Focal length of the objective lens, f₁ = 2.0 cm
Focal length of the eyepiece, f₂ = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision,  d' = 25 cm
∴ Image distance for the eyepiece, v₂ = -25 cm
Object distance for the eyepiece = u₂
According to the lens formula, we have the relation
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { f }_{ 2 } } \)
= \(\frac { 1 }{ -25 } -\frac { 1 }{ 6.25 } =\frac { -1-4 }{ 25 } =\frac { -5 }{ 25 } \)
∴ u₂ = -5 cm
Image distance for the objective lens,
v₁ = d + u₂ = 15 - 5 = 10 cm
Object distance for the objective lens = u₁
According to the lens formula, we have the relation:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
\(\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { f }_{ 1 } } \)
= \(\frac { 1 }{ 10 } -\frac { 1 }{ 2 } =\frac { 1-5 }{ 10 } =\frac { -4 }{ 10 } \)
\(\therefore { u }_{ 1 }=\)-2.5 cm
Magnitude of the object distance, |u₁| = 2.5 cm
The magnifying power of a compound microscope is given by the relation:
\(m=\frac { { v }_{ 1 } }{ \left| { u }_{ 1 } \right| } \left( 1+\frac { { d }^{ ' } }{ { f }_{ 2 } } \right) \)
= \(\frac { 10 }{ 2.5 } \left( 1+\frac { 25 }{ 6.25 } \right) =4(1+4)=20\)
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
∴ Image distance for the eyepiece, v₂ = ∞
Object distance for the eyepiece = u₂
According to the lens formula, we have the relation:
\(\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ \infty } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ 6.25 } \)
∴ u₂ = -6.25 cm
Image distance for the objective lens,
v₁ = d + u₂ = 15 - 6.25 = 8.75 cm
Object distance for the objective lens = u₁
According to the lens formula, we have the relation:
\(\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { f }_{ 1 } } \)
\(\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { f }_{ 1 } } \)
\(=\frac { 1 }{ 8.75 } -\frac { 1 }{ 2.0 } =\frac { 2-8.75 }{ 17.5 } \)
Magnitude of the object distance,  |u₁| = 2.59 cm
The magnifying power of a compound microscope is given by the relation:
\(m=\frac { { v }_{ 1 } }{ \left| { u }_{ 1 } \right| } \left( \frac { { d }^{ ' } }{ \left| { u }_{ 2 } \right| } \right) \)
\(=\frac { 8.75 }{ 2.59 } \times \frac { 25 }{ 6.25 } =13.51\)
Hence, the magnifying power of the microscope is 13.51.

Angle of minimum deviation, \({ \delta }^{ ' }_{ m }\) = 40°
Angle of the prism, A = 60°
Refractive index of water, µ = 1.33
Refractive index of the material of the prism = µ'
The angle of deviation is related to refractive index (µ') as:
\({ \mu }^{ ' }=\frac { sin\frac { (A+{ \delta }_{ m }) }{ 2 } }{ sin\frac { A }{ 2 } } \)
\(=\frac { sin\frac { ({ 60 }^{ o }+{ 40 }^{ o }) }{ 2 } }{ sin\frac { { 60 }^{ o } }{ 2 } } \)\(=\frac { sin{ 50 }^{ o } }{ sin{ 30 }^{ o } } =1.532\)
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is given by the relation:
\({ \mu }_{ g }^{ w }=\frac { { \mu }^{ ' } }{ \mu } =\frac { sin\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } }{ sin\frac { A }{ 2 } } \)
=\(sin\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } =\frac { { \mu }^{ ' } }{ \mu } sin\frac { A }{ 2 } \)
= \(sin\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } =\frac { 1.532 }{ 1.33 } \times sin\frac { { 60 }^{ o } }{ 2 } =\)0.5759
= \(\frac { (A+{ \delta }^{ ' }_{ m }) }{ 2 } ={ sin }^{ -1 }{ 0.5759=35.16 }^{ o }\)o
= \({ 60 }^{ o }+{ \delta }^{ ' }_{ m }={ 70.32 }^{ o }\)
\(\therefore { \delta }^{ ' }_{ m }={ 70.32 }^{ o }-{ 60 }^{ o }={ 10.32 }^{ o }\)
Hence, the new minimum angle of deviation is 10.32°.

Image formed by the first lens
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { -30 } } =\frac { 1 }{ 10 } \)
or v₁ = 15 cm
The image formed by the first lens serves as the object for the second.
This is at a distance of (15 - 5) cm = 10 cm to the right of the second lens. Though the image is real, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens.
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ 10 } =\frac { 1 }{ -10 } \)
or v₂ = ∞
The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.
\(\frac { 1 }{ { v }_{ 3 } } -\frac { 1 }{ { u }_{ 3 } } =\frac { 1 }{ { f }_{ 3 } } \)
or \(\frac { 1 }{ { v }_{ 3 } } =\frac { 1 }{ \infty } +\frac { 1 }{ 30 } \)
or v₃ = 30 cm
The final image is formed 30 cm to the right of the third lens.

In the given diagram,

OP is the incident ray, which makes the angle i₁ normal, and \(\angle N ' Q R\) is the angle of emergence, which is represented by \(\mathrm{i}_2\).
A is the prism angle, and \(\mu\) is the refractive index of the prism.
A = Prism angle, \(\delta=\) Angle of deviation, i₁ = Angle of incidence,i₂= Angle of emergence .
In the case of minimum deviation,
\( \angle r_1=\angle r_2=\angle r\)
\(\mathrm{~A}=\angle r_1+\angle r_2\)
So,\(A=\angle r+\angle r=\angle 2 r\)
\(\angle \mathrm{r}=\frac{A}{2}\)
Now, again
\(A+\delta=i_1+i_2 \ldots\left(\because\right.\) In the case of minimum deviation \(i_1=i_2=i\) and \(\left.\delta=\delta_m\right)\)
So, \(A+\delta_m=i+i=2 i\)
Now, \(i=\frac{\left(A+\delta_m\right)}{2}\)
Now, from Snell's rule,
\(\mu=\frac{\sin i}{\sin r}\)
\(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\)

The angle of incidence corresponding to an angle of refraction of \(90^{\circ}\) is called the critical angle for the given pair of media. If the angle of incidence of light, when travelling from a denser medium to a rarer medium, is greater than the critical angle then total internal reflection takes place.
Let the angle of incidence i and C be the critical angle =C.
Let the angle of refraction \(r=90^{\circ}\).
The refractive index of the rarer medium is \(\mu_{\mathrm{a}}\).
The refractive index of the denser medium is \(\mu_{\mathrm{b}}\).
Applying Snell's law, \(\frac{\sin i}{\sin r}=\frac{\mu_a}{\mu_b} \)
\(\mu_{\mathrm{b}} \sin C=\mu_{\mathrm{a}} \sin 90^{\circ} \ldots\left[\because \mathrm{i}=\mathrm{C} \text { and } \mathrm{r}=90^{\circ}\right]\)
\(\frac{\mu_a}{\mu_b}=\frac{1}{\sin C}\)
Thus, we arrive at a formula expressing the critical angle and refractive index relation:
\({ }_a \mu_b=\frac{1}{\sin C}\)

(i) According to the mirror equation, we have
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
where, U = distance of the object from the mirror, v = distance of the image from the mirror and f = focal length of the mirror.
From the mirror equation, we have
\(v=\frac{u f}{u-f}\)  ..........(i)
Applying new cartesian sign convention, we get
f = - ve and u = - ve
Given, f < u < 2f
\(\Rightarrow\) v = - ve        [from Eq. (i)]
Magnification is given by m = \(-\left(\frac{-v}{-u}\right)=-v e\)
Hence, the image formed is real.
From the mirror formula, whenu = - 2f
\( \Rightarrow \frac{1}{-2 f}+\frac{1}{v}=\frac{1}{-f} \)
\(\Rightarrow \frac{1}{v}=\frac{1}{2 f}-\frac{1}{f}=\frac{-1}{2 f} \)
When the object is at f, then image is formed at infinity.
This shows that when f < u < 2f, then \(\infty\) > v > 2f.
(ii) For convex mirror, f > 0
Also, u < 0
But from mirror equation,
\( \frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{v}+\frac{1}{|-u|} \)  [taking u with sign]
\(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)
If f and u to be positive. then
\(\frac{1}{v}>0 \Rightarrow v>0\)
Hence, virtual image is formed.
(iii) For concave mirror,
f < 0, u < 0, | f | > | u | > 0
But from mirror equation,
\( \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \quad \Rightarrow \frac{-1}{|f|}=\frac{1}{v}-\frac{1}{|u|} \)
∵ \(\frac{1}{v}=\frac{1}{|u|}-\frac{1}{|f|} \)
⇒ \(|v|<|f| \quad \Rightarrow \frac{1}{|u|}>\frac{1}{|f|} \)
\(\frac{1}{v}>0 \Rightarrow v>0 \)
Image is formed on RHS of mirror; i.e. virtual image.
Also, \(\frac{1}{f}=\frac{1}{|v|}-\frac{1}{|u|}\)
For concave mirror, f is negative.
\(\Rightarrow \quad \frac{1}{|v|}<\frac{1}{|u|} \quad \Rightarrow\frac{|v|}{|u|}>1 \Rightarrow m>1\)
Enlarged virtual image formed on the other side of mirror.

Given, object distance, u = -3 cm
Radius of curvature, R = -24 cm
∴ \(f=\frac{R}{2}=\frac{-24}{2}=-12 \mathrm{~cm}\)
According to mirror formula,
\( \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \)
⇒ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{(-12)}-\frac{1}{-3} \)
⇒ \(\frac{1}{v}=\frac{1}{-12}+\frac{1}{3}=\frac{-1+4}{12} \)
⇒  \(v=4 \mathrm{~cm}\)
∵ Magnifiiccaatitoin, m \(=-\frac{v}{u}=\frac{-4}{-3}=+1.33\)
i.e. The image formed is virtual, erect and magnified.

Focal length of the objective lens, f_o = 144 cm
Focal length of the eyepiece, f_e = 6.0 cm
The magnifying power of the telescope is given as:
m = \(\frac{f_o}{f_c}\)
= \(\frac{144}{6}\) = 24
The separation between the objective lens and the eyepiece is calculated as:
f_o + f_e
= 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

Refractive index of glass, μ
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = R₁
Radius of curvature of the other face of the lens = R₂
Radius of curvature of the double-convex lens = R The value of R can be calculated as:
∴ R₁ = R and R₂ = -R
The value of R can be calculated as:
\(\frac { 1 }{ f } =(\mu -1)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
\(\frac { 1 }{ 20 } =(1.55)\left[ \frac { 1 }{ R } +\frac { 1 }{ R } \right] \)
\(\frac { 1 }{ 20 } =0.55\times \frac { 2 }{ R } \)
∴ R = 0.55 x 2 x 20 = 22 cm
Hence, the radius of curvature of the double-convex lens is 22 cm.

Power of thin lens,
P = + 5D
\(\mu_g=15 \)
Focal length of thin lens, in air,
\(f_{\text {air }}=\frac{1}{P}=\frac{1}{5}\) = 0.2m = 20cm
By using lens Maker formula,
\( \frac{1}{f_{\text {air }}} =\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
\(=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \frac{1}{f_{\text {air }}} =\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
when lens is dipped in liquied, then focal length in liquid.
\(f_l=-100 \mathrm{~cm}\)
Using lens Maker's formula,
\(\frac{1}{f_l}=\left(h \mathrm{k}_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
\( =\left(\frac{\mu_g}{\mu_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
\(\frac{1}{100}=\left(\frac{15-\mu_l}{\mu_l}\right) \cdot \frac{2}{f_{\text {atr }}} \)
\(\Rightarrow \frac{1}{-100}=\frac{15-\mu_i}{\mu_l} \cdot \frac{2}{20} \)
\(-\frac{1}{10}=\frac{15-\mu_l}{\mu_l} \Rightarrow 15-10 \mu_l=\mu_l \)
\( \Rightarrow \quad \mu_t=\frac{15}{9} \equiv \frac{5}{3} \text { [from Eq. (i)] } \)

Given \(i=\theta, r=\frac{\theta}{2}\)
\( \because \quad \frac{\sin i}{\sin r}=\frac{n_2}{n_1} \\ \text { i.e. } \quad \sin r<\sin i \Rightarrow n_2>n_1 \)
Hence, 2^nd medium is optically denser.

(i) When the final image is at infinity,
\(m=-\frac{f_{0}}{f_{e}}=\frac{-20}{5}\)
⇒ m = -4
(ii) When the final image is at 25 cm from the eye,
i.e. D = 25 cm
\(m=\frac{-f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)=\frac{-20}{5}\left(1+\frac{5}{25}\right)\)
⇒ m = -4.8

When lens is immersed in a liquid, then
\(\frac{1}{f_{L}}=\left({ }^{L} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
where, ^lμ_g = refractive index of lens material (glass) w.r.t. liquid.
\(\therefore \quad \frac{\mu_{g}}{\mu_{L}}=\frac{1.5}{1.5}=1\)
Hence, \(\frac{1}{f_{l}}=(1-1)\left(\frac{l}{R_{1}}-\frac{1}{R_{2}}\right)=0 \Rightarrow f_{L}=\infty\)

We know that, \(\mu=\frac{1}{\sin C}\)
\( \Rightarrow \ \sin C=\frac{1}{\mu}=\frac{1}{\sqrt{2}} \\ \therefore C=45 \)

The distance of normal vision is 25 cm. So if a book is at u = -25 cm, its image should be formed at v = -50 cm. Therefore, the desired focal length is given by
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\text { or } \frac{1}{f}=\frac{1}{-50}-\frac{1}{-25}=\frac{1}{50}\)
or f = + 50 cm (convex lens).

(a): Microscope is an optical instrument which forms a magnified image of a small nearby object and thus, increases the visual angle sub tended by the image at the eye so that the object is seen to be bigger and distinct. Therefore, angular magnification for image is more than object.