BITSAT 2024 Shift 1 Paper

In Young's double slit experiment within a medium with a refractive index of 1.33 , the maximum intensity is represented as $I_0$.To find the intensity at a point on the screen where the path difference between the light from the slits is $\;\frac{\lambda}{4}$, follow these steps:Path Difference: Given as $\Delta x = \;\frac{\lambda}{4}$.Phase Difference Calculation:$\varphi = \left( \;\frac{2\pi}{\lambda} \right) \times \Delta x$Substituting the given path difference:$\varphi = \;\frac{2\pi}{\lambda} \times \;\frac{\lambda}{4} = \;\frac{\pi}{2}$Intensity Calculation: The intensity at any point on the screen with a phase difference $\varphi$ is given by:$I = I_0 \cos^2 \left( \;\frac{\varphi}{2} \right)$Substituting the phase difference:$I = I_0 \cos^2 \left( \;\frac{\pi/2}{2} \right)$Simplifying further:$I = I_0 \cos^2 \left( \;\frac{\pi}{4} \right)$$I = I_0 \left( \;\frac{1}{\sqrt{2}} \right)^2$$I = \;\frac{I_0}{2}$Thus, the intensity at the specified point on the screen is $\;\frac{I_0}{2}$.