Question

A layer of ice $(\mu =1.33)$ lies on a glass plate $(\mu =1.5)$. A ray of light makes an angle of incidence $60^{\circ }$ on the surface of ice as shown in the figure below.

With reference to the above figure, match the items in Column I with terms in Column II and choose the correct option from the codes given below.

Column I		Column II
A	$\sin r_1$	1	$(1/\sqrt{3})$
B	$\sin r_2$	2	$(3\sqrt{3})/8$
C	Refractive Index of ice with respect to glass	3	$8/9$

• A. A - (1) , B - (2) , C - (3)
• B. A - (2) , B - (1) , C - (3)
• C. A - (2) , B - (3) , C - (1)
• D. A - (1) , B - (3) , C - (2)

Answer 1

Answer: (B)

In this case, the light gets refracted twice from air to ice at $A$ and from ice to glass at $B$.
We have find angle $r_2$. At $A$, Air-ice interface
Applying Snell's law $\frac{\sin i_1}{\sin r_1}={}^{\text{air }}{\mu }_{\text{ice }}$

$\Rightarrow \frac{\sin 60^{\circ }}{\sin r_1}=\frac{{\mu }_{\text{ice }}}{1}...(i)$
$\Rightarrow \sin r_1=\frac{\sin 60^{\circ }}{{\mu }_{\text{ice }}}=\frac{\sqrt{3}/2}{4/3}=\frac{3\sqrt{3}}{8}...(ii)$
At $B$, Ice - glass interface
Applying Snell's law,
$\frac{\sin r_1}{\sin r_2}={}^{\text{ice }}{\mu }_{\text{glass }}=\frac{{\mu }_{\text{glass }}}{{\mu }_{\text{ice }}}...(A)$
On multiplying Eqs. (i) and (ii), we get
$\frac{\sin 60^{\circ }}{\sin r_1}\times \frac{\sin r_1}{\sin r_2}=\frac{{\mu }_{\text{ice }}}{1}\times \frac{{\mu }_{\text{glass }}}{{\mu }_{\text{ice }}}$
$\Rightarrow \frac{\sin 60^{\circ }}{\sin r_2}={\mu }_{\text{glass }}$
$\Rightarrow \sin r_2=\frac{\sin 60^{\circ }}{{\mu }_{\text{glass }}}=\frac{(\sqrt{3/2})}{1.5}=\frac{\sqrt{3/2}}{3/2}=\left(\frac{1}{\sqrt{3}}\right)...(B)$
Also, refractive index of ice w.r.t. glass
$\Rightarrow {}^{\text{glass }}{\mu }_{\text{ice }}=\frac{{}_{\text{air }}{\mu }_{\text{ice }}}{\text{ air }{\mu }_{\text{glass }}}=\frac{4/3}{3/2}=\frac{8}{9}...(C)$