JEE MainPhysics

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Optics: JEE Main Physics Question with Solution

Q: Do we need to convert nm and mm into SI units (meters)?

No, because we are taking the ratio of wavelengths , the unit conversion factors cancel out.

The width of fringe is

2\text{ mm}

on the screen in a double slits experiment for the light of wavelength of

400\text{ nm}

. The width of the fringe for the light of wavelength

600\text{ nm}

will be:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

4\text{ mm}

1.33\text{ mm}

3\text{ mm}

2\text{ mm}

Step-by-step solutionView

Correct answer

Since fringe width is directly proportional to wavelength ( $\beta = \frac{\lambda D}{d}$ ), increasing the wavelength from $400\text{ nm}$ to $600\text{ nm}$ scales the fringe width from $2\text{ mm}$ to $3\text{ mm}$ .

Option analysis

Why each option works or fails

A ·

4\text{ mm}

Doubling the fringe width instead of scaling by the actual ratio of $600/400 = 1.5$ . Compute the exact ratio $\frac{\lambda_2}{\lambda_1} = \frac{600}{400} = 1.5$ , giving $\beta_2 = 1.5 \times 2\text{ mm} = 3\text{ mm}$ .

B ·

1.33\text{ mm}

Assuming fringe width is inversely proportional to wavelength and calculating $\beta_2 = 2\text{ mm} \times \frac{400}{600} = 1.33\text{ mm}$ . Recall that $\beta = \frac{\lambda D}{d}$ , so $\beta$ is directly proportional to $\lambda$ , not inversely proportional.

C ·

3\text{ mm}

Correct option. Correctly applied $\beta_2 = \beta_1 \left(\frac{\lambda_2}{\lambda_1}\right) = 2\text{ mm} \times \frac{600}{400} = 3\text{ mm}$ .

D ·

2\text{ mm}

Believing fringe width depends only on geometry ( $D$ and $d$ ) and remains unchanged when wavelength changes. Fringe width directly depends on the wavelength of light used: $\beta \propto \lambda$ .

Reviewed route

Solution

StepWorking

01given

Initial fringe width $\beta_1 = 2\text{ mm}$ , initial wavelength $\lambda_1 = 400\text{ nm}$ , new wavelength $\lambda_2 = 600\text{ nm}$ . The slit separation $d$ and screen distance $D$ remain constant.

02find

Find the new fringe width $\beta_2$ for wavelength $\lambda_2 = 600\text{ nm}$ .

03strategise

Fringe width in YDSE is given by $\beta = \frac{\lambda D}{d}$ . Since $D$ and $d$ are unchanged, $\beta \propto \lambda$ , which gives the ratio relation $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$ .

04execute

Substitute the values to calculate $\beta_2$ : $\beta_2 = \beta_1 \times \frac{\lambda_2}{\lambda_1} = 2\text{ mm} \times \frac{600}{400} = 2 \times 1.5 = 3\text{ mm}$ .

✓verify

Since $\lambda$ increases by a factor of 1.5, the fringe width must also increase proportionally from $2\text{ mm}$ to $3\text{ mm}$ . The units match (mm).

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 5 September 2026

Quick checks

Students also ask

Do we need to convert nm and mm into SI units (meters)?

No, because we are taking the ratio of wavelengths $\frac{\lambda_2}{\lambda_1}$ , the unit conversion factors cancel out.

Answer

Since fringe width is directly proportional to wavelength ( $\beta = \frac{\lambda D}{d}$ ), increasing the wavelength from $400\text{ nm}$ to $600\text{ nm}$ scales the fringe width from $2\text{ mm}$ to $3\text{ mm}$ .

Why each option works or fails

A: $4\text{ mm}$ - Doubling the fringe width instead of scaling by the actual ratio of $600/400 = 1.5$ . Compute the exact ratio $\frac{\lambda_2}{\lambda_1} = \frac{600}{400} = 1.5$ , giving $\beta_2 = 1.5 \times 2\text{ mm} = 3\text{ mm}$ .
B: $1.33\text{ mm}$ - Assuming fringe width is inversely proportional to wavelength and calculating $\beta_2 = 2\text{ mm} \times \frac{400}{600} = 1.33\text{ mm}$ . Recall that $\beta = \frac{\lambda D}{d}$ , so $\beta$ is directly proportional to $\lambda$ , not inversely proportional.
C · correct: $3\text{ mm}$ - Correct option. Correctly applied $\beta_2 = \beta_1 \left(\frac{\lambda_2}{\lambda_1}\right) = 2\text{ mm} \times \frac{600}{400} = 3\text{ mm}$ .
D: $2\text{ mm}$ - Believing fringe width depends only on geometry ( $D$ and $d$ ) and remains unchanged when wavelength changes. Fringe width directly depends on the wavelength of light used: $\beta \propto \lambda$ .

Step-by-step solution

given: Initial fringe width $\beta_1 = 2\text{ mm}$ , initial wavelength $\lambda_1 = 400\text{ nm}$ , new wavelength $\lambda_2 = 600\text{ nm}$ . The slit separation $d$ and screen distance $D$ remain constant.
find: Find the new fringe width $\beta_2$ for wavelength $\lambda_2 = 600\text{ nm}$ .
strategise: Fringe width in YDSE is given by $\beta = \frac{\lambda D}{d}$ . Since $D$ and $d$ are unchanged, $\beta \propto \lambda$ , which gives the ratio relation $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$ .
execute: Substitute the values to calculate $\beta_2$ : $\beta_2 = \beta_1 \times \frac{\lambda_2}{\lambda_1} = 2\text{ mm} \times \frac{600}{400} = 2 \times 1.5 = 3\text{ mm}$ .
verify: Since $\lambda$ increases by a factor of 1.5, the fringe width must also increase proportionally from $2\text{ mm}$ to $3\text{ mm}$ . The units match (mm).

Shortcut:

When to use it: Direct inspection using $\beta \propto \lambda$ .

quick_kill: Since $\beta \propto \lambda$ , scaling $\lambda$ by $\frac{600}{400} = 1.5$ scales $\beta$ by $1.5$ : $\beta_2 = 2\text{ mm} \times 1.5 = 3\text{ mm}$ (Option 2).

Optics: JEE Main Physics Question with Solution

What feels right?

Since fringe width is directly proportional to wavelength ( $\beta = \frac{\lambda D}{d}$ ), increasing the wavelength from $400\text{ nm}$ to $600\text{ nm}$ scales the fringe width from $2\text{ mm}$ to $3\text{ mm}$ .

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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