(a) When the final image is formed at infinity, the telescope is said to be in normal adjustment and its magnifying power is given as
\(m=\frac{f_{o}}{-f_{e}}\)

The given figure shows the formation of an image by a refracting telescope in normal adjustment. In this arrangement, the separation between the objective and the eyepiece is (f_o + f_e).
(b) (i) It is not free from chromatic aberration.
(ii) Lenses oflarge aperture are big and heavy. So, they are difficult to make and support by their edges.
(c) Magnifying power of a telescope, \(m=\frac{f_{o}}{f_{e}}\)
So, for a higher magnifying power, the telescope must have an eyepiece of shorter focal length and an objective of larger focal length.

Given: u = -8 cm,f = +12 cm
\(\because \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{12}-\frac{1}{8} \Rightarrow v=-24 \mathrm{~cm} \)
\((i) m=\frac{v}{u}=\frac{-24}{-8}=3 \)
\((ii) A_{o}=1 \mathrm{~mm}^{2}\)
∵ Areal magnification \(=\frac{A_{I}}{A_{o}}=m^{2}\)
\(\Rightarrow A_{I}=A_{o} \times m^{2}=1 \mathrm{~mm}^{2} \times(3)^{2}=9 \mathrm{~mm}^{2}\)
∴ The area of each square in the virtual image is equal to 9 mm².

From Snell's law,
\(\frac{\sin 30^{\circ}}{\sin e}=\frac{1}{\mu}\)

\( \Rightarrow \sin e =\mu \sin 30^{\circ}=\sqrt{3} \times \frac{1}{2} \)
\(\therefore \quad e =\sin ^{-1}\left(\sqrt{\frac{3}{2}}\right)=60^{\circ} \)

The refractive index of a transparent medium decreases with increase in wavelength of the incident light.
We have \(\mu_{g a}=\frac{v_{a}}{v_{g}}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=\frac{3}{2}=1.5\)
\( \therefore \quad \mu_{g a}=\frac{1}{\sin i_{c}} \)
\(\Rightarrow i_{c}=\sin ^{-1}\left(\frac{1}{\mu_{g a}}\right) \)
\(\Rightarrow \quad i_{c}=\sin ^{-1}\left(\frac{2}{3}\right)=41.8\)

Given: \(|u|=f \pm \frac{f}{N}=f\left(\frac{N \pm 1}{N}\right)\)
As we know \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\( \therefore v=f(N \pm 1) \)
\(\text { and } |m|=\left|\frac{v}{u}\right|=N \)
\(\therefore \quad f=\frac{100}{2.5}=40 \mathrm{~cm} \quad(\because P=2.5 \mathrm{D}) \)
\(\therefore \quad m=\frac{v}{u}=N \)
\( \Rightarrow u=\frac{v}{N}=\frac{f(N \pm 1)}{N}=f+\frac{f}{N} \)
\(\text { and } u=f-\frac{f}{N} \quad(\because N=4) \)
\(\text { i.e. } u=40+\frac{40}{4}=50 \mathrm{~cm} \)
\(\text { or } u=40-\frac{40}{4}=30 \mathrm{~cm} \)

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

For compound microscope, 10 = 4 cm, f_e = 10 cm,
u_o = - 6 cm, v_e = D = -25 cm
For objective lens, \(\frac{1}{f_{0}} \equiv \frac{1}{v_{o}}-\frac{1}{u_{o}} \Rightarrow \frac{1}{4}=\frac{1}{v_{o}}+\frac{1}{6}\)
\(\Rightarrow \quad \frac{1}{v_{o}}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12} \Rightarrow v_{o}=12 \mathrm{~cm}\)
∴ magnifying power, m = -2 \(-\left(\frac{v_{o}}{u_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)
\(=-\left(\frac{12}{6}\right)\left(1+\frac{25}{10}\right)=-2\left(\frac{7}{2}\right)=-7\)
Length of microscope = | v_o | + | u_e |
where, v_o = 12 cm
for eye lens, v_e = -25 cm, f_e = 10 cm, u_e = ?
\( \therefore \quad \frac{1}{f_{e}}=\frac{1}{v_{e}}-\frac{1}{u_{e}} \)
\(\Rightarrow \quad \frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}=\frac{1}{-25}-\frac{1}{10} \)
\(\Rightarrow \quad \frac{1}{u_{e}}=\frac{-2-5}{50}=-\frac{7}{50}\)
⇒ u_e = -7.14 cm
∴ Length of microscope = | v_o | + | u_e |
= 12 + 7.14 = 19.14 cm

For lens \(L_{1}, \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Given, u = - 15 cm, f = +10 cm, v =?
\(\therefore \quad \frac{1}{10}=\frac{1}{v}+\frac{1}{15} \Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15} \Rightarrow \frac{1}{v}=\frac{1}{30}\)
Distance of image from lens L₁, v = 30 cm
For lens L₃, \(\frac{1}{f^{\prime \prime}}=\frac{1}{v^{\prime \prime}}-\frac{1}{u^{\prime \prime}}\)
Distance of image from lens L₃, v" = 10 cm
\(\therefore \quad \frac{1}{10}=\frac{1}{10}-\frac{1}{u^{\prime \prime}} \Rightarrow \frac{1}{u^{\prime \prime}}=0 \Rightarrow u^{\prime \prime}=\infty\)
The refracted rays from lens L₂ becomes parallel to principal axis. It is possible only when image formed by L₁ lies at first focus of L₂ , i.e. at a distance of 10 cm from L₂
Separation between L₁ and L₂ = 30 + 10 = 40 cm
The distance between L₂ and L₃ may take any value.

The diameter of objective of the telescope
= 150 x 10^-3 m, f_o = 4 m
f_e = 25 x 10^-3 m and D = 0.25 m
Magnifying power, m = \(-\frac{f_{0}}{f_{e}}\left(1+\frac{D}{f_{e}}\right)\)
\(=-\frac{4}{25 \times 10^{-3}}\left(1+\frac{0.25}{25 \times 10^{-3}}\right)=-1760\)
Now, \(d \theta=\frac{1.22 \lambda}{D}=\frac{1.22 \times 6 \times 10^{-7}}{0.25}\)
= 2.9 x 10^-6 rad
∴ Resolving power  \(=\frac{1}{d \theta}=\frac{1}{2.9 \times 10^{-6}}\)
= 0.34 x 10⁶

First measurement gives the focal length (f_eq = x) combination of the convex lens and the plano-convex liquid lens. Second measurement gives the focal length (f₁ = y) of the convex lens.
Focal length (f₂) of plano-convex lens is given by
\(\frac{1}{f_{2}}=\frac{1}{f_{\mathrm{eq}}}-\frac{1}{f_{1}}=\frac{1}{x}-\frac{1}{y}\)
\(\Rightarrow \quad f_{2}=\frac{x y}{y-x}\)  .........(i)
For equiconvex glass lens using Lens Maker's formula, we get
\(\frac{1}{f_{1}}=\left(n_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\frac{1}{y}=(1.5-1)\left(\frac{2}{R}\right)\)
(As R₁ = R and R₂ = -R)
\(\Rightarrow \quad \frac{1}{y}=\frac{1}{2} \times \frac{2}{R} \Rightarrow R=y\)

Now, we apply lens Maker's formula for plano-convex lens.
Here, R₁ = R and R₂ = \(\infty\) and let n_l = refractive index of liquid.
\(\frac{1}{f_{2}}=(n_{l}-1)\left ( \frac{1}{R}-\frac{1}{\infty } \right )\)
\(\Rightarrow \frac{1}{f_{2}}=(n_{l}-1)\left ( \frac{1}{R} \right )\)
\(\Rightarrow n_{l}=1+\frac{R}{f_{2}}=1+\frac{y}{\frac{xy}{y-x}}\) {using Eq. (i)}
\(=1+\frac{y-x}{x}=\frac{y}{x}\)

Distance of object from mirror
\(=15+\frac{33.25}{4} \times 3=39.93 \mathrm{~cm}\)
Distance of image from the mirror
\(=15+\frac{25}{4} \times 3=33.75 \mathrm{~cm}\)
Using mirror formula, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\Rightarrow \frac{1}{-33.75}-\frac{1}{39.93}=\frac{1}{f}\)
∴ f = -18.3 cm

Any ray entering at an angle i shall be guided along AC, if the ray makes an angle Φ with the face AC greater than the critical angle as per the principle of total internal reflection, Φ + r = 90°, therefore sin Φ = cos r.
\(\Rightarrow \quad \sin \phi \geq \frac{1}{\mu} \Rightarrow \cos r \geq \frac{1}{\mu}\)

\(\text { or } 1-\cos ^{2} r \leq 1-\frac{1}{\mu^{2}} \Rightarrow \sin ^{2} r \leq 1-\frac{1}{\mu^{2}}\)  [∵ 1 - cos² r = sin² r]
Since sin i = μ sin r
\(\frac{1}{\mu^{2}} \sin ^{2} i \leq 1-\frac{1}{\mu^{2}}\)
or sin² i > μ² -1
When i = \(\frac{\pi}{2}\) then we have smallest angle Φ.
If the angle Ф is greater than the critical angle, then all other angles of incidence shall be more than the critical angle.
Thus, 1 < μ² -1 or μ² > 2
\(\Rightarrow\) μ > \(\sqrt{2}\)
This is the required result.

Given, velocity of light in glass, v = 2 x 10⁸ m/s
Velocity of light in air, c = 3 x 10⁸ m/s
∴ Refractive index of glass with respect to air,
\({ }^{a} \mu_{g}=\frac{c}{v}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=1.5\)
∴ Normal shift in the position of ink dot,
\( d =t\left(1-\frac{1}{{ }^{a} \mu_{g}}\right) \quad[\because t=6 \mathrm{~cm}] \)
\(=6\left(1-\frac{1}{1.5}\right)=\frac{6 \times 0.5}{1.5}=2 \mathrm{~cm} \)

(i) The ray diagram for the formation of the image of the mobile phone is shown below. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e. B'C = BC

(ii) We may think that the image will now show only half of the object, but considering the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object.

However, as the area of the reflecting surface has been reduced, the intensity of the image will be low, i.e. half.

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 \(n=\frac{\mu}{\sqrt{2}}\)

\( \text { Given: } \angle A=60^{\circ}, \angle i=\angle e, \angle i=\frac{3}{4} \angle A=45^{\circ} \text { and } \)
\(\angle r=\frac{\angle A}{2}=30^{\circ} \)
\(\mu=\frac{\sin i}{\sin r}=\frac{2}{\sqrt{2}}=1.414\)

Given: D = 80 cm, d = 10 cm
\( \therefore \quad f =\frac{D^{2}-d^{2}}{4 D} \)
\(=\frac{6400-100}{320}=\frac{6300}{320}=19.7 \mathrm{~cm} \)

It is a diverging (concave lens) and the refracted rays are bending away from the principal axis.

Total internal reflection.

The spectrum of visible light whose frequency is less than that of green light will come out of the water into the air.

(i) Refractive index is minimum for vacuum (μ = 1).
(ii) Refractive index is maximum for diamond.

A ray of light which is incident normally on a mirror, is reflected along the same path, i.e. the angle of incidence as well as the angle of reflection is zero.

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

Given, f_o = 1.25 cm, f. = - 5 cm
Magnification, m = 30 , D = 25 cm
If the object is very close to the principal focus of the objective and the image formed by the objective is very close to eyepiece, then magnifying power of a microscope is given by
\( m=-\frac{L}{f_{0}} \cdot \frac{D}{f_{e}} \)
\(\Rightarrow 30=\frac{L}{1.25} \cdot \frac{25}{5} \)
\(\Rightarrow L=\frac{125 \times 30 \times 5}{25 \times 100} \)
\(\Rightarrow L=\frac{25 \times 30}{100} \Rightarrow L=\frac{30}{4} \)
⇒  L = 7.5 cm
This is a required separation between the objective and the eyepiece.

The least distance of distinct vision of an average person, (i.e. D) is 25 cm, in order to view an object with magnification of 10.
Here, v = D = 25 cm and u = f
But the magnification, m = v/u = D/f
\(\therefore \quad m=\frac{D}{f}\)
\(\Rightarrow \quad f=\frac{D}{m}=\frac{25}{10}=25=0.025 \mathrm{~m}\)
and \(P=\frac{1}{0.025}=40 \mathrm{D}\)  \(\left[\because P=\frac{1}{f}\right]\)
This is the required power of lens.

(i) i = 52\(\unicode{xb0} \), when prism is adjusted at an angle of minimum deviation, then angle of incidence is equal to the angle of emergence.
Hence, r = 0
This ray pass unrefracted at AC interface and reaches AB interface. Here, we can see angle of incidence becomes 30\(\unicode{xb0} \).
Thus, applying snell's law, \(\frac{\sin 30^{\circ}}{\sin e}=\frac{\mu_{a}}{\mu_{g}}=\frac{1}{\sqrt{3}}\)
\(\sin e=\sqrt{3} \times \sin 30^{\circ}=\frac{\sqrt{3}}{2}\)
Thus, e = 60\(\unicode{xb0} \)

When a glass lens (\(\mu\)_g = 1.5) is immersed in water (\(\mu\)_w = 1.33), then focal length of the lens increases but its nature remains unchanged. due to increase in focal length, its power reduces.

Here, \(\mu=\frac{\text { real depth } / \text { thickness of object }}{\text { apparent depth }}\)
Now, due to refraction at two different boundaries, the apparent depth of object,
apparent depth = \(\frac{\text { thickness of glass }}{\mu_{\text {glass }}}+\frac{\text { thickness of water }}{\mu_{\text {water }}}\)
\(=\frac{6}{1.5}+\frac{4}{1.3}=3+4=7 \mathrm{~cm}\)

Snell's law of refraction fails in two situations
(i) When both media have same value of refractive index.
(ii) When light is incident normally on a surface, as i = 0, r = 0.