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 a distance $f_1$ from the film, while rays of light traversing from glass to air get focused at a distance $f_2$ from the film. Then -

• A. $|f_1|=3R$ and $|f_2|=2R$
• B. $|f_1|=2R$ and $|f_2|=3R$
• C. $|f_1|=4R$ and $|f_2|=5R$
• D. $|f_1|=5R$ and $|f_2|=4R$

Answer 1

The correct option is A $|f_1|=3R$ and $|f_2|=2R$


We know that for refraction at curved surface,

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

For air to glass :

At air-film surface -

$\frac{n_1}{v}-\frac{1}{\infty }=\frac{n_1-1}{R}...\left(1\right)$

At film-glass surface -

$\frac{n_2}{f_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

$\Rightarrow \frac{n_2}{f_1}-\frac{n_1}{v}=\frac{n_2-n_1}{R}[\text{Here}u=v]...\left(2\right)$

On adding equations $\left(1\right)$ and $\left(2\right)$, we get,

$\frac{n_2}{f_1}=\frac{n_1-1}{R}+\frac{n_2-n_1}{R}$

$\Rightarrow \frac{1.5}{f_1}=\frac{1.4-1}{R}+\frac{1.5-1.4}{R}$

$\Rightarrow f_1=3R$

Similarly, for glass to air,

$\frac{1}{f_2}=\frac{n_1-n_2}{-R}+\frac{1-n_1}{-R}$

$\Rightarrow \frac{1}{f_2}=\frac{1.4-1.5}{-R}+\frac{1-1.4}{-R}$

$\Rightarrow f_2=2R$

Hence, option $\left(A\right)$ is the correct answer.