Question

The figure shows an arrangement of two transparent slabs A and B with their respective thickness and refractive indices as shown. Find the vertical shift of the image of a point object placed at the bottom of a due to both the slabs. Also find the effective refractive index of the compound slab.

Answer 1

Concept used:
$D_a=D_r\frac{n_{\text{observer}}}{n_{\text{object}}}$
Here,
$D_a$ is a apparent depth
$D_r$ is real depth
$n_{observer}$ is refractive index of media of observer
$n_{object}$ is refractive index of media of object
The shift due to both the slabs will be sum of individual shifts. i.e., shift, $\delta ={\delta }_1+{\delta }_2$
and $\delta =\alpha (1-\frac{1}{\mu })$
Solution:
For this equation

$d_1=2,{\mu }_1=\frac{4}{3}$
$d_2=3,{\mu }_2=\frac{3}{2}$
So,
$\delta ={\delta }_1+{\delta }_2$
=$2(1-\frac{1}{\frac{4}{3}})+3(1-\frac{1}{\frac{3}{2}})$
=$2(1-\frac{3}{4})+3(1-\frac{2}{3})$
=$2\left(\frac{1}{4}\right)+3\left(\frac{1}{3}\right)$
$\delta =1.5cm$
And effective refractive index of compound slab can be
$1.5=5(1-\frac{1}{{\mu }_e})$
$0.3=1-\frac{1}{{\mu }_e}$
${\mu }_e=\frac{10}{7}$
Hence effective refractive index of the compound slab is $\frac{10}{7}$