Question

At a certain point on a screen the path difference for the two interfering rays is ${\left(1/8\right)}^{th}$ of a wavelength. Find the ratio of the intensity at this point to that at the centre of a bright fringe.

Answer 1

$d_1=\frac{d}{8}$

$I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$ [$\varphi$ is the phase difference.]

$\varphi =\frac{2\pi }{d}\times pathdifference$

$=\frac{2\pi }{d}\times \frac{d}{S}=\frac{\pi }{4}$.

$\frac{I}{I_0}={\cos }^2\left(\frac{\varphi }{2}\right)={\cos }^2\left(\frac{\pi }{2}\right)$

$\frac{I}{I_0}=\frac{1}{2}$.