Question

In $YDSE$ shown in the Figure, a parallel beam of light is incident on the slits from a medium of refractive index $n_1$. The wavelength of light in the medium is ${\lambda }_1$. A transparent slab of thickness $t$ and refractive index $n_3$ is put in front of one slit. The refractive index of the medium between the beam and the plane of the slits is $n_2$. The phase difference between the light waves reaching point $O$ (symmetrical, relative to the slit) 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. $\frac{2\pi n_1}{n_2{\lambda }_1}\left(\frac{n_3}{n_2}\right)t$
• D. $\frac{2\pi n_1}{{\lambda }_1}(n_3-n_2)t$

Answer 1

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

Optical path of light from $S_1$ to $O$ is given by,
$x_1=(S_{1}O-t)n_2+t{n}_3$

Optical path of light from $S_2$ to $O$ is given by,
$x_1=\left(S_{2}O\right)n_2$

Path difference in air at point $O$:

$\Rightarrow \Delta x=\left[\right(S_{1}O-t)n_2+t{n}_3-(S_{2}O\left)n_2\right]$

$\Delta x=\left[\right(S_{1}O-S_{2}O)n_2+(n_3-n_2\left)t\right]$


Due to symmetry, $S_{1}O=S_{2}O$

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

The corresponding phase difference is given by,

$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$

If ${\lambda }_1$ is the wavelength of light in the medium of refractive index $n_1$, then
The wavelength of light in air is given by,

$\lambda =n_1{\lambda }_1$

$\Rightarrow \Delta \varphi =\frac{2\pi }{n_1{\lambda }_1}\Delta x$

$\therefore \Delta \varphi =\frac{2\pi }{n_1{\lambda }_1}(n_3-n_2)t$

Hence, $\left(A\right)$ is the correct answer.