Question

In Young's double-slit experiment, both the slits produce equal intensities on a screen. A $100\%$ transparent thin film of refractive index $\mu =1.5$ is kept in front of one of the slits, due to which the intensity at the point $O$ on the screen becomes $75\%$ of its initial value. If the wavelength of monochromatic light is $720nm$ , then what is the minimum thickness (in $nm$ ) of the film?

Answer 1

Answer: 240

The path difference at the point $O$ is
$\Delta x=(\mu -1)t$
$I=4l_0{\cos }^2\frac{\Phi }{2}$
$0.75(4I_0)=4I_0{\cos }^2\frac{\Phi }{2}$
$\Rightarrow \cos \frac{\Phi }{2}=\pm \frac{\sqrt{3}}{2}$
$\Rightarrow {\Phi }_{\min }=60^{\circ }$
${\Phi }_{\min }=\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\mu -1t=\frac{\pi }{3}$
$t=\frac{\lambda }{6\mu -1}=\frac{{\lambda }_0}{61.5-1}=240nm$