AIIMS 2019 Physics Young's Double Slit Experiment MCQ Question

In Young's Double Slit Experiment (YDSE), the intensity $I$ at a point on the screen is given by the formula $I = I_0 \cos^2 \left( \frac{\pi d}{\lambda} \right)$, where $I_0$ is the maximum intensity, $d$ is the path difference, and $\lambda$ is the wavelength. For the intensity to be 50% of the central maximum, we set $I = \frac{I_0}{2}$, leading to $\cos^2 \left( \frac{\pi d}{\lambda} \right) = \frac{1}{2}$. This results in $\frac{\pi d}{\lambda} = \frac{\pi}{4}$, giving $d = \frac{\lambda}{4}$. Given $\lambda = 500 \, \text{nm} = 0.5 \, \mu m$, the distance $d$ calculates to $125 \, \mu m$, which is not listed among the options, confirming that option D (N/A) is correct. The other options (1000 μm, 500 μm, and 250 μm) are incorrect as they do not correspond to the calculated distance for 50% intensity.