Question

In YDSE the parallel light of wavelength 600 nm is incident on the slits making an angle ${60}^0$ with $S_1{S}_2$ line. The separation between the slits $S_1$ & $S_2$ is 0.3mm. A thin glass plate of refractive index 1.5 is placed infront of one slit such that central maxima obtained at symmetric point of YDSE arrangement. The thickness of the glass plate is $x\times {10}^{-4}m$. Find the value of x.

Answer 1

When a thin glass plate is placed anywhere in the path of the beam, the net path difference is given by,
$(\mu -1)t=n\lambda =dcos\theta$
Given $\mu =1.5$, $d=0.3mm=0.3\times {10}^{-3}m$ and $t=X^x\times {10}^{-4}m$
Hence we have
$(1.5-1)x\times {10}^{-4}=0.3\times {10}^{-3}\times cos{60}^o$
$⟹x=3$