Q 12-10-015NEETJEE MainMedium
In Young's experiment with two slits of equal intensity $I_0$, the intensity at a point where the path difference is $\dfrac{\lambda}{3}$ is
Answer: (D) $I_0$
Concept. $I = 4I_0\cos^2\dfrac{\phi}{2}$ with $\phi = \dfrac{2\pi}{\lambda}\Delta x$.
Step 1. Phase. $\Delta x = \dfrac{\lambda}{3} \Rightarrow \phi = \dfrac{2\pi}{3}$.
Step 2. Intensity. $I = 4I_0\cos^260° = I_0$.
Answer. $I_0$ (Option D).
Solution by Sreeraj P, M.Sc Physics