Question

In a $YDSE$ experiment with $d=1\text{mm}$ and $D=1\text{m}$, two slabs with refractive index and thickness $(t_1=1\mu \text{m},{\mu }_1=3)$ and $(t_2=0.5\mu \text{m},{\mu }_2=2)$ are introduced in front of upper and lower slits respectively, The shift in the fringe pattern will be

• A. $1.5\text{mm}$ downwards
• B. $1.5\text{mm}$ upwards
• C. $2.5\text{mm}$ upwards
• D. $2\text{mm}$ downwards

Answer 1

The correct option is B $1.5\text{mm}$ upwards

The presence of slabs will affect the optical path of light coming from both slits.


Optical path for light coming from upper slit $S_1$ will be,
$x_1=S_{1}P+({\mu }_1-1)t_1=S_{1}P+(3-1)1\mu \text{m}$

$x_1=S_{1}P+2\mu \text{m}$

Optical path for light coming from lower slit $S_2$ will be,

$x_2=S_{1}2P+({\mu }_2-1)t_2=S_{2}P+(2-1)0.5\mu \text{m}$

$x_2=S_{2}P+0.5\mu \text{m}$

Path difference between light rays arriving from both slits:

$\Delta x=x_2-x_1=(S_{2}P+0.5\mu \text{m})-(S_{1}P+2\mu \text{m})$

$\Rightarrow \Delta x=(S_{2}P-S_{1}P)-1.5\mu \text{m}$

We know that, $S_{2}P-S_{1}P=\frac{yd}{D}$,

where $y$ is the distance of point $P$ from centre of screen

$\Rightarrow \Delta x=\frac{yd}{D}-1.5\mu \text{m}$

For central maxima position,
$\Delta x=0$

$\Rightarrow \frac{yd}{D}=1.5\mu \text{m}$

$\Rightarrow y=\frac{1.5\mu \text{m}\times D}{d}$

$y=\frac{1.5\mu \text{m}\times 1\text{m}}{1\text{mm}}$

$\therefore y=1.5\times {10}^{-3}\text{m}=1.5\text{mm}$

Thus, shift in fringe pattern $=$ shift in central maxima position

$\Rightarrow \Delta y=1.5\text{mm}$

Since, the refractive index and thickness of upper slab is greater, the whole pattern will shift $1.5\text{mm}$ upwards.

Hence, option $\left(B\right)$ is correct.

Why this question ? Note: Placement of the slab in front of a slit will increase the optical path of light wave arriving from that slit.