(a)
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
\( \frac{1}{v}-\frac{1}{u} =\frac{1}{f} \)
\(\therefore \quad \text { Power } P =\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \)
(b) Given: Angle of prism, A = 60°,

Refractive index,  \(n=\sqrt{3}\)
In this case refracted ray is going parallel to base, hence it is the case of minimum deviation. We know
\( r_{1}+r_{2} =A \)
\(i+e =A+D \)       ........(i)
For minimum deviation r₁ = r₂, i = e and D = D_m
\( \therefore \quad 2 r =A \Rightarrow r=\frac{60}{2}=30^{\circ} \)
\(\because \quad \frac{\sin i}{\sin r}=\sqrt{3} \)
\(\Rightarrow \quad \sin i =\frac{\sqrt{3}}{2} \Rightarrow i=60^{\circ} \)
\(\text { From equation (i) } 2 i =A+D_{m} \Rightarrow D_{m}=60^{\circ}\)

(ii) Since, \(\angle\)A = 60°              [given]
\(\begin{aligned} \therefore \quad \mu & =\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \left(60^{\circ} / 2\right)} \\ \end{aligned}\)
\(\begin{aligned} =\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sqrt{3}}{2} \times \frac{2}{1}=\sqrt{3}=1.732 \end{aligned}\)

Focal length of the convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = -20 cm
Distance between the two lenses, d = 8.0 cm
(a) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
Where,
u₁ = Object distance = ∞
v₁ = Image distance
\(\frac { 1 }{ { v }_{ 1 } } =\frac { 1 }{ 30 } -\frac { 1 }{ \infty } =\frac { 1 }{ 30 } \)
∴ v₁ = 30 cm
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
Where,
u₂ = Object distance
= (30 - d) = 30 - 8 = 22 cm
v₂= Image distance
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ 22 } -\frac { 1 }{ 20 } =\frac { 10-11 }{ 220 } =\frac { -1 }{ 220 } \)
∴ v₂ = -220 cm
The parallel incident beam appears to diverge from a point that is \(\left( 220-\frac { d }{ 2 } =220-4 \right) 216\) cm from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula, we have:
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } +\frac { 1 }{ { u }_{ 2 } } \)

Where,
u₂ = Object distance = -∞
v₂ = Image distance
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ -20 } +\frac { 1 }{ -\infty } =-\frac { 1 }{ 20 } \)
∴ v₂ = -20 cm
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
Where,
u₁ = Object distance
= -(20 + d) = -(20 + 8) = -28 cm
v₁ = Image distance
\(\frac { 1 }{ { v }_{ 1 } } =\frac { 1 }{ 30 } +\frac { 1 }{ -28 } =\frac { 14-15 }{ 420 } =\frac { -1 }{ 420 } \)
∴ v₂ = - 420 cm
Hence, the parallel incident beam appear to diverge from a point that is (420 - 4) 416 cm from the left of the centre of the combination of the two lenses
The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.
(b) Height of the image, h₁ = 1.5 cm
Object distance from the side of the convex lens, u₁ = -40 cm
|u₁| = 40 cm
According to the lens formula:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
Where,
v₁ = Image distance
\(\frac { 1 }{ { v }_{ 1 } } =\frac { 1 }{ 30 } +\frac { 1 }{ -40 } =\frac { 4-3 }{ 120 } =\frac { 1 }{ 120 } \)
∴ v₁ = 120 cm
 \(m=\frac { { v }_{ 1 } }{ \left| { u }_{ 1 } \right| } \)
= \(\frac { 120 }{ 40 } =3\)
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula:
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
Where,
u₂ = Object distance
= +(120 - 8) = 112 cm.
v₂ = Image distance
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ -20 } +\frac { 1 }{ 112 } =\frac { -112+20 }{ 2240 } =\frac { -92 }{ 2240 } \)
∴ v₂ = \(\frac { 2240 }{ 92 } \) cm
Magnification, \({ m }^{ ' }=\left| \frac { { v }_{ 2 } }{ { u }_{ 2 } } \right| \)
\(=\frac { 2240 }{ 92 } \times \frac { 1 }{ 112 } =\frac { 20 }{ 92 } \)
Hence, the magnification due to the concave lens is \(\frac { 20 }{ 92 } \)
The magnification produced by the combination of the two lenses is calculated as:
m x m'
\(=3\times \frac { 20 }{ 92 } =\frac { 60 }{ 92 } =0.652\)
The magnification of the combination is given as:
\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } =0.652\)
h₂ = 0.652 x h₁
Where,
h₁ = Object size = 1.5 cm
h₂ = Size of the image
∴  h₂ = 0.652 x 1.5 = 0.98 cm
Hence, the height of the image is 0.98 cm

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 final image, formed by the combination, is coinciding with the object itself. This implies that the rays, from the object, are retracing their path, after refraction from the lens and reflection from the mirror.
The (refracted) rays are, therefore, falling normally on the mirror. It follows that the rays AB, and A'B' when produced, are meeting at the centre of curvature, C of the mirror. Hence, O₂O = 20 cm, i.e. the radius of curvature of the
mirror.
From the figure, we then see that for the convex lens, u = - 25 cm and v = + (30 + 20) cm = + 50 cm. from the focal length of the lens, we have
\( \frac{1}{50}-\frac{1}{(-25)} =\frac{1}{f} \Rightarrow \frac{1}{f}=\frac{1+2}{50} f =\frac{50}{3} \mathrm{~cm}=16.67 \mathrm{~cm} \)

If a point source S is placed at the common focus of both the mirrors, then the rays after reflection from mirror 1 will become parallel to the principal axis. When these parallel rays fall on mirror 2, the rays will get focussed at S.
\(\therefore \quad d=2 f\)

If a source S is placed at the centre of curvature of the two mirrors, the image of the source will be formed at C only.

∴ d = 4 f, where f = focal length of each mirror.

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

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,
\( d =t\left(1-\frac{1}{{ }^{a} \mu_{g}}\right) \quad[\because t=6 \mathrm{~cm}] \)
\(=6\left(1-\frac{1}{1.5}\right)=\frac{6 \times 0.5}{1.5}=2 \mathrm{~cm} \)

For a convex lens of focal length f₁,
\(\frac{1}{f_{1}}=\frac{1}{v^{\prime}}-\frac{1}{u}\)       ...........(i)
For a concave lens of focal length f₂, 
\(\frac{1}{-f_{2}}=\frac{1}{v}-\frac{1}{v^{\prime}}\)            .............(ii)
The image of the 1st lens acts as a virtual object for the 2nd lens. From equations (i) and (ii), we get
\(\frac{1}{f_{1}}-\frac{1}{f_{2}}=\frac{1}{v^{\prime}}-\frac{1}{u}+\frac{1}{v}-\frac{1}{v^{\prime}}\)

\( \Rightarrow \ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(\text {Thus, } \frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{-f_{2}} \Rightarrow F=\frac{f f_{2}}{f_{2}-f_{1}} \)

\(\mathrm{m}=1+\frac{D}{f}\)
\(\because f_V<f_R\)
So, the magnifying power is larger when the object is seen in violet light.

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

When a lens is placed in a liquid, where refractive index is more than that of the material of lens, then the nature of the lens changes. So, when a biconvex lens of refractive index 1.25 is immersed in water (refractive index 1.33), i.e. in the liquid of higher refractive index, its nature will change. So, biconvex lens will act as converging or diverging lens.

Since, light travels in vacuum with a constant velocity, i.e. 3 x 10⁸ m/s, hence ratio of velocities of all wavelengths remains same.