Question

In a thick glass slab of thickness $\ell$, and refractive index $n_1$, a cuboidal cavity of thickness $m$ is carved as shown in the figure and is filled with a liquid of refractive index $n_2$ $(n_1>n_2$). Find the ratio $\frac{l}{m}$, such that the net shift produced by this slab is zero when an observer $\text{A}$ , observes an object $\text{B}$ with paraxial rays is

• A. $\frac{n_1-n_2}{n_2-1}$
• B. $\frac{n_1-n_2}{n_2(n_1-1)}$
• C. $\frac{n_1-n_2}{n_1-1}$
• D. $\frac{n_1-n_2}{n_1(n_2-1)}$

Answer 1

The correct option is B

$\frac{n_1-n_2}{n_2(n_1-1)}$

Shift in object position due to cuboidal cavity,

${\delta }_1=m(1-\frac{1}{n_2})$

Shift in object position due to the glass slab

${\delta }_2=(\ell -m)(1-\frac{1}{n_1})$

From the data given in the question, Net shift is equal to zero.

$\therefore {\delta }_1+{\delta }_2=0$

$\Rightarrow m(1-\frac{1}{n_2})+(\ell -m)(1-\frac{1}{n_1})=0$

$\Rightarrow \ell -\frac{\ell }{n_1}=\frac{m}{n_2}-\frac{m}{n_1}$

$\Rightarrow \frac{\ell }{m}=\frac{(n_1-n_2)}{n_1{n}_2}\times \frac{n_1}{n_1-1}\Rightarrow \frac{\ell }{m}=\frac{n_1-n_2}{n_2(n_1-1)}$

Hence, option (b) is the correct answer.