Q. A thin oil layer floats on water. A ray of light making an angle of incidence of $40{^\circ}$ shines on oil layer. The angle of refraction of light ray in water is $(\mu_{oil} = 1.45,\ \mu_{water} = 1.33)$
A) $36.1{^\circ}$
B) $44.5{^\circ}$
C) $26.8{^\circ}$
D) $28.9{^\circ}$
Explanation
(d) Refraction at air-oil point $\mu_{oil} = \frac{\sin i}{\sin r_{1}}$
$\therefore\sin{r_{1} = \frac{\sin 40}{1.45}} = 0.443$
Refraction at oil-water point$\ _{oil}\mu_{water} = \frac{\sin r_{1}}{\sin r}$
$\therefore\frac{1.33}{1.45} = \frac{0.443}{\sin r}\ or\sin r = \frac{0.443 \times 1.45}{1.33} \Rightarrow r = 28.9{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.