Question

A young's double slit experiment is conducted in the water $\left({\mu }_1\right)$ as shown in the figure. A glass plate of thickness $t$ and refractive index ${\mu }_2$ is placed on the path of $S_2$. Find the magnitude of optical path difference at $O$.

• A. $\left|(\frac{{\mu }_2}{{\mu }_1}-1)t\right|$
  
• B. $\left|(\frac{{\mu }_1}{{\mu }_2}-1)t\right|$
  
• C. $\left|\right({\mu }_2-{\mu }_1\left)t\right|$
  
• D. $\left|\right({\mu }_2-1\left)t\right|$
  

Answer 1

The correct option is A $\left|(\frac{{\mu }_2}{{\mu }_1}-1)t\right|$


The waves arriving from slits $S_2$ will have increased optical path.

The path difference between arriving waves at $O$,
$\Rightarrow \Delta x=[S_{2}O+({\mu }_{rel}-1\left)t\right]-S_{1}O$
$\Rightarrow \Delta x=(S_{2}O-S_{1}O)+({\mu }_{rel}-1)t$
From symmetry of figure,
$S_{2}O=S_{1}O$
$\Rightarrow \Delta x=({\mu }_{rel}-1)t$

or, $\Delta x=\left|(\frac{{\mu }_2}{{\mu }_1}-1)t\right|$

$$\begin{array}{c}\text{Why this question?}\\ \text{Tip : When YDSE setup is placed in liquid and a slab is placed infront of one of the slits, then path difference caused between interfering waves will be based on relative refractive index of the slab.}\end{array}$$