Q 12-10-064JEE MainJEE Main 2024 (29 Jan, Shift 2)Easy
In Young's double slit experiment, light from two identical sources is superimposing on a screen. The path difference between the two lights reaching at a point on the screen is $\dfrac{7\lambda}{4}$. The ratio of intensity of the fringe at this point with respect to the maximum intensity of the fringe is:
Answer: (A) $\dfrac12$
Phase difference $\phi = \dfrac{2\pi}{\lambda}\times\dfrac{7\lambda}{4} = \dfrac{7\pi}{2}$.
$$\frac{I}{I_{max}} = \cos^2\frac\phi2 = \cos^2\frac{7\pi}{4} = \frac12$$
Solution by Sreeraj P, M.Sc Physics