MHTCET 19 Apr 2026 Shift 2 Paper

Concept:In a biprism experiment, the fringe width is directly proportional to the wavelength of light and the distance of the screen from the slits.Explanation:The fringe width is given by $\beta = \frac{\lambda D}{d}$, where $D$ is the screen distance and $d$ is the separation between the virtual images.Since $d$ remains unchanged, we can write $\frac{\beta_2}{\beta_1} = \frac{\lambda_2 D_2}{\lambda_1 D_1}$.Given $\lambda_1 = 5000 \ \text{Å}$, $\lambda_2 = 6400 \ \text{Å}$, and $D_1 = 100 \ \text{cm}$.The new fringe width is reduced by $20\%$ of its initial value, so $\beta_2 = 0.8 \beta_1$.Substituting the values: $0.8 = \frac{6400 \times D_2}{5000 \times 100}$.Solving for $D_2$: $D_2 = \frac{0.8 \times 5000 \times 100}{6400} = 62.5 \ \text{cm}$.Since $D_2 < D_1$, the screen must be moved towards the source by $100 - 62.5 = 37.5 \ \text{cm}$.Answer:The screen should be moved towards the source by $37.5 \ \text{cm}$.Hence, the correct option is A.