Question

A narrow pencil of parallel light is incident normally on a solid transparent sphere of radius r. What should be the refractive index is the pencil is to be focussed (a) at the surface of the sphere, (b) at the centre of the sphere.

Answer 1

Given,
The radius of the transparent sphere = r
Refraction at convex surface.
As per the question,
u = −∞, μ₁ = 1, μ₂ = ?

(a) When image is to be focused on the surface,
Image distance (v) = 2r, Radius of curvature (R) = r
We know that,

\[\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\] 

\[ \Rightarrow \frac{\mu_2}{2r} - \left( \frac{1}{- \infty} \right) = \frac{\mu_2 - 1}{r}\] 

\[ \Rightarrow \frac{\mu_2}{2r} = \frac{\mu_2 - 1}{r}\] 

\[ \Rightarrow  \mu_2  = 2 \mu_2  - 2\] 

\[ \Rightarrow  \mu_2  = 2\]

(b) When the image is to be focused at the centre,
Image distance (v) = r, Radius of curvature (R) = r

\[\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\] 

\[ \Rightarrow \frac{\mu_2}{r} - \left( \frac{1}{- \infty} \right) = \frac{\mu_2 - 1}{r}\] 

\[ \Rightarrow \frac{\mu_2}{r} = \frac{\mu_2 - 1}{r}\] 

\[ \Rightarrow  \mu_2  =  \mu_2  - 1\] 
The above equation is impossible.
Hence, the image cannot be focused at centre.