A monochromatic light wave with… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Optics

2023mcqeasy

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:

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

Options

A
${\lambda }_{2}=\sqrt{2}{\lambda }_{1},{\nu }_{2}={\nu }_{1}$
B
${\lambda }_{2}={\lambda }_{1},{\nu }_{2}=\frac{1}{\sqrt{2}}{\nu }_{1}$
C
${\lambda }_{2}={\lambda }_{1},{\nu }_{2}=\sqrt{2}{\nu }_{1}$
D
${\lambda }_{2}=\frac{1}{\sqrt{2}}{\lambda }_{1},{\nu }_{2}={\nu }_{1}$

Did you get this right?

Solution

By using Snell's law, ${n}_{1}\mathrm{sin}{\theta }_{1}={n}_{2}\mathrm{sin}{\theta }_{2}$ , the refractive index is

$\mu =\frac{\mathrm{sin}45^{\circ}}{\mathrm{sin}30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}$

Also using the relation $n=\frac{c}{v}$ where $n$ is the refractive index, $c$ is the velocity of light in air and $v$ is the velocity in the medium

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

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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Options

${\lambda }_{2}=\sqrt{2}{\lambda }_{1},{\nu }_{2}={\nu }_{1}$

${\lambda }_{2}={\lambda }_{1},{\nu }_{2}=\frac{1}{\sqrt{2}}{\nu }_{1}$

${\lambda }_{2}={\lambda }_{1},{\nu }_{2}=\sqrt{2}{\nu }_{1}$

${\lambda }_{2}=\frac{1}{\sqrt{2}}{\lambda }_{1},{\nu }_{2}={\nu }_{1}$

Solution

By using Snell's law, ${n}_{1}\mathrm{sin}{\theta }_{1}={n}_{2}\mathrm{sin}{\theta }_{2}$ , the refractive index is

$\mu =\frac{\mathrm{sin}45^{\circ}}{\mathrm{sin}30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}$

Also using the relation $n=\frac{c}{v}$ where $n$ is the refractive index, $c$ is the velocity of light in air and $v$ is the velocity in the medium

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

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.

Options

Did you get this right?

Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics