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, $i=40^{\circ },{\mu }_1={\mu }_{\text{oil }}=1.45$,
${\mu }_2={\mu }_w=1.33$
$\therefore \sin r=\frac{1.45}{1.33}\sin 40^{\circ }$
$=\frac{1.45}{1.33}\times 0.6428$
$\Rightarrow \sin r=0.7007$
$\Rightarrow r={\sin }^{-1}(0.7007)=44.5^{\circ }$