Question

In a Young's double slit experiment with $d=1\text{mm}$ and $D=1\text{m}$, slabs of $(t=1\mu \text{m},\mu =3)$ and $(t=0.5\mu \text{m},\mu =2)$ are introduced in front of upper and lower slits, respectively. Find the shift in the fringe pattern.

• A. $1.5\text{mm}$
• B. $2\text{mm}$
• C. $3\text{mm}$
• D. $2.5\text{mm}$

Answer 1

The correct option is A $1.5\text{mm}$

Optical path for light coming from the upper slit, $S_1$ is,
$S_{1}P+t(\mu -1)=S_{1}P+1(3-1)=S_{1}P+2\mu \text{m}$

Similarly optical path for light coming from $S_2$ is
$S_{2}P+t(\mu -1)=S_{2}P+0.5(2-1)=S_{2}P+0.5\mu \text{m}$

Path difference, $\Delta x=(S_{2}P+0.5\mu \text{m})-(S_{1}P+2\mu \text{m})$
$=(S_{2}P-S_{1}P)-\left(1.5\mu \text{m}\right)=\frac{yd}{D}-1.5\mu \text{m}$

The shift in interference pattern will be,
$\beta =\frac{D}{d}\times 1.5\times {10}^{-6}$
$=\frac{1}{{10}^{-3}}\times 1.5\times {10}^{-6}=1.5\text{mm}$

So, the whole pattern is shifted by $1.5\text{mm}$.

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