Question

A thin oil layer floats on water. A ray of light making an angle of incidence 40${}^o$ shines an oil layer. The angle of refraction of Iight ray in water is $(m{u}_{oil}=1.45,{\mu }_{water}=1.33)$
$[sin{40}^o=0.6428,sin{44.5}^o=0.7007]$

• A. 28.9${}^o$
• B. 26.5${}^o$
• C. 44.5${}^o$
• D. 40${}^o$

Answer 1

Answer: (C)

44.5${}^o$

$\frac{sini}{sinr}=\frac{{\mu }_{water}}{{\mu }_{oil}}=\frac{1.33}{1.45}$
$\frac{sin40}{sinr}=\frac{1.33}{1.45}$
$\Rightarrow \frac{0.6428}{sinr}=\frac{1.33}{1.45}$
$\Rightarrow sinr=\frac{0.6428\times 1.45}{1.33}$
$\Rightarrow sinr=0.7007=sin{44.5}^o$
$\Rightarrow r={44.5}^o$