JEE Physics 2025 Question with Solution

Given: Wavelength $\lambda_1 = 600 \, \text{nm}$, position of the 10th bright fringe $y_1 = 10 \, \text{mm}$, and new wavelength $\lambda_2 = 660 \, \text{nm}$.

Find: The new distance of the same 10th bright fringe from the central maximum.

For a double-slit interference pattern, the position of the $n$-th bright fringe is

where $\lambda$ is the wavelength, $D$ is the distance between slit and screen, and $d$ is the slit separation.

Since the same 10th bright fringe is considered, and $D$ and $d$ remain unchanged, the fringe position is directly proportional to the wavelength:

Substituting the given values:

Therefore, the distance of the 10th bright fringe from the central maximum is $11 \, \text{mm}$.

Given: The 10th bright fringe is at $10 \, \text{mm}$ for $\lambda_1 = 600 \, \text{nm}$.

Find: Its new position for $\lambda_2 = 660 \, \text{nm}$.

For the same fringe order in Young's double-slit experiment, the fringe position scales directly with wavelength. Therefore, increase the old position by the same factor as the wavelength increase:

So the new position is

This works because only the wavelength changes; the slit separation and screen distance stay constant. Hence the correct numerical value is $11$.