Question

A thin oil layer floats on water. A ray of light making an angle of incidence $40^{\circ}$ shines on oil layer. The angle of refraction of light ray in water is (Given, $\mu_{\text {oil }}=1.45, \mu_{\text {water }}=1.33$ )

• A. $ 28.9^{\circ} $
• B. $ 26.8^{\circ} $
• C. $ 44.5^{\circ} $
• D. $ 36.1^{\circ}$

Answer 1

Answer: (C)

From Snell's law
${ }_{1} \mu_{2}=\frac{\sin i}{\sin r}$ ...(i)
where ${ }_{1} \mu_{2} =\frac{\mu_{2}}{\mu_{1}}$ ...(ii)
From Eqs. (i) and (ii), we get
$\sin r=\frac{\mu_{1}}{\mu_{2}} \sin i$
Given, $i=40^{\circ}, \mu_{1}=\mu_{\text {oil }}=1.45$,
$\mu_{2}=\mu_{w}=1.33$
$\therefore \sin r=\frac{1.45}{1.33} \sin 40^{\circ}$
$=\frac{1.45}{1.33} \times 0.6428$
$\Rightarrow \sin r=0.7007$
$\Rightarrow r=\sin ^{-1}(0.7007)=44.5^{\circ}$