Question 377: Ray Optics

Question

A transparent thin film of uniform thickness and refractive index n
1
= 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n
2
= 1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f
1
from the film, while rays of light traversing from glass to air get focused at distance f
2
from the film. Then
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ky4odbpq/e6daa16d-70e6-4172-b782-fd4ecf06701e/9f772ae0-6fdf-11ec-bb8f-b3be3f2269a8/file-1ky4odbpr.png?format=png)

Accepted Answer

As the film has R
1
= R
2
= R (say), its focal length
${1 \over f} = ({n_1} - 1)\left( {{1 \over R} - {1 \over R}} \right)$
${1 \over f} = 0$ or $f = \infty $. It will not cause refraction.
Refraction will take place at air and glass cylinder. Given u = $\infty$
$ - {{{n_1}} \over u} + {{{n_2}} \over v} = {{{n_2} - {n_1}} \over R}$
$ - {{{n_1}} \over \infty } + {1.5 \over {{f_1}}} = {{1.5 - 1} \over R}$ or ${f_1} = 3R$
For glass-air refraction, we have
$ - {{{n_2}} \over u} + {{{n_1}} \over v} = {{{n_1} - {n_2}} \over R}$
$ - {{{n_2}} \over \infty } + {1 \over {{b_2}}} = {{1 - 1.5} \over R}$
or ${f_2} =  - 2R$ or $\left| {{f_2}} \right| = 2R$.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ky4ofhpk/b4a20297-f3b7-40f8-804e-bf5cfdadeaa0/dbb70e80-6fdf-11ec-bb8f-b3be3f2269a8/file-1ky4ofhpl.png?format=png)

Answer

$\left| {{f_1}} \right| = 3R$

Answer

$\left| {{f_1}} \right| = 2.8R$

Answer

$\left| {{f_2}} \right| = 2R$

Answer

$\left| {{f_2}} \right| = 1.4R$

Ray Optics

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