JEE Physics 2025 Question with Solution

Given: The two wavelengths are $\lambda_1 = 480 \, \text{nm}$ and $\lambda_2 = 600 \, \text{nm}$.

Find: The least number of bright fringes of $480 \, \text{nm}$ light required for the first coincidence with bright fringes of $600 \, \text{nm}$ light.

In Young's double-slit experiment, the condition for bright fringes is

$$d \sin \theta = n\lambda$$

For the first coincidence of bright fringes of two wavelengths, the path difference must be the same for both. Therefore,

$$n_1 \lambda_1 = n_2 \lambda_2$$

Substituting the given wavelengths,

$$n_1 \times 480 = n_2 \times 600$$

So,

$$\frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}$$

Hence,

$$n_1 = 5k, \qquad n_2 = 4k$$

where $k$ is an integer. For the least coincidence, take $k = 1$. Thus,

$$n_1 = 5$$

Therefore, the least number of bright fringes of $480 \, \text{nm}$ light is $5$. The correct option is D.

Given: The wavelengths are $480 \, \text{nm}$ and $600 \, \text{nm}$.

Find: The first common bright fringe corresponding to the $480 \, \text{nm}$ pattern.

The position of the $m$-th bright fringe in Young's double-slit experiment is

$$y_m = \frac{m\lambda D}{d}$$

For coincidence of bright fringes,

$$y_m(480) = y_n(600)$$

So,

$$\frac{m \times 480 \, D}{d} = \frac{n \times 600 \, D}{d}$$

Cancelling the common factors $D$ and $d$,

$$m \times 480 = n \times 600$$

Therefore,

$$\frac{m}{n} = \frac{600}{480} = \frac{5}{4}$$

The smallest integer solution is

$$m = 5, \qquad n = 4$$

Therefore, the first coincidence occurs at the 5th bright fringe of $480 \, \text{nm}$ light. The correct option is D.