}$$.

Also, $$\mathrm{\angle }i=\mathrm{\angle }e$$.

Now, $$r_1=r_2=r$$ (say).

We know that: 

$$\mathrm{\angle }i=\mathrm{\angle }e={\delta }_m+A$$  ......(1)

Also,

$$r_1+r_2=A$$

$$\Rightarrow 2r=A$$

$$\Rightarrow r={\displaystyle \frac{A}{2}}$$

Applying minimum deviation condition is equation (1);

$$2i={\delta }_m+A$$

$$\Rightarrow i={\displaystyle \frac{{\delta }_m+A}{2}}$$

Applying Snell’s law,

$$\mu ={\displaystyle \frac{\sin i}{\sin r}}$$

$$\Rightarrow \mu ={\displaystyle \frac{\sin \left({\displaystyle \frac{A+\delta {}_m}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}}$$

16. An object of size $\mathbf{3}.\mathbf{0}\mathbf{cm}$ is placed $\mathbf{14}\mathbf{cm}$ in front of a concave lens of focal length $\mathbf{21}\mathbf{cm}$. Describe the image produced by the lens. What happens if the object is moved further away from the lens? 

Ans: Given that,

Size of the object, $$h_1=3cm$$

Object distance, $$u=-14cm$$

Focal length of the concave lens, $$f=-21cm$$

Let image distance $$=v$$

According to the lens formula, we have the relation:

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

Substituting the given values,

$$\Rightarrow {\displaystyle \frac{1}{v}}=-{\displaystyle \frac{1}{21}}-{\displaystyle \frac{1}{14}}={\displaystyle \frac{-2-3}{42}}={\displaystyle \frac{-5}{42}}$$

$$\therefore v=-{\displaystyle \frac{42}{5}}=-8.4cm$$

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual. 

The magnification of the image is given as: 

$$m={\displaystyle \frac{\mathrm{Im}ageheight\left(h_2\right)}{Objectheight\left(h_1\right)}}={\displaystyle \frac{v}{u}}$$

$$\Rightarrow h_2={\displaystyle \frac{-8.4}{-14}}\times 3=0.6\times 3=1.8cm$$

Hence, the height of the image is $1.8cm$.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. Also, the size of the image will decrease with the increase in the object distance.

17. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam $\mathbf{12}\mathbf{cm}$ from P. At what point does the beam converge if the lens is

1. A convex lens of focal length $\mathbf{20}\mathbf{cm}$

Ans: In the given situation, the object is virtual and the image formed is real.

Object distance, $$u=+12cm$$

Focal length of the convex lens, $$f=20cm$$

Image distance $$=v$$

According to the lens formula, we have the relation:

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

$$\Rightarrow {\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{12}}={\displaystyle \frac{1}{20}}$$

$$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{20}}+{\displaystyle \frac{1}{12}}={\displaystyle \frac{3+5}{60}}={\displaystyle \frac{8}{60}}$$

$$\therefore v={\displaystyle \frac{60}{8}}=7.5cm$$

Hence, the image is formed $7.5cm$ away from the lens, toward its right. 

2. A concave lens of focal length $\mathbf{16}\mathbf{cm}$. 

Ans: Focal length of the concave lens, $$f=16cm$$

Image distance $$=v$$

Object distance, $$u=+12cm$$

According to the lens formula, we have the relation: 

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

$$\Rightarrow {\displaystyle \frac{1}{v}}=-{\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{12}}={\displaystyle \frac{-3+4}{48}}={\displaystyle \frac{1}{48}}$$

$$\therefore v=48cm$$

Hence, the image is formed $48cm$ away from the lens, toward its right.

18. 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 $20cm$? 

Ans: Given that,

Refractive index of glass, $$\mu =1.55$$

Focal length of the double-convex lens, $$f=20cm$$

Radius of curvature of one face of the lens $$=R_1=R$$ 

Radius of curvature of the other face of the lens $$=R_2=-R$$ 

Let radius of curvature of the double-convex lens $$=R$$

The value of R can be calculated as:

$${\displaystyle \frac{1}{f}}=(\mu -1)[{\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}}]$$

$$\Rightarrow {\displaystyle \frac{1}{20}}=(1.55-1)[{\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}}]$$

$$\Rightarrow {\displaystyle \frac{1}{20}}=0.55\times {\displaystyle \frac{2}{R}}$$

$$\Rightarrow R=0.55\times 2\times 20=22cm$$

Hence, the radius of curvature of the double-convex lens is $22cm$.

19. A small telescope has an objective lens of focal length $\mathbf{144}\mathbf{cm}$ and an eyepiece of focal length $\mathbf{6}.\mathbf{0}\mathbf{cm}$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Ans: Given,

Focal length of the objective lens, $$f_o=144cm$$

Focal length of the eyepiece, $$f_e=6.0cm$$

The magnifying power of the telescope is given as: 

$$m={\displaystyle \frac{f_o}{f_e}}={\displaystyle \frac{144}{6}}=24$$

The separation between the objective lens and the eyepiece is calculated as:

$$D=f_o+f_e=144+6=150cm$$

Hence, the magnifying power of the telescope is $24$and the separation between the objective lens and the eyepiece is $150cm$.

20. 

1. A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

Ans: Given that,

Focal length of the objective lens, 

$$f_o=15m=15\times {10}^{2}cm$$

Focal length of the eyepiece,

$$f_e=1.0cm$$

The angular magnification of a telescope is given as: 

$$\alpha ={\displaystyle \frac{f_o}{f_e}}={\displaystyle \frac{15\times {10}^2}{1.0}}=1500$$

Hence, the angular magnification of the given refracting telescope is 1500. 

2. 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 $$\mathbf{3}.\mathbf{48}\times \mathbf{1}{\mathbf{0}}^{\mathbf{6}}\mathbf{m}$$, and the radius of lunar orbit is $$\mathbf{3}.\mathbf{8}\times \mathbf{1}{\mathbf{0}}^{\mathbf{8}}\mathbf{m}$$.

Ans: Given that,

Diameter of the moon, $$d=3.48\times {10}^{6}m$$

Radius of the lunar orbit, $$r_o=3.8\times {10}^{8}m$$

Let $$d^{\prime }$$ be the diameter of the image of the moon formed by the objective lens. 

The angle subtended by the diameter of the moon is equal to the angle subtended by the image. 

$$\Rightarrow {\displaystyle \frac{d}{r_0}}={\displaystyle \frac{d^{\prime }}{d_0}}$$

$$\Rightarrow {\displaystyle \frac{3.48\times {10}^6}{3.8\times {10}^8}}={\displaystyle \frac{d^{\prime }}{15}}$$

$$\Rightarrow d^{\prime }={\displaystyle \frac{3.48}{3.8}}\times {10}^{-2}\times 15$$

$$\Rightarrow d^{\prime }=13.74\times {10}^{-2}m=13.74cm$$

Hence, the diameter of the moon's image formed by the objective lens is $13.74cm$.

21. 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 $$\mu =\mathbf{1}.\mathbf{5}$$. Does the answer depend on the location of the slab? 

Ans: Given that, actual depth of the pin, $$d=15cm$$.

Refractive index of glass, $$\mu =1.5$$

Let the apparent depth of the pin $$=d^{\prime }$$.

Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.,

$$\mu ={\displaystyle \frac{d}{d^{\prime }}}$$

$$\Rightarrow d^{\prime }={\displaystyle \frac{d}{\mu }}$$

$$\Rightarrow d^{\prime }={\displaystyle \frac{15}{1.5}}=10cm$$

Now, the distance by which the pin appears to be raised $$=d-d^{\prime }=15-10=5cm$$.

For a small angle of incidence, this distance does not depend upon the location of the slab.

22. A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm. 

1. What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

Ans: Given that,

Focal length of the magnifying glass, $$f=5cm$$

Least distance of distance vision, $$d=25cm$$

Let the object distance $=u$ 

Image distance, $$v=-d=-25cm$$

According to the lens formula, we have:

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

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

$$\Rightarrow {\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{-25}}-{\displaystyle \frac{1}{5}}={\displaystyle \frac{-5-1}{25}}={\displaystyle \frac{-6}{25}}$$

$$\Rightarrow u=-{\displaystyle \frac{25}{6}}=-4.167cm$$

Hence, the closest distance at which the person can read the book is $4.167cm$. 

For the object at the farthest possible distance, the image distance $$(v^{\prime })=\mathrm{\infty }$$. 

According to the lens formula, we have: 

$${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v^{\prime }}}-{\displaystyle \frac{1}{u^{\prime }}}$$