JEE MainPhysics

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Refraction of Light: Frequency Invariance and Snell's Law for Wavelength

Q: Why does the frequency of light remain unchanged when entering a new medium?

Frequency is determined solely by the oscillation frequency of the source charges creating the wave. At any boundary, the oscillations on both sides must remain continuous and phase-locked, keeping frequency invariant.

A monochromatic light wave with wavelength λ_1 and frequency v_1 in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45° and 30° respectively, then the wavelength λ_2 and frequency v_2 of the refracted wave are:

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Aλ_2 = λ_1, v_2 = √(2)v_1Bλ_2 = 1/(√(2))λ_1, v_2 = v_1Cλ_2 = √(2)λ_1, v_2 = v_1Dλ_2 = λ_1, v_2 = 1/(√(2))v_1

Step-by-step solutionView

Correct answer

Because frequency does not change across an interface ( $v_{2} = v_{1}$ ) and Snell's law gives $μ = \sin 45^{\circ} / \sin 30^{\circ} = \sqrt{2}$ , the new wavelength is $λ_{2} = λ_{1} / \sqrt{2}$ .

Option analysis

Why each option works or fails

A · λ_2 = λ_1, v_2 = √(2)v_1

Believing that wavelength remains invariant across media while frequency scales with refractive index. Frequency is determined solely by the source and does not change when entering a new medium; wavelength changes in inverse proportion to the refractive index.

B · λ_2 = 1/(√(2))λ_1, v_2 = v_1

This is the correct option. Frequency is constant ( $v_{2} = v_{1}$ ), and $λ_{2} = λ_{1} \cdot \frac{\sin r}{\sin i} = λ_{1} \cdot \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = \frac{λ_{1}}{\sqrt{2}}$ .

C · λ_2 = √(2)λ_1, v_2 = v_1

Inverting Snell's ratio when calculating wavelength, leading to multiplying by $\sqrt{2}$ instead of dividing by $\sqrt{2}$ . Light slows down and its wavelength compresses in an optically denser medium ( $r < i$ ), so $λ_{2} = λ_{1} \frac{\sin r}{\sin i} = \frac{λ_{1}}{\sqrt{2}}$ .

D · λ_2 = λ_1, v_2 = 1/(√(2))v_1

Confusing frequency with wavelength by assuming frequency decreases while wavelength stays constant. Frequency is invariant upon refraction, whereas wave speed and wavelength both decrease.

Reviewed route

Solution

StepWorking

01Given

Incident wave: wavelength $λ_{1}$ , frequency $ν_{1}$ , angle of incidence $i = 45^{\circ}$ in air (refractive index $μ_{1} \approx 1$ ). Refracted wave: angle of refraction $r = 30^{\circ}$ in medium of refractive index $μ_{2}$ .

02Strategise

Frequency $ν$ is a characteristic of the source, so it remains invariant during refraction: $ν_{2} = ν_{1}$ . Apply Snell's law $μ_{1} \sin i = μ_{2} \sin r$ along with the relation $μ \propto 1 / λ$ (since $v = ν λ = c / μ$ ) to find $λ_{2}$ .

03Execute

From Snell's law: $\frac{μ_{1}}{μ_{2}} = \frac{\sin r}{\sin i} = \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = \frac{1 / 2}{1 / \sqrt{2}} = \frac{1}{\sqrt{2}}$ . Since $μ_{1} λ_{1} = μ_{2} λ_{2}$ , we have $λ_{2} = (\frac{μ_{1}}{μ_{2}}) λ_{1} = \frac{1}{\sqrt{2}} λ_{1}$ . Thus, $λ_{2} = \frac{1}{\sqrt{2}} λ_{1}$ and $ν_{2} = ν_{1}$ .

✓Verify

The ray bends towards the normal ( $r < i$ ), meaning the medium is optically denser (higher $μ$ ). Speed and wavelength must decrease in a denser medium ( $λ_{2} < λ_{1}$ ), which matches $λ_{2} = λ_{1} / \sqrt{2}$ . Frequency remains constant.

Hints that build this answer step by step

What happens to the frequency of a light wave when it passes from air into another medium?
The frequency remains unchanged ( $v_{2} = v_{1}$ ) because it is determined by the source.
Using Snell's law, what is the relation between the refractive index of the second medium $μ_{2}$ (relative to air, $μ_{1} = 1$ ) and the angles of incidence ( $i = 45^{\circ}$ ) and refraction ( $r = 30^{\circ}$ )?
$μ_{2} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \sqrt{2}$
How does the wavelength in the second medium ( $λ_{2}$ ) relate to the wavelength in air ( $λ_{1}$ ) and the refractive index $μ_{2} = \sqrt{2}$ ?
$λ_{2} = \frac{λ_{1}}{μ_{2}} = \frac{1}{\sqrt{2}} λ_{1}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Students also ask

Why does the frequency of light remain unchanged when entering a new medium?

Frequency is determined solely by the oscillation frequency of the source charges creating the wave. At any boundary, the oscillations on both sides must remain continuous and phase-locked, keeping frequency invariant.

Answer

Because frequency does not change across an interface ( $v_{2} = v_{1}$ ) and Snell's law gives $μ = \sin 45^{\circ} / \sin 30^{\circ} = \sqrt{2}$ , the new wavelength is $λ_{2} = λ_{1} / \sqrt{2}$ .

Why each option works or fails

A: λ_2 = λ_1, v_2 = √(2)v_1 - Believing that wavelength remains invariant across media while frequency scales with refractive index. Frequency is determined solely by the source and does not change when entering a new medium; wavelength changes in inverse proportion to the refractive index.
B · correct: λ_2 = 1/(√(2))λ_1, v_2 = v_1 - This is the correct option. Frequency is constant ( $v_{2} = v_{1}$ ), and $λ_{2} = λ_{1} \cdot \frac{\sin r}{\sin i} = λ_{1} \cdot \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = \frac{λ_{1}}{\sqrt{2}}$ .
C: λ_2 = √(2)λ_1, v_2 = v_1 - Inverting Snell's ratio when calculating wavelength, leading to multiplying by $\sqrt{2}$ instead of dividing by $\sqrt{2}$ . Light slows down and its wavelength compresses in an optically denser medium ( $r < i$ ), so $λ_{2} = λ_{1} \frac{\sin r}{\sin i} = \frac{λ_{1}}{\sqrt{2}}$ .
D: λ_2 = λ_1, v_2 = 1/(√(2))v_1 - Confusing frequency with wavelength by assuming frequency decreases while wavelength stays constant. Frequency is invariant upon refraction, whereas wave speed and wavelength both decrease.

Step-by-step solution

Given: Incident wave: wavelength $λ_{1}$ , frequency $ν_{1}$ , angle of incidence $i = 45^{\circ}$ in air (refractive index $μ_{1} \approx 1$ ). Refracted wave: angle of refraction $r = 30^{\circ}$ in medium of refractive index $μ_{2}$ .
Strategise: Frequency $ν$ is a characteristic of the source, so it remains invariant during refraction: $ν_{2} = ν_{1}$ . Apply Snell's law $μ_{1} \sin i = μ_{2} \sin r$ along with the relation $μ \propto 1 / λ$ (since $v = ν λ = c / μ$ ) to find $λ_{2}$ .
Execute: From Snell's law: $\frac{μ_{1}}{μ_{2}} = \frac{\sin r}{\sin i} = \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = \frac{1 / 2}{1 / \sqrt{2}} = \frac{1}{\sqrt{2}}$ . Since $μ_{1} λ_{1} = μ_{2} λ_{2}$ , we have $λ_{2} = (\frac{μ_{1}}{μ_{2}}) λ_{1} = \frac{1}{\sqrt{2}} λ_{1}$ . Thus, $λ_{2} = \frac{1}{\sqrt{2}} λ_{1}$ and $ν_{2} = ν_{1}$ .
Verify: The ray bends towards the normal ( $r < i$ ), meaning the medium is optically denser (higher $μ$ ). Speed and wavelength must decrease in a denser medium ( $λ_{2} < λ_{1}$ ), which matches $λ_{2} = λ_{1} / \sqrt{2}$ . Frequency remains constant.

Refraction of Light: Frequency Invariance and Snell's Law for Wavelength

What feels right?

Because frequency does not change across an interface ( $v_{2} = v_{1}$ ) and Snell's law gives $μ = \sin 45^{\circ} / \sin 30^{\circ} = \sqrt{2}$ , the new wavelength is $λ_{2} = λ_{1} / \sqrt{2}$ .

Why each option works or fails

Solution

01Given

02Strategise

03Execute

✓Verify

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