Q 12-10-143JEE MainJEE Main 2019 (11 Jan, Shift 1)Easy
In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $\dfrac18$th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:
Answer: (B) $0.85$
Phase difference $\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac\lambda8 = \dfrac\pi4$.
$$\frac{I}{I_{\max}} = \cos^2\frac\phi2 = \cos^2\frac\pi8 = \frac{1 + \cos(\pi/4)}{2} = \frac{1 + 0.707}{2} \approx 0.85$$
Solution by Sreeraj P, M.Sc Physics