A ray of light is incident on one face of a rectangular glass slab of thickness $0.1\text{m}$ and refractive index $1.5$ at an angle of ${60}^{\circ }$ with the normal. Calculate the lateral shift produced. (Given $\sin {35.3}^{\circ }=0.5773,\cos {35.3}^{\circ }=0.816,\sin {24.7}^{\circ }=0.148)$

The correct option is D $0.0513\text{m}$
Given
Angle of incidence, $i_1={60}^{\circ }$
Thickness of the slab, $t=0.1\text{m}$
Refractive index of the slab. $\mu =1.5$.
Let the angle of refraction is $r_1$
We know that $\mu =\frac{\sin i_1}{\sin r_1}$
$\Rightarrow \sin r_1=\frac{\sin i_1}{\mu }$
$\Rightarrow \sin r_1=\frac{\sin {60}^{\circ }}{1.5}$
$\Rightarrow \sin r_1=0.5773$
$r_1={\sin }^{-1}\left(0.5773\right)={35.3}^{\circ }$

The lateral shift ($d$) by glass slab is given by
$d=\frac{t\sin (i_1-r_1)}{\cos r_1}$
$\Rightarrow d=\frac{0.1\sin ({60}^{\circ }-{35.3}^{\circ })}{\cos {35.3}^{\circ }}$
$\Rightarrow d=\frac{0.1\sin {24.7}^{\circ }}{\cos {35.3}^{\circ }}$
$\Rightarrow d=\frac{0.1\times 0.418}{0.816}$
$\Rightarrow d=0.0513\text{m}$

Hence, option (d) is correct.