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.
(a) 1.65 x108 m/s; 410 nm
(b) 1.75x108 m/s; 363 nm
(c) 1.97 x108 m/s; 210 nm
(d) 1.97x108 m/s; 210 nm​

Answer 1

Given

$\theta_a$ = Angle of incidence = $40^{\circ}$

$\theta_o$ = Angle of refraction = $22^{\circ}$

$\lambda_a$ = Wavelength of light in air = 623.8 nm

To find

The speed and wavelength of this light in the oil

Solution

$n_a$ = Refractive index of air = 1

$n_o$ = Refractive index of oil

From Snell's law we have

$n_a\sin\theta_a=n_o\sin\theta_o\\\Rightarrow n_o=\dfrac{n_a\sin\theta_a}{\sin\theta_o}\\\Rightarrow n_o=\dfrac{1\times \sin40^{\circ}}{\sin22^{\circ}}\\\Rightarrow n_o=1.72$

Refractive index is given by

$n_o=\dfrac{c}{v}\\\Rightarrow v=\dfrac{c}{n_o}\\\Rightarrow v=\dfrac{3\times 10^8}{1.72}\\\Rightarrow v=174418604.65\ \text{m/s}\approx 1.75\times 10^8\ \text{m/s}$

Wavelength is given by

$\lambda_o=\dfrac{\lambda_a}{n_o}\\\Rightarrow \lambda_o=\dfrac{623.8}{1.72}\\\Rightarrow \lambda_o=362.67\approx 363\ \text{nm}$

The speed and wavelength of the light in oil is $1.75\times 10^8\ \text{m/s}$ and $363\ \text{nm}$ respectively.

Correct option is (b) $1.75\times 10^8\ \text{m/s}$ ; $363\ \text{nm}$.