Question

In a Young ’s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $\frac{1}{8}th$ of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is

Answer 1

Answer: 0.85

Given, path difference,$\Delta x=\frac{\lambda }{8}$
Phase difference$\left(\Delta \varphi \right)$ is given by
$\Delta \varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)$
$\Delta \varphi =\frac{\left(2\pi \right)}{\lambda }\frac{\lambda }{8}=\frac{\pi }{4}$
For two sources in different phases,
$I=I_0$ cos${}^2\left(\frac{\pi }{8}\right)$
$\frac{I}{I_0}=co{s}^2\left(\frac{\pi }{8}\right)$
$=\frac{1+cos\frac{\pi }{4}}{2}=\frac{1+\frac{1}{\sqrt{2}}}{2}=0.85$