Question

A light ray coming from the Sun falls on the water surface from air at an angle of 45° as shown. If the ray got refracted by 30° on passing through the air-water interface, then calculate:

i. Refractive index of water

ii. Speed of light inside water

Answer 1

Given: Angle of incidence $\angle$i = 45°, Angle of refraction ∠r = 30°

(i) Let refractive index of water $n_2$ , refractive index of air $n_1$ =1

Use snell's law : $n_{1}sin\left(i\right)=n_{2}sin\left(r\right)$

Using values we get $\Rightarrow 1\times sin\left({45}^o\right)=n_2\times sin\left({30}^o\right)$

$\Rightarrow n_2=\frac{sin\left({45}^o\right)}{sin\left({30}^o\right)}=\frac{\left(\frac{1}{\sqrt{2}}\right)}{\frac{1}{2}}=\sqrt{2}$

Hence refractive index of water will be $\sqrt{2}$

(ii) Using refractive index formula: $n=\frac{Speedoflightinvacuum}{Speedoflightinmedium}=\frac{c}{v}$

Where , n is refractive index and c = $3\times {10}^{8}m{s}^{-1}$

For speed of light in water: $n_{water}=\frac{c}{v_{water}}$

Put values: $v_{water}=\frac{c}{n_{water}}=3\times \frac{{10}^{8}m{s}^{-1}}{\sqrt{2}}=2.25\times {10}^{8}m{s}^{-1}$

Hence speed of light in water is 2.25 × ${10}^8$ m/s.