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Wave Optics question

2023 · 8 Apr · Shift 2 · Q56

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Wave Optics
2023 · 8 Apr · Shift 2 · Q56

Wave Optics question

2023 · 8 Apr · Shift 2 · Q56

JEE MainPhysicsWave OpticsMCQ+4 / −1

The width of fringe is

2 \mathrm{~mm}

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

400 \mathrm{~nm}

. The width of the fringe for the light of wavelength 600

\mathrm{nm}

will be:

A
4 mm
B
1.33 mm
C
2 mm
D
3 mm

View written solutionFree

Correct answer: D

In Young’s double slit experiment, the fringe width is given by

$\beta = \frac{\lambda D}{d}$

where:

$\beta$ = fringe width
$\lambda$ = wavelength of light
$D$ = distance of screen from slits
$d$ = slit separation

Since $D$ and $d$ remain unchanged, fringe width is directly proportional to wavelength:

$\beta \propto \lambda$

Given:

For $\lambda_1 = 400\,\text{nm}$ , fringe width $\beta_1 = 2\,\text{mm}$
For $\lambda_2 = 600\,\text{nm}$ , fringe width $\beta_2 = ?$

Using proportionality,

$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$

$\beta_2 = \beta_1 \cdot \frac{\lambda_2}{\lambda_1}$

Substitute the values:

$\beta_2 = 2 \times \frac{600}{400}$

$\beta_2 = 2 \times 1.5 = 3\,\text{mm}$

Therefore, the correct option is:

$\boxed{\text{D: } 3\,\text{mm}}$

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