IIT JEE Advanced 2009 Paper 2

(A)→(P), (S) Since the path difference is $S_1P_0$ = $S_2P_0$ = 0 the phase difference becomes $\delta(P_0)$ = 0 The intensity at any point is I = $I_{\max} \cos^2 \left(\frac{8}{2}\right)$ ow, $I(P_0)$ = $I_{\max}$ [Since $\delta(P_0)$ = 0] $I(P_1)$ = $I_{\max} \cos^2 \left(\frac{\pi}{4}\right)$ = $\frac{I_{\max}}{2}$ Therefore, $I(P_0)$ > $I(P_1)$ (B)→(Q) In this case, the path difference at $P_1$ is $P_1$ = $\frac{\lambda}{4} - (\mu - 1)t$ = $\frac{\lambda}{4} - \frac{\lambda}{4}$ = 0 Therefore, $\delta(P_1)$ = 0. (C)→(T) In this case, the path difference at $P_1$ is $P_1$ = $\frac{\lambda}{4} - (\mu - )t$ = $\frac{\lambda}{4} - \frac{\lambda}{2}$ = $-\frac{\lambda}{4}$ Therefore, the phase difference at $P_1$ is $P_1$ = $\frac{2\pi}{\lambda} \times \left(-\frac{\lambda}{4}\right)$ = - $\frac{\pi}{2}$ Therefore, $I(P_1)$ = $I_{\max} \cos^2 \left(\frac{\pi}{4}\right)$ = $\frac{I_{\max}}{2}$ The path difference at $P_2$ is $P_2$ = $\frac{\lambda}{3} - (\mu - 1)t$ = $\frac{\lambda}{3} - \frac{\lambda}{2}$ = $-\frac{\lambda}{6}$ Therefore, the phase difference at $P_1$ is $P_2$ = $\frac{2\pi}{\lambda} \times \left(-\frac{\lambda}{6}\right)$ = $-\frac{\pi}{3}$ Therefore, $I(P_2)$ = $I_{\max} \cos^2 \left(\frac{\pi}{6}\right)$ = $\frac{3}{4} I_{\max}$ Therefore, $I(P_2)$ > $I(P_1)$. (D)→(R), (S), (T) In this case, the path difference at $P_1$ is $P_1$ = $\frac{\lambda}{4} - (\mu - 1)t$ = $\frac{\lambda}{4} - \frac{3\lambda}{4}$ = $-\frac{\lambda}{2}$ Therefore, the phase difference at $P_1$ is $P_1$ = $\frac{2\pi}{\lambda} \times \left(-\frac{\lambda}{2}\right)$ = - π Therefore, $I(P_1)$ = $I_{\max} \cos^2 \left(\frac{\pi}{2}\right)$ = 0 The path difference at $P_0$ is $P_0$ = 0 - (µ - 1)t = $\frac{-3\lambda}{4}$ Therefore, the phase difference at $P_0$ is $P_0$ = $\frac{2\pi}{\lambda} \times \left(\frac{3\lambda}{4}\right)$ = $\frac{-3\pi}{2}$ Therefore, $I(P_0)$ = $I_{\max} \cos^2 \frac{3\pi}{4}$ = $\frac{I_{\max}}{2}$ Now, the path difference at $P_1$ is $P_1$ = $\frac{\lambda}{4} - (\mu - 1)t$ = $\frac{\lambda}{-}\frac{3\lambda}{4}$ = $-\frac{\lambda}{2}$ The phase difference at $P_1$ is $P_1$ = $\frac{2\pi}{\lambda} \times \left(-\frac{\lambda}{2}\right)$ = - π Therefore, $I(P_1)$ = $I_{\max} \cos^2 \left(\frac{\pi}{2}\right)$ = 0