Question

A ray of sunlight enters a spherical water droplet ( $n \, =4/3$ ) at an angle of incidence $53^\circ $ measured with respect to the normal to the surface. It is reflected from the back surface of the droplet and re-enters into the air. The angle between the incoming and outgoing ray is [Take $sin \, 53^\circ \, = \, 0.8$ ]

• A. $15^\circ $
• B. $34^\circ $
• C. $138^\circ $
• D. $30^\circ $

Answer 1

Answer: (C)

$\text{Applying law of refraction}\Rightarrow 1\times \text{sin }53^\circ =\frac{4}{3}\times \text{sin }\textit{r}$
$\frac{4}{5}=\frac{4}{3}\text{sin }\textit{r}\Rightarrow \text{sin }\textit{r}=\frac{3}{5}\Rightarrow \text{sin }\textit{r}=0\text{.}6\Rightarrow r=37^\circ $
$\text{δ}_{1}=53^\circ -37^\circ \Rightarrow \text{δ}_{1}=16^\circ \text{C.W.}$
$\text{δ}_{2}=180-2\times 37^\circ \Rightarrow \text{δ}_{2}=106^\circ \text{C.W.}$
$\text{δ}_{3}=\text{5}\text{3°}-37^\circ \Rightarrow \text{δ}_{3}=16^\circ \text{C.W.}$
$\text{δ}_{\text{net}}=\text{δ}_{1}+\text{δ}_{2}+\text{δ}_{3}=138^\circ $