Question

In a thick glass slab of thickness $l$, 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 $n_2(n_1>n_2)$. The ratio $\frac{l}{m}$, so that shift produced by this slab is zero when an observer $A$ observes an object $B$ with paraxial rays is

• A. $\frac{n_1-n_2}{n_2-n_1}$
• 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 = thickness $\times \frac{1}{\text{R.I of the slab}}$

Shift due to length $x=x(1-\frac{1}{n_1})$

Shift due to length $l-m-x=(l-m-x)(1-\frac{1}{n_1})$

Shift due to length $m=m(1-\frac{1}{n_2})$

Sum of the shifts should be zero

$x(1-\frac{1}{n_1})+(l-m-x)(1-\frac{1}{n_1})+m(1-\frac{1}{n_2})=0$

$\Rightarrow (m-l)(1-\frac{1}{n_1})=m(1-\frac{1}{n_2})$

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

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

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