Question

A thin lens made of a material of refractive index μ₂ has a medium of refractive index μ₁on one side and a medium of refractive index μ₃ on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium μ₁ and (b) from the medium μ₃?

Answer 1

Given,
A biconvex lens with two radii of curvature that have equal magnitude R.
Refractive index of the material of the lens is μ₂.
First medium of refractive index is μ_1.
Second medium of refractive index is μ₃.
As per the question, the light beams are travelling parallel to the principal axis of the lens.
i.e., u (object distance) = ∞

(a) The light beam is incident on the lens from first medium μ₁.
Thus, refraction takes place at first surface
Using equation of refraction,
\[\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\]
Where, v is the image distance.
Applying sign convention, we get:

\[\frac{\mu_2}{v} - \frac{\mu_1}{\left( - \infty \right)} = \frac{\mu_2 - \mu_1}{R}\] 

\[ \Rightarrow \frac{1}{v} = \frac{\mu_2 - \mu_1}{\mu_2 R}\] 

\[ \Rightarrow  \Rightarrow v = \frac{\mu_2 R}{\mu_2 - \mu_1}\] 
Now, refraction takes place at 2^nd surface

Thus, 

\[\frac{\mu_3}{v} - \frac{\mu_2}{u} = \frac{\mu_3 - \mu_2}{R}\] 
Here, the image distance of the previous case becomes object distance:

\[\frac{\mu_3}{v} =  - \left[ \frac{\mu_3 - \mu_2}{R} - \frac{\mu_2}{\mu_2 R}\left( \mu_2 - \mu_1 \right) \right]\] 

\[=  - \left[ \frac{\mu_3 - \mu_2 - \mu_2 + \mu_1}{R} \right]\] 

\[v =  - \left[ \frac{\mu_3 R}{\mu_3 - 2 \mu_2 + \mu_1} \right]\] 
Therefore, the image is formed at

\[\frac{\mu_3 R}{2 \mu_2 - \mu_1 - \mu_3}\]
(b) The light beam is incident on the lens from second medium μ₃.
Thus, refraction takes place at second surface.
Using equation of refraction,
\[\frac{\mu_2}{v} - \frac{\mu_3}{u} = \frac{\mu_2 - \mu_3}{R}\]
Where, v is the image distance
Applying sign convention, we get:

\[\frac{\mu_2}{v} - \frac{\mu_3}{\left( - \infty \right)} = \frac{\mu_2 - \mu_3}{R}\] 

\[ \Rightarrow \frac{1}{v} = \frac{\mu_2 - \mu_3}{\mu_2 R}\] 

\[ \Rightarrow  \Rightarrow v = \frac{\mu_2 R}{\mu_2 - \mu_3}\] 

Now, refraction takes place at 2nd surface.

Thus,

\[\frac{\mu_1}{v} - \frac{\mu_2}{u} = \frac{\mu_1 - \mu_2}{R}\] 
Here, the image distance of the previous case becomes object distance.

\[\frac{\mu_1}{v} =  - \left[ \frac{\mu_1 - \mu_2}{R} - \frac{\mu_2}{\mu_2 R}\left( \mu_2 - \mu_3 \right) \right]\] 

\[ =  - \left[ \frac{\mu_1 - \mu_2 - \mu_2 + \mu_3}{R} \right]\] 

\[v =  - \left[ \frac{\mu_1 R}{\mu_3 - 2 \mu_2 + \mu_1} \right]\] 
Therefore, the image is formed at \[\frac{\mu_1 R}{2 \mu_2 - \mu_1 - \mu_3}\]