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 $b=0.04 \,m$ 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$