MHTCET 23 Apr 2025 Shift 2 Paper

Concept:In Young's double slit experiment, the intensity $I$ at a point depends on the phase difference $\phi$ between the two interfering waves. The relation is $I = I_0 \cos^2(\phi/2)$, where $I_0$ is the maximum intensity.Explanation:The phase difference $\phi$ is calculated from the path difference $\Delta x$ using the formula:$$\phi = \frac{2\pi}{\lambda} \Delta x$$Given the path difference is $\frac{\lambda}{4}$, we get:$$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$$Using the intensity formula:$$I = I_0 \cos^2\left(\frac{\phi}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right)$$Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, we have:$$\cos^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$Therefore:$$I = \frac{I_0}{2}$$Now, the required ratio is:$$\frac{I_0}{I} = \frac{I_0}{I_0/2} = 2$$Thus, $I_0 : I = 2 : 1$.Answer:The ratio of $I_0$ to $I$ is $2:1$. The correct option is A.