Question

In YDSE shown in figure, a parallel beam of light is incident on the slits from a medium of refractive index $n_1$. The wavelength of light in this medium is ${\lambda }_1$. A transparent slab of thickness $t$ and refractive index $n_3$ is put in front of one slit. The medium between the screen and the plane of the slits is $n_2$. The phase difference between the light waves reaching point $O$ (symmetrical, relative to the slits) is :

• A. $\frac{2\pi }{n_1{\lambda }_1}(n_3-n_2)t$
• B. $\frac{2\pi }{{\lambda }_1}(n_3-n_2)t$
• C. $\frac{2\pi n_1}{n_2{\lambda }_1}(\frac{n_3}{n_2}-1)t$
• D. $\frac{2\pi n_1}{{\lambda }_1}(n_3-n_2)t$

Answer 1

The correct option is A $\frac{2\pi }{n_1{\lambda }_1}(n_3-n_2)t$

Path difference in air at point O $\Delta x=[(S_{1}O-t)n_2+t{n}_3-\left(S_{2}O\right)n_2]=[(S_{1}O-S_{2}O)n_2+(n_3-n_2)t]$

$Duetosymmetry,(S_{1}O-S_{2}O)=0$

$\Delta x=(n_3-n_2)t$

Phase difference $=\frac{2\pi }{n_1\lambda }\times$ Path difference in air

$\Delta \varphi =\frac{2\pi }{n_1\lambda }\times \Delta x=\frac{2\pi }{n_1\lambda }(n_3-n_2)t$