Q 12-10-118JEE MainJEE Main 2020 (8 Jan, Shift 2)Easy
In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves is $\dfrac{1}{8}$th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is:
Answer: (A) $0.853$
Phase difference $\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{8} = \dfrac{\pi}{4}$.
$$\frac{I}{I_{\max}} = \cos^2\frac{\phi}{2} = \cos^2\frac{\pi}{8} = \frac{1 + \cos(\pi/4)}{2} = \frac{1 + 0.707}{2} \approx 0.853$$
Solution by Sreeraj P, M.Sc Physics