Q132 / 150 · MHT CET 2026 (16 April Shift 1) · Physics

In Young's double slit experiment, the distance between the slits is 3 mm and the slits are 2m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength 480 nm and 600 nm respectively. The separation on the screen between the $5^{\text{th}}$ order bright fringes on two interference patterns is

  • A
    $2 \times 10^{-4}\,\text{m}$
  • B
    $1 \times 10^{-4}\,\text{m}$
  • C
    $8 \times 10^{-4}\,\text{m}$
  • Correct answer
    D
    $4 \times 10^{-4}\,\text{m}$

Solution

Given:

$d = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}$

$D = 2 \text{ m}$

$\lambda_1 = 480 \text{ nm} = 480 \times 10^{-9} \text{ m}$

$\lambda_2 = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$

The position of the $n^{\text{th}}$ order bright fringe is given by $y_n = \dfrac{n \lambda D}{d}$.

The separation between the $5^{\text{th}}$ order bright fringes of the two patterns is:

$\Delta y = y_{5,2} - y_{5,1} = \dfrac{5 D}{d} (\lambda_2 - \lambda_1)$

Substituting the given values:

$\Delta y = \dfrac{5 \times 2}{3 \times 10^{-3}} (600 - 480) \times 10^{-9}$

$\Delta y = \dfrac{10}{3 \times 10^{-3}} \times 120 \times 10^{-9}$

$\Delta y = 400 \times 10^{-6} \text{ m}$

$\Delta y = 4 \times 10^{-4} \text{ m}$

Answer: $4 \times 10^{-4}\,\text{m}$

Hence, the correct answer is option: 'D'.

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