(i) Derive the mathematical relation between refractive indices n₁ and n₂ of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n₁ and a real image formed in the denser medium of refractive index n₂. Hence, derive lens maker's formula.

(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed?

OR

(a) Draw a labeled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.

(b) You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.

(i) Which lenses should he used as objective and eyepiece? Justify your answer.

(ii) Why is the aperture of the objective preferred to be large?

(i) We are given a convex spherical surface with a radius of curvature R and the refractive indices of two media as n₁ and n₂.

We can write

tan∠SPT = ; ;

We assume that the incident ray is very close to the principle axis , this cause the angles to be very small. Thus, we get,

tanx = x = sin x

Therefore,

i = ∠SPT + ∠SCT

⇒ ……(a)

Similarly, r = ∠SCT - ∠SIT

⇒ ……(b)

We use snell’s law i.e. n₁sini = n₂ sinr

And for small angles, n₁i = n₂r

On substituting these values from (a) and (b) we get,

⇒ n₁ = n₂

⇒ (n₁ / PT) + (n₂/TI) = (n₂ – n₁) /MC

We apply the Cartesian sign conventions to get,

PT = -u ; TI = + v ; TC = + R

On substituting these values we get,

For the derivation of Len’s maker formula, we make the following assumptions:

(i) The lens is thin .So when we compare the distances measured from the pole of its two surfaces , it is equal to the distance from its optical center.

(ii) The lens’ aperture is small.

(iii) The object is taken as a point object placed at the principle axis of the lens.

(iv) The incident ray and the refracted ray extend very small angles with the principle axis.

From the first surface refraction, we have:

……….(i)

From the second surface refraction we have:

Since the less is so thin, we have d ≪ v’ And thus d can be ignored compared to v’

Thus we have,

……….(ii)

Adding both the equation (i) and (ii) we have:

Also, we put n₂/n₁ = n , the refractive index of the lens in comparison to the surrounding medium, we have :

If the object is placed at infinite distance i.e. u = ∞, from the lens, the image is formed at the principle focus of the lens, .i.e. v = f, then we have:

Therefore, we get:

This formula is known as the lens makers’ formula.

(ii) The lens makers’ formula to find the image distance can be given as:

Where,

n₂ is the refractive index of the lens

n₁ is the refractive index of the surrounding medium

R is the radius of curvature

‘v’ is the image distance and ‘u’ is the object distance.

Given,

Refractive index of lens, n₂ = 1.5,

Refractive index of the surrounding air,n₁ = 1

Radius of curvature, R = 20 cm

Distance of the object, u = -100 cm

Thus , we use the formula to find the image distance:

⇒

⇒

The image is thus formed at a distance of 100 cm from the optical center.

OR

(a) The real image formed by an astronomical telescope in normal adjustment position is given as:

Magnifying power: It is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at the eye.

(b) (i) The telescope with higher magnification have least power, thus the focal length is greater, f₀>f , So the objective lens with less power should be used, i.e. 0.5 D

(ii)The resolving power of the telescope increases with the larger aperture of the objective lens , as it receives a larger amount of light from the object kept at a distance.