Question

ABCD represents a glass slab and PQRS indicates the path of a light ray passing through the slab. If the refractive index of the glass is $\sqrt{3}$, find the lateral displacement (x).

• A. $15$ cm
• B. $16.33$ cm
• C. $8.66$ cm
• D. $12.33$ cm

Answer 1

The correct option is C $8.66$ cm

Lateral shift due to a glass slab is given by $\delta =\frac{t}{\cos r}\sin (i-r)$ here t is the thickness of the slab.
and it is given $t=15cmr={30}^0$
From Snell's Law
$\mu =\frac{\sin i}{\sin r}$
$\sqrt{3}=\frac{sini}{sin{30}^0}$
$\sin i=\frac{\sqrt{3}}{2}=\sin {60}^0$
$i={60}^0$
So, $\delta =\frac{15}{\cos 300}\sin ({60}^0-{30}^0)$
$=\frac{15}{\frac{\sqrt{3}}{2}}/\frac{1}{2}$
$\delta =8.66cm$