Q136 / 150 · MHT CET 2025 (26 Apr Shift 1) · Physics
In Young's double slit experiment, the distance between the slits is 2 mm and the slits are 1 m away, from the screen. Two interference patterns can be obtained on the screen due to light of wavelength ' $\lambda_1$ ' and ' $\lambda_2$ ' respectively. The separation on the screen between the $3^{\text {rd }}$ order bright fringes on the two interference patterns is ( $\lambda_2=1 \cdot 5 \lambda_1$ )
- Correct answerA$\left(0.75 \times 10^{+3}\right) \lambda_1$
- B$\left(1 \cdot 75 \times 10^{+3}\right) \lambda_1$
- C$\left(2 \cdot 00 \times 10^{+3}\right) \lambda_1$
- D$\left(2 \cdot 25 \times 10^{+3}\right) \lambda_1$
Solution
Position of the $n$th order bright fringe: $y_n = \frac{n \lambda D}{d}$.
Given $\lambda_2 = 1.5\lambda_1$, the positions for $n = 3$ are:
$y_{3,1} = \frac{3\lambda_1 D}{d}$,
$y_{3,2} = \frac{3(1.5\lambda_1) D}{d} = \frac{4.5\lambda_1 D}{d}$.
The separation is
$\Delta y = y_{3,2} - y_{3,1} = \frac{(4.5 - 3)\lambda_1 D}{d} = \frac{1.5\lambda_1 D}{d}$.
Substituting $D = 1 \text{ m}$ and $d = 2 \times 10^{-3} \text{ m}$,
$\Delta y = \frac{1.5 \cdot 1}{2 \times 10^{-3}} \lambda_1 = (0.75 \times 10^{+3}) \lambda_1$.
The result corresponds to option A.
Hence, the correct answer is option: 'A'.