Q 12-10-033JEE MainJEE Main 2026 (5 Apr, Shift 2)Medium
The maximum intensity in a Young's double slit experiment is $I_0$. Distance between the slits ($d$) is $5\lambda$, where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D = 10d$ is ______.
Answer: (B) $\dfrac{I_0}{2}$
The point opposite a slit is at $y = \dfrac{d}{2}$ from the centre. Path difference:
$$\Delta = \frac{yd}{D} = \frac{d^2}{2D} = \frac{d}{20} = \frac{5\lambda}{20} = \frac{\lambda}{4}$$
Phase difference $\phi = \dfrac{\pi}{2}$, so $I = I_0\cos^2\dfrac{\phi}{2} = \dfrac{I_0}{2}$.
Solution by Sreeraj P, M.Sc Physics