Question

Figure shows a ray of light entering and passing through a dense glass slab and emerging from the other side. If the angle $i=60^{\circ }$, slab thickness $b=0.04m$ and the refractive index of glass $=\sqrt{3}$, the parallel shift $d$ between the emerging and entering rays in $mm$ is

• A. $\sqrt{\frac{3}{4}}$
• B. $\sqrt{\frac{4}{3}}$
• C. $\frac{40}{\sqrt{3}}$
• D. $15\sqrt{3}$

Answer 1

Answer: (C)

We have, $\sin r=\frac{\sin i}{\mu }=\frac{\sin 60^{\circ }}{\sqrt{3}}=\frac{1}{2}$
$\Rightarrow$ Angle of refraction, $r=30^{\circ }$
Now, lateral shift $d$ is given by
$d=t\cdot \sec (r)\cdot \sin (i-r)$
$=0.04\times \sec \left(30^{\circ }\right)\cdot \sin \left(60^{\circ }-30^{\circ }\right)=0.04\times \tan 30^{\circ }$
$=\frac{0.04}{\sqrt{3}}m$
$=\frac{0.04\times 1000}{\sqrt{3}}mm$
$=\frac{40}{\sqrt{3}}mm$