Given: \({ }^{a} \mu_{g}=1.5=3 / 2 ; f=+18 \mathrm{~cm}\)
\( \frac{1}{f}=\left({ }^{a} \mu_{g}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)   .......(i)
\(\frac{1}{f^{\prime}}=\left({ }^{w} \mu_{g}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] \)    ............(ii)
Dividing equation (i) by (ii), we get
\( \frac{f^{\prime}}{f}=\frac{\mu_{g} / \mu_{a}-1}{\mu_{g} / \mu_{w}-1} \)
\(\Rightarrow \quad f^{\prime} =\left(\frac{3 / 2-1}{9 / 8-1}\right) f=4 f \)
∴f' = 4 x 18 = 72 cm
Change in focal length = 3 f = 54 cm

According to Snell's law

\( \mu_{1} =\sqrt{2}, \mu_{2}=1 \)
\(\mu_{1} \sin i =\mu_{2} \sin r \)
\(\sqrt{2} \sin i =1 \sin 90^{\circ}=1 \)
\(\sin i =\frac{1}{\sqrt{2}}=\sin 45^{\circ} \)
\(i =45^{\circ} \)

\(\because \quad r_{1}+r_{2}=A\)
At minimum deviation,
\( r_{1}=r_{2} \)
\(\text {i.e. } r=\frac{A}{2}=30^{\circ} \)
\( \text {As } i=\frac{3}{4} \times A=45^{\circ} \)
\(\therefore \quad \ \mu=\frac{c_{1}}{c_{2}}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}} \)
\(\Rightarrow \ c_{2}=2.12 \times 10^{8} \mathrm{~ms}^{-1}\)

(i) Cassegrainian telescope

(ii) The following are the two advantages of a reflecting type telescope over a refracting type telescope:
1. As there is no refraction, it is free from the chromatic aberration.
2. The light gathering power of the objective must be higher to get better resolution. It is easier to handle and cheaper to make mirrors of larger diameters.
(iii) Magnifying power of a telescope,
\(m=-\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)\)
So, the focal length of objective must be larger for higher magnification.
With the larger aperture of object, light gathering capacity ofthe telescope increases. Hence, a better resolution is obtained.

Power of a lens is the measure of convergence or divergence which a lens can introduce in beam of light falling on it. The SI unit of power is dioptre (D).

For the 1^st lens, we have relation
\(\frac{1}{f_{1}}=\frac{1}{v_{1}}-\frac{1}{u}\)        .........(i)
For the 2^nd lens, the relation is
\(\frac{1}{f_{2}}=\frac{1}{v}-\frac{1}{v_{1}}\)          .........(ii)
Here an image formed by the 1^st lens acts as a virtual object for the 2^nd lens.
Adding equations (i) and (ii), we get
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
If this two lens system is considered as an equivalent single lens of focal length f, we have
\( \frac{1}{v}-\frac{1}{u} =\frac{1}{f} \)
\(\therefore \quad \text { Power } P =\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \)

(a) As there is no refraction in case of mirror, the focal length will not change.
(b) Given: \(f_{a}=15 \mathrm{~cm}, \mu_{w}=1.33, \mu_{g}=1.5\)
\( \therefore \quad f_{w} =\frac{\left(\mu_{g}-1\right)}{\left(\frac{\mu_{g}}{\mu_{w}}-1\right)} \times f_{a} \)
\(=\frac{(1.5-1)}{\left(\frac{1.5}{1.33}-1\right)} \times 15 \)
\(=\frac{0.5 \times 1.33}{0.17} \times 15 \approx 59 \mathrm{~cm} \)

Here, \(\mu_{1}=1, \mu_{2}=1.5, u=-\infty, R=\frac{15}{2}=7.5 \mathrm{~cm}\)
∴ Using  \(\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}\)

As magnification, m = \(\frac{I}{O}=\frac{v}{u} \Rightarrow I=4 \times\) length of object
\(\Rightarrow \frac{I}{O}=4\Rightarrow \frac{v}{u}=4 \Rightarrow v=4 u\)
Using lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{(-4 u)}-\frac{1}{(-u)}\)
\( \Rightarrow \frac{1}{f}=-\frac{1}{4 u}+\frac{1}{u} \Rightarrow \frac{1}{20}=\frac{4-1}{4 u}=\frac{3}{4 u} \)
\(\Rightarrow u=\frac{20 \times 3}{4}=15 \mathrm{~cm} \)
⇒ v = 4u = 15 x 4 = 60 cm
Distance of the object, u = 15 cm
Distance of the image, v = 60 cm
The image is on the same side of the object.

By geometry, angle of incidence (i) at face AC for all three rays is 45°. Light suffers total internal reflection for which this angle of incidence is greater than critical angle.
i > i_c ⇒  sin i > sin i_c
or sin 45° > sin i_c
\(\Rightarrow \quad \frac{1}{\sin 45^{\circ}}<\frac{1}{\sin i_{c}} \Rightarrow \sqrt{2}<\mu\)
Total internal reflection takes place on AC for rays with μ > \(\sqrt {2}\) = 1.414, i.e. green and blue colour suffer total internal reflection, whereas red undergoes refraction.

The power of the spectacles used by the myopic person, P = -1.0 D
Focal length of the spectacles, \(f=\frac{1}{P}=\frac{1}{-1 \times 10^{-2}}=-100 \mathrm{~cm}\)
Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.
During old age, the person uses reading glasses of power, P′ = +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 25 cm.

By Snell's law,


\(\begin{aligned} \frac{\sin i}{\sin r} & =\frac{n}{\mu} \end{aligned}\)
\(\begin{aligned} \frac{\sin 45^{\circ}}{\sin 90^{\circ}} & =\frac{n}{\mu} \end{aligned}\)
n = \(\mu\) sin 45°            [\(\because\) sin 90° = 1]
\(n=\frac{\mu}{\sqrt{2}}\)                 \(\left[\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
Thus, refractive index ofthe transparent medium is \(n=\frac{\mu}{\sqrt{2}}\)

Magnifying power: It is the ratio of the angle subtended by the image formed at infinity to the angle subtended on the eye by the object placed at least distance of distinct vision.
\(\because \quad m=\frac{-D}{f_{e}} \cdot \frac{L}{f_{o}}\)
Therefore, to increase angular magnification, f_o and f_e should be small.

As the speed of light is maximum in vacuum, therefore absolute value of refractive index cannot be less than unity as it is given by the relation \(n=\frac{c}{v}\)

Since, \(m=\frac{\tan \beta}{\tan \alpha} \approx \frac{\beta}{\alpha}=\frac{f_{e}}{f_{e}}\)
\(\therefore \quad \frac{\beta}{0.5^{\circ}}=\frac{0.3}{0.03}=5^{\circ}\)

Blue colour of the sky is due to scattering of light from atmosphere's particles. Light of shorter wavelength is scattered more than the light of longer wavelength.

The relation between the angle of incidence i, angle of prism A and the angle of minimum deviation \(\delta \)_m, for a triangular is given as i \(=\frac{A+\delta_m}{2}\)

Given, u = -f
∴ Lens equation is, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}+\frac{1}{f}=\frac{1}{f} \Rightarrow \frac{1}{v}=0\)
\(\Rightarrow \quad v=\frac{1}{0}\) infinity

From Snell's law, \(\mu=\frac{\sin i}{\sin r}=\frac{c}{v}\)
⇒  v ∝ sin r, for given value of i.
Smaller the angle of refraction, smaller the velocity of light in medium.
Velocity of light is minimum in medium A as the angle of refraction is minimum, i.e. 15°.

A convex mirror produces a real image of a virtual object. Therefore, if a beam of light from a virtual object converges to a point behind the convex mirror, then its real image will be formed in front of the mirror.

A thick plana mirror consist of two surfaces (top and bottom), where the reflection takes place. The images are formed after reflection from both the surfaces, except for the first image, The second image is the brightest of all as minimum absorption takes place and bounces of the silvery layers which makes the bottom surface.

(a) A wavefront is defined as the locus of all the particles of a medium vibrating in the same phase at a given instant. The shape of a wavefront depends upon the shape of the source of disturbance and it is normal to the direction of propagation of wave.
A line drawn perpendicular to the plane wavefront gives the direction of propagation of a wave and is called ray of light.
(b) Huygens' Principle and Reflection at a Plane Surface
(c) Position of nth maxima, \(y_n=\frac{n D \lambda}{d}\) (Bright fringe) and position of mth minima,
\(y_m^{\prime}=\left(m-\frac{1}{2}\right) \frac{D \lambda}{d}\) (Dark fringe)
Given, d = 3 mm = 3 \(\times\) 10^-3 m
D = 1 m
and y₄ - y₂' = 5 mm = 5 \(\times\) 10^-3m
Now, \(\frac{4 D \lambda}{d}-\left(2-\frac{1}{2}\right) \frac{D \lambda}{d}=5 \times 10^{-3} \mathrm{~m}\)
\(\Rightarrow\) \(\frac{4 D \lambda}{d}-\frac{3}{2} \frac{D \lambda}{d}=5 \times 10^{-3}\)
\(\begin{aligned} \Rightarrow \left(4-\frac{3}{2}\right) \frac{D \lambda}{d}=5 \times 10^{-3} \end{aligned}\)
\(\begin{aligned} \Rightarrow \frac{5}{2}\left(\frac{(1) \lambda}{3 \times 10^{-3}}\right)=5 \times 10^{-3} \end{aligned}\)
\(\Rightarrow\)  \(\lambda=\frac{3}{2} \times 10^{-6} \mathrm{~m}\)
\(\Rightarrow\) \(\lambda=1.5 \times 10^{-6} \mathrm{~m}\)
\(\Rightarrow\) \(\lambda=1.5 \mu \mathrm{m}\)

Mirror Formula
Mirror formula (or equation) is a relation between focal length of the mirror, distances of object and image from the mirror.
In principle, we can take any two rays originating from a point on an object, trace their paths, find their point of intersection and thus, obrain the image of the point due to reflection at a spherical mirror.
However, in practice, it is convenient to choose any two of the following rays
(i) The ray from the point, which is parallel to the principal axis after reflection goes through the focus of the mirror.
(ii) The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a convex mirror simply retraces the path.
(iii) The ray passing through (or directed towards) the focus of the concave mirror or appearing to pass through (or directed towards) the focus of a convex mirror after reflection is parallel to the principal axis.
(iv) The ray incident at any angle at the pole is reflected following the laws of reflection.

In the above figure, the ray diagram is considering three rays for image formation by a concave mirror. In the figure, triangles A'B'F and NEF are similar.
Then, \(\frac{A^{\prime} B^{\prime}}{N E}=\frac{A^{\prime} F}{N F}\)
As, the aperture of the concave mirror is small and the points N and P lie very close to each other, then
NF \(\approx\) PF and NE = AB
\(\Rightarrow \quad \frac{A^{\prime} B^{\prime}}{A B}=\frac{A^{\prime} F}{P F}\)
Since, all the distances are measured from the pole of the concave mirror, we have
A' F = PA' - PF
\(\therefore \quad \frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}-P F}{P F}\)       ....(i)
Also, triangles ABP and A'B'P are similar, then
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A}\)                    ...(ii)
From Eqs. (i) and (ii), we get
\(\frac{P A^{\prime}-P F}{P F}=\frac{P A^{\prime}}{P A}\)               ...(iii)
Applying new Cartesian sign convention, we have
PA = - u
[\(\because\) distance of object is measured against incident ray]
PA' = -v
[\(\because\) distance of image is measured against incident ray]
PF = -f
[\(\because\) focal length of concave mirror is measured agaist incident ray]
Substituting these values in Eq. (iii), we have
\(\begin{aligned} \frac{-v-(-f)}{-f}=\frac{-v}{-u} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{v-f}{f} & =\frac{v}{u} \Rightarrow \frac{v}{f}-1=\frac{v}{u} \end{aligned}\)
Dividing both sides by v, we get
\(\therefore\) \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
The above relation is called mirror formula. Relation between u, v and R
\(\because\) Focal length of the mirror, \(f=\frac{R}{2}\)
\(\therefore\)\(\frac{1}{u}+\frac{1}{v}=\frac{1}{R / 2} \Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{2}{R}\)

A compound microscope consists of two convex lenses. The lens facing the object to be seen is called an objective lens. The lens facing the eye is called an eye lens. Generally, the eye lens is a combination of lenses and is called an eyepiece. The aperture and the focal length of the objective lens are smaller as compared with those of eye lens. The object to be magnified is placed just beyond the focal point of the objective lens which forms its real, magnified and inverted image. This image acts as the object for the eye lens whose position is so adjusted that the final image formed by the eye lens is at the distance of distinct vision as shown in the diagram.
(b) (i) To achieve a large magni fication of small object, the eyepiece and the objective must have short focal lengths.
(ii) If we place our eyes too close to the eyepiece, the area of the pupil of the eye is less than the area of the eye-ring. So, our eyes will not collect much of the light and our field of view will get reduced.

(ii) Given, the emergent ray grazes along the face AC,

e = 90\(\unicode{xb0} \)
μ = \(\sqrt {2}\)
\(\frac{\sin i}{\sin r_{1}}=\mu=\frac{\sin e}{\sin r_{2}}\)
\(\Rightarrow \ \frac{\sin 90^{\circ}}{\sin r_{2}}=\sqrt{2}\)
\(\text { i.e. } \ \sin r_{2}=\frac{1}{\sqrt{2}} \text { or } r_{2}=45^{\circ}\)
⇒ r₁ + r₂ = ∠A = 60\(\unicode{xb0} \)
r₁ = 60 - r₂ = 15\(\unicode{xb0} \)
\(\Rightarrow \ \frac{\sin i}{\sin 15^{\circ}}=\sqrt{2}\)
⇒ i = 21.47\(\unicode{xb0} \)

(ii) Given, magnification, m = 20
Magnification of eyepiece, m_e = 5
Least distance vision, D = 20 cm
Distance between the object and eyepiece,
L = 14 cm
We know that, magnification, m = m_e x m_o
\(\Rightarrow \quad m_{o}=\frac{m}{m_{e}}=\frac{20}{5}=4\)
As, \(m_{e}=1+\frac{D}{f_{e}}\)
where, fe is focal length of eyepiece.
\(\Rightarrow \quad 5=1+\frac{20}{f_{e}} \Rightarrow f_{e}=5 \mathrm{~cm}\)
Using lens formula for eyepiece,
\(\frac{1}{u_{e}}=\frac{-1}{20}-\frac{1}{5}=\frac{-5}{20}=\frac{-1}{4}\)
⇒ u_e = -4 cm (object distance for eyepiece)
⇒ L = v_o + | u_e |
⇒ v_o = L - | u_e |
⇒ v_o = L - | u_e |
= 14 - 4 = 10 cm
Magnification produced by object, m_o = \(-\frac{v_{o}}{u_{o}}\)
Object distance for object,
\(u_{o}=\frac{-v_{o}}{m_{o}}=\frac{-10}{4}=-2.5 \mathrm{~cm}\)
Using lens formula for object,
\(\frac{1}{f_{0}}=\frac{1}{v_{o}}-\frac{1}{u_{0}}=\frac{1}{10}-\frac{1}{-2.5}=\frac{1}{10}+\frac{1}{2.5}\)
f_o = 2 cm

(i) (d): Total internal reflection is the basis for following phenomenon:
(a) Sparkling of diamond.
(b) Optical fibre communication.
(c) Instrument used by doctors for endoscopy.
(ii) (d): Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met:The light is in the more denser medium and approaching the less denser medium.The angle of incidence is greater than the critical angle.
(iii) (c) : If incidence of angle, i = critical angle e, then angle of refraction, r = \(90^{\circ}\)
(iv) (b): In optical fibres, core is surrounded by cladding, where the refractive index of the material of the core is higher than that of cladding to bound the light rays inside the core.
(v) (b): From Snell's law, \(\sin C={ }_{1} n_{2}=\frac{v_{1}}{v_{2}}\)
where, c = critical angle = 30° and V₁ and V₂ are speed oflight in medium and vacuum, respectively.
We know that, v² = 3 x 108 m s^-l
\(\therefore \quad \sin 30^{\circ}=\frac{v_{1}}{3 \times 10^{8}} \)
\(\Rightarrow \quad v_{1}=3 \times 10^{8} \times \frac{1}{2} \Rightarrow v_{1}=1.5 \times 10^{8} \mathrm{~ms}^{-1} \)

(i) (b): Each half lens will form an image in the same plane. The optic axes of the lenses are displaced
\(\frac{1}{v}-\frac{1}{(-30)}=\frac{1}{20} ; v=60 \mathrm{~cm}\)
(ii) (a): Here \(f_{1}=20 \mathrm{~cm} ; f_{2}=?\)
F= 80 cm
\(\text { As } \frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{F} \Rightarrow \frac{1}{f_{2}}=\frac{1}{F}-\frac{1}{f_{1}}\)
\(\frac{1}{f_{2}}=\frac{1}{80}-\frac{1}{20}=\frac{-3}{80}\)
\(f_{2}=\frac{-80}{3}=-26.7 \mathrm{~cm}\)
(iii) (b): The bubble behaves libe a diverging lens
(iv) (b): Convex lens is used in magnifying glass.
(v) (d)

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(a): Due to the variation of the refractive index of the material of the lens, the focal length also varies accordingly. Now as white light is composed of different colours of light, each colour will produce its own image based on the focal length for that colour.This particular phenomenon for a single lens is known as chromatic aberration.