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 $sin53^{\circ }=0.8$ ]

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

Answer 1

Answer: (C)

$\text{Applying law of refraction}\Rightarrow 1\times \text{sin }53^{\circ }=\frac{4}{3}\times \text{sin }\text{r}$
$\frac{4}{5}=\frac{4}{3}\text{sin }\text{r}\Rightarrow \text{sin }\text{r}=\frac{3}{5}\Rightarrow \text{sin }\text{r}=0\text{.}6\Rightarrow r=37^{\circ }$
${\text{δ}}_1=53^{\circ }-37^{\circ }\Rightarrow {\text{δ}}_1=16^{\circ }\text{C.W.}$
${\text{δ}}_2=180-2\times 37^{\circ }\Rightarrow {\text{δ}}_2=106^{\circ }\text{C.W.}$
${\text{δ}}_3=\text{5}\text{3°}-37^{\circ }\Rightarrow {\text{δ}}_3=16^{\circ }\text{C.W.}$
${\text{δ}}_{\text{net}}={\text{δ}}_1+{\text{δ}}_2+{\text{δ}}_3=138^{\circ }$