Question

At $t=1s$, a slab of thickness $2\times {10}^{-3}mm$ and refractive index $1.5$ is placed just in front of $S_1$. The central maximum is formed at

• A. $1mm$ above $C$
• B. $1mm$ below $C$
• C. $2mm$ above $C$
• D. $2mm$ below $C$

Answer 1

The correct option is B $1mm$ below $C$

Path Difference Due to Slab:

${\delta }_{slab}=t\times (\mu -1)=1\times {10}^{-3}mm$

Path Difference Due to Source Position:

${\delta }_{source}=2\times {10}^{-3}mm$

As we know that both are applied on different slits, we will consider their difference:

${\delta }_{effective}=1\times {10}^{-3}mm$

for Central Maxima, Path difference should be 0.

Therefore,

${\delta }_{effective}=\frac{x\times D}{d}$

which gives $x=1mm$

Position will be below C as we know that At C, $S_2$ has more Path Difference than $S_1$. To Equalize Path Difference, below is the only option

Answer is B: 1mm Below C