Q179 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
In Young's double slit experiment, the slits are $2\ \text{mm}$ apart and are illuminated by photons of two wavelengths $\lambda_1 = 6000\ Å$ and $\lambda_2 = 5000\ Å$ one after the other. At what minimum distance from the common central bright fringe on the screen $2\ \text{m}$ from the slit will, a bright fringe from one interference pattern coincide with a bright fringe from the other?
- A$8\ \text{mm}$
- B$6\ \text{mm}$
- C$4\ \text{mm}$
- Correct answerD$3\ \text{mm}$
Solution
Given:
$d = 2\ \text{mm} = 2 \times 10^{-3}\ \text{m}$
$D = 2\ \text{m}$
$\lambda_1 = 6000\ \text{\AA} = 6 \times 10^{-7}\ \text{m}$
$\lambda_2 = 5000\ \text{\AA} = 5 \times 10^{-7}\ \text{m}$
Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$. The position of the $n$-th bright fringe from the central maximum is given by $y = \dfrac{n \lambda D}{d}$.
For coincidence:
$\dfrac{n_1 \lambda_1 D}{d} = \dfrac{n_2 \lambda_2 D}{d}$
$n_1 \lambda_1 = n_2 \lambda_2$
$\dfrac{n_1}{n_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{5000}{6000} = \dfrac{5}{6}$
For the minimum distance from the central maximum, we take the smallest integer values for $n_1$ and $n_2$, which are $n_1 = 5$ and $n_2 = 6$.
Substituting $n_1 = 5$ into the position formula:
$y = \dfrac{5 \times 6 \times 10^{-7} \times 2}{2 \times 10^{-3}}$
$y = 30 \times 10^{-4}\ \text{m}$
$y = 3 \times 10^{-3}\ \text{m} = 3\ \text{mm}$
Answer: $3\ \text{mm}$
$d = 2\ \text{mm} = 2 \times 10^{-3}\ \text{m}$
$D = 2\ \text{m}$
$\lambda_1 = 6000\ \text{\AA} = 6 \times 10^{-7}\ \text{m}$
$\lambda_2 = 5000\ \text{\AA} = 5 \times 10^{-7}\ \text{m}$
Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$. The position of the $n$-th bright fringe from the central maximum is given by $y = \dfrac{n \lambda D}{d}$.
For coincidence:
$\dfrac{n_1 \lambda_1 D}{d} = \dfrac{n_2 \lambda_2 D}{d}$
$n_1 \lambda_1 = n_2 \lambda_2$
$\dfrac{n_1}{n_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{5000}{6000} = \dfrac{5}{6}$
For the minimum distance from the central maximum, we take the smallest integer values for $n_1$ and $n_2$, which are $n_1 = 5$ and $n_2 = 6$.
Substituting $n_1 = 5$ into the position formula:
$y = \dfrac{5 \times 6 \times 10^{-7} \times 2}{2 \times 10^{-3}}$
$y = 30 \times 10^{-4}\ \text{m}$
$y = 3 \times 10^{-3}\ \text{m} = 3\ \text{mm}$
Answer: $3\ \text{mm}$
Hence, the correct answer is option: 'D'.