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Ray Optics

JEE Main level, single correct, medium

92 of 295

Question 92 of 295

A monochromatic light wave with wavelength ${\lambda }_{1}$ and frequency ${\nu }_{1}$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are $45^{\circ}$ and $30^{\circ}$ respectively, then the wavelength ${\lambda }_{2}$ and frequency ${\nu }_{2}$ of the refracted wave are:

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Answer: (D) ${\lambda }_{2}=\dfrac{1}{\sqrt{2}}{\lambda }_{1},{\nu }_{2}={\nu }_{1}$

By using Snell's law,

{n}_{1}\sin {\theta }_{1}={n}_{2}\sin {\theta }_{2}

, the refractive index is

\mu =\dfrac{\sin 45^{\circ}}{\sin 30^{\circ}}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}}=\sqrt{2}

Also using the relation

n=\dfrac{c}{v}

where

n

is the refractive index,

c

is the velocity of light in air and

v

is the velocity in the medium

\Rightarrow \dfrac{{c}_{\text{air}}}{\text{v} }=\sqrt{2}=\dfrac{{\lambda }_{\text{1}}}{{\lambda }_{2}}

The velocities are given by

v=\nu {\lambda }_{2} c=\nu {\lambda }_{1}

\Rightarrow {\lambda }_{1}=\sqrt{2}{\lambda }_{2}

and

{\nu }_{1}={\nu }_{2}

.
The frequency of light does not change when the medium changes.

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Ray Optics (Refraction of light from plane surface), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.