Sol. Since colour A is bent more towards the normal. Hence refractive index medium

For colour A is more than that for colour B, As $v \propto \frac{1}{\mu},$ so colour A travels slowly in medium.

Sol. At a distance $\frac{f}{2}$ from the optical centre and on the same side of boject.

Sol. Power of the lens increases

Sol. The point is convergence will be shifted away from the glass plate.

Sol. According to lens makers formula

$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \Rightarrow f \propto \frac{1}{(\mu-1)}$

As the refractive index of glass w.r.t. water is lesser than the refractive index of glass w.r.t. air, so focal length of converging lens will be more in water than that in air.

Sol. (i) Focal length of concave mirror remains unchanged.

(ii) For lens $\frac{1}{f} \propto(\mu-1)$ As $\mu_{V}>\mu_{R} \Rightarrow f_{R}>f_{V}$

Sol.

(i)

(ii)

Sol.

The image will still be of full size but the intensity of the image will be lesser, when the lower half of the lens in painted black.

Sol. (a) There are no water vapours on the moon. Hence there is neither dispersion no total internal reflection of sun rays. Consequently, rainbow is never observed on the moon.

(b) During sunrise on sunset, the sun is near the horizon. The shorter waves of blue region are scattered away by the greater thickness of the atmosphere. The least scattered red rays reach the observer. So the sun appears red.

Sol. By using lens makers formula

(i) If $n_{1}>n_{2}$ then $f$ will be positive. It means that lens will show opposite nature $i . e .$ it will converge the light rays

(ii) If $n_{1}=n_{2}$ then $f$ will be infinite. It means that lens behaves like a plane glass plate i.e.n light raysn passes undeviated.

(iii) If $n_{1}<n_{2}$ then $f$ will be negative. It means nthat lens will show it’s normal behaviour i.e. itn will diverges the light rays.

Sol. By using lens makers formula

$\quad \frac{1}{f}=\left(\frac{\mu_{2}}{\mu_{1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

Where $\mu_{1}=n_{1}=$ refractive index of medium $\mu_{2}=n_{2}=$ refractive index of lens

Hence $\frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

(i) If $n_{1}>n_{2}$ then $f$ will be positive. It means thatlens will show normal behaviour i.e. it will converge the light rays

(ii) If $n_{1}=n_{2}$ then $f$ will be infinite. It means that lens behaves like a plane glass plate i.e. light rays passes undeviated

(iii) If $n_{1}<n_{2}$ then $f$ will be negative. It means that lens will show opposite behaviour and diverges the light rays

Sol.

In the position of minimum deviation

$i=e, r_{1}=r_{2}=r$

Hence by using $i+e=A+\delta \Rightarrow i+i=A+D_{m}$

$\Rightarrow i=\frac{A+D_{m}}{2}$ …(i)

Also $A=r_{1}+r_{2}=r+r=2 r \Rightarrow r=\frac{A}{2}$ …(ii)

Hence by using snell’s

$\operatorname{law} \mu=\frac{\sin i}{\sin r} \Rightarrow \mu=\frac{\sin \frac{A+D_{m}}{2}}{\sin A / 2}$

Sol. For a real image (on wall), minimum distance

between the object and image should be $4 f .$

$\therefore 4 f=3 m \quad \therefore f=\frac{3}{4} m=0.75 m$

Sol. Here, $\mu=1.55, R_{1}=R$ and $R_{2}=-R, f=20 \mathrm{cm}$

As $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$\therefore \frac{1}{20}=(1.55-1)\left(\frac{1}{R}+\frac{1}{R}\right)=\frac{1.10}{R}$$\Rightarrow \quad R=20 \times 1.1=22 \mathrm{cm}$

Sol. (a) Ray diagram

(b) Magnification : Area of each square (i.e. object)

$O=1 \mathrm{mm}^{2}$

Distance of object $u=-9 \mathrm{cm}$

Focal length of lens $f=+10 \mathrm{cm}$

By using lens formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}-\frac{1}{(-9)} \Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{9}=-\frac{1}{90} \Rightarrow v=-90 \mathrm{cm}$

Hence magnification $=\frac{v}{u}=\frac{90}{9}=10$

also $m^{2}=\frac{A_{\text {image }}}{A_{\text {object }}} \Rightarrow A_{\text {image}}=m^{2} \times A_{\text {object}}$

$=(10)^{2} \times 1 m m^{2}=100 m m^{2}$

(c) Magnification power $=\frac{D}{4}=\frac{25}{9}=2.8=2.8$

Sol. By using lens makers formula

For the lens in air, $\mu_{2}=1.6, \mu_{1}=1,$ and $f=+20 \mathrm{cm}$

When lens is immersed in a liquid of $\mu=1.8$ then

$\mu_{l e n s}<\mu_{m e d i u m} \mathrm{SO}$ lens behaves oppositely i.e like a diverging lens.

Sol. Here, size of candle,

$h_{1}=2.5 \mathrm{cm}, u=-27 \mathrm{cm}, R=-36 \mathrm{cm}$

$\quad f=\frac{R}{2}=-18 \mathrm{cm}, v=?,$ As $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\therefore \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-18}+\frac{1}{27}=\frac{-3+2}{54}=\frac{1}{54}$ or $v=-54 \mathrm{cm}$

$\therefore$ Screen should be placed at $54 \mathrm{cm}$ from the

mirror on the same side as the object.

If $h_{2}$ is size of image, then as the image is real

$m=\frac{-h_{2}}{h_{1}}=\frac{-v}{-u}=\frac{v}{u} \Rightarrow \frac{-h_{2}}{2.5}=\frac{(-54)}{(-27)}=2$

$\Rightarrow h_{2}=-5 \mathrm{cm} .$

Minus sign indicates that image is inverted. When the candle is moved closer to mirror, the screen has to be moved away from the mirror. However, when candle is at a distance less than 18cm from the mirror, image formed would be virtual and screen is not required as image cannot be taken on the screen.

Sol. Here $h_{1}=3 c m, u=-14 c m, f=-21 c m, v=?$

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

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-21}+\frac{1}{-14}=\frac{-5}{42} \Rightarrow v=\frac{-42}{5}=-8.4 \mathrm{cm}$

$\therefore$ Image is erect, virtual and at $8.4 \mathrm{cm}$ from the

lens on the same side of the object.

As $\frac{I}{O}=\frac{v}{u} \Rightarrow \frac{I}{3}=\frac{-8.4}{-14} \Rightarrow I=0.6 \times 3=1.8 \mathrm{cm}$

As the object is moved away from the lens, virtual image moves towards focus of lens (but never beyond focus). The size of image goes on decreasing.

Sol. Since light rays incident normally on face ab, so no refraction takes place and light rays passes undeviated and then incident on face ac at an angle of $49^{\circ}$.

For total internal reflection at $a c$

$i>c \Rightarrow \sin i>\sin c \Rightarrow \sin 45^{\circ}>\frac{1}{\mu}$

$\Rightarrow \frac{1}{\sqrt{2}}>\frac{1}{\mu} \Rightarrow \mu>\sqrt{2}$ or $\mu>1.41$

As $\mu_{R}<\mu$ but $\mu_{G}>\mu$ and $\mu_{B}>\mu,$ so only red

colour will be transmitted through face $a c$ while

green and blue while internally reflected.

Sol. (a) For convex lens $u=+12 \mathrm{cm}, f=+20 \mathrm{cm}$

$\therefore \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{20}=\frac{1}{v}-\frac{1}{12} \Rightarrow \frac{1}{20}+\frac{1}{12} \Rightarrow v=7.5 \mathrm{cm}$

Thus the beam converges at a point $7.5 \mathrm{cm}$ to the

right of the lens.

(b) For concave lens $u=+12 c m, f=-16 c m$

$\therefore \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{-16}=\frac{1}{v}-\frac{1}{12} \Rightarrow \frac{1}{v}=\frac{-1}{16}+\frac{1}{12}$

$\Rightarrow v=48 c m$

Thus the beam converges at a point 48 cm to the right of the lens.

Sol.  A convex lens is made up of two convex spherical refracting surfaces. The final image is formed after two refractions. In figure (A) $P_{1}, P_{2}$ are the poles, of a thin biconvex lens with optical centre at $O$. Let $\mu_{2}$ be the refractive index of material of lens and $\mu_{1}$ be the refractive index of the surrounding.

Consider a point object $O^{\prime}$ lying on the principle axis of the lens. The first refracting surface $A P_{1} B$ forms the image $I_{1}$ of the object at a distance $v_{1}$ from the surface. The image $I_{1}$ acts as a virtual object for the second surface that forms the final image at $I$. The general equation for refraction at spherical surface is

$\frac{\mu_{2}}{\text { distance of image }}-\frac{\mu_{1}}{\text { distance of object }}$

$=\frac{\mu_{2}-\mu_{1}}{\text { Radius of curvature }}$ …(i)

To use this equation for the two surfaces $\left(A P_{1} B\right.$ and $\left.A P_{2} B\right)$ origin’s are at $P_{1}$ and $P_{2}$ respectively. We should take the origin at O, because lens is Thin Refraction at first surface $A P_{1} B:$ For the first refraction object is at $O^{\prime}$ the image is at $I_{1}$ and centre of curvature of $C_{1}$. If $u, v_{1}$ and $R$ denotes their respective distance from $O$ then by using equation ( i )

$\frac{\mu_{2}}{v_{1}}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{2}}{R_{1}}$ …(ii)

Refraction at second surface $A P_{2} B:$ For the second refraction from surface $A P_{2} B, I_{1}$ acts as virtual object, $I$ is the final image and $C_{2}$ is the centre of curvature of this surface. If $v_{1}, v$ and $R_{2}$ denotes object distance, and image distance and radius of curvature for this surface then again by using equation ( i)

$\frac{1 / \mu_{2}}{v}-\frac{1 / \mu_{1}}{v_{1}}=\frac{\mu_{2}-\mu_{2}}{R_{1}}$

(Here $1 / \mu$ is taken in place of because refraction is taking place from denser to rarer medium)

$\Rightarrow \frac{\mu_{1}}{v}-\frac{\mu_{2}}{v_{1}}=\frac{\mu_{1}-\mu_{2}}{R_{2}}=-\frac{\left(\mu_{2}-\mu_{1}\right)}{R_{2}} \ldots$ (iii)

on adding equation (ii) and (iii) we get

$\frac{\mu_{1}}{v}-\frac{\mu_{1}}{u}=\mu_{2}-\mu_{1}\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ …(ii)

If $\mu_{1}=1$ and $\mu_{2}=\mu$ then

$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ this is the required result

Assumptions : (i) The lens is thin so that distances measured from the poles of it’s surface can be taken as equal to the distance from the optical centre of the lens

(ii) The aperture of the lens is small.

(iii) The object is point object

(iv) The incident ray and refracted ray makes small angles with the principle axis of the lens.

Sign convention : (i) All distances are measured from the optical centre of the lens

(ii) In the direction of light all the distances are taken positive while opposite to the direction of light all the distances are taken negative.

(iii) Above the principal axis all the distances are taken positive while below the principal axis all distances are taken negative.