Q71 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
In two separate Young's double-slit experimental set-ups and two monochromatic light sources of different wavelengths are used to get fringes of equal width. The ratios of the slits separations and that of the wavelengths of light used are $2: 1$ and $1: 2$ respectively. The corresponding ratio of the distances between the slits and the respective screens ($D_{1} / D_{2}$) is $\_\_\_\_$.
Correct answer: 4
Solution
Fringe width in double slit: $\beta = \frac{\lambda D}{d}$
Given: $\frac{d_1}{d_2} = 2:1$ and $\frac{\lambda_1}{\lambda_2} = 1:2$, with equal fringe widths
For equal fringe widths: $\frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2}$
$\frac{\lambda_1}{\lambda_2} \times \frac{D_1}{D_2} = \frac{d_1}{d_2}$
$\frac{1}{2} \times \frac{D_1}{D_2} = 2$
$\frac{D_1}{D_2} = 4$
Given: $\frac{d_1}{d_2} = 2:1$ and $\frac{\lambda_1}{\lambda_2} = 1:2$, with equal fringe widths
For equal fringe widths: $\frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2}$
$\frac{\lambda_1}{\lambda_2} \times \frac{D_1}{D_2} = \frac{d_1}{d_2}$
$\frac{1}{2} \times \frac{D_1}{D_2} = 2$
$\frac{D_1}{D_2} = 4$
Hence, the correct answer is 4.