Question

A monochromatic light of wavelength 623.8 nm
is incident at an angle of 40° from the
vertical
into the oil and it refracted to 22°
from the
vertical. Find the speed and wavelength of
this light in the oil.​

Answer 1

Given

$n_1$ = Air's refractive index = 1

$\theta_1$ = Angle of incidence on the air oil interface = $40^{\circ}$

$\theta_2$ = Angle of refraction in the oil = $22^{\circ}$

$\lambda_1$ = Wavelength of the given light in air = 623.8 nm

To find

Speed and wavelength of this light in the oil.

Solution

$n_2$ = Oil's refractive index

$\lambda_2$ = Wavelength of the given light in the oil

c = Speed of light in air = $3\times 10^8\ \text{m/s}$

We obtain the following relation from Snell's law

$n_1\sin\theta_1=n_2\sin\theta_2\\\Rightarrow 1\sin40^{\circ}=n_2\sin22^{\circ}\\\Rightarrow n_2=\dfrac{\sin40^{\circ}}{\sin22^{\circ}}\\\Rightarrow n_2=1.7159$

Refractive index is given by

$n_2=\dfrac{c}{v}\\\Rightarrow 1.7159=\dfrac{3\times 10^8}{v}\\\Rightarrow v=\dfrac{3\times 10^8}{1.7159}\\\Rightarrow v=174835363.366\ \text{m/s}$

Speed of the given light in the oil is $174835363.366\ \text{m/s}$.

Wavelength is given by

$\lambda_2=\dfrac{\lambda_1}{n_2}\\\Rightarrow \lambda_2=\dfrac{623.8\times 10^{-9}}{1.7159}\\\Rightarrow \lambda_2=3.635\times 10^{-7}\ \text{m}$

The wavelength of light in the oil is $3.635\times 10^{-7}\ \text{m}$.