Question

In Young's double-slit experiment, a slab of thickness $1.2\mu m$ and refractive index $1.5$ is placed in front of one slit and another slab of thickness $t$ and refractive index $2.5$ is placed in front of the second slit. If the position of the central fringe remains unaltered, then the thickness $t$ is-

• A. $0.4µm$
• B. $0.8µm$
• C. $1.2µm$
• D. $7µm$

Answer 1

Answer: (A)

$\because $ Central fringe at same position.
$\therefore $ Path difference $\left(\right.\Delta x\left.\right)=0$
Now $\left(\left(\mu \right)_{1} - 1\right)t_{1}=\left(\left(\mu \right)_{2} - 1\right)t_{2}$
$\left(\right.1.5-1\left.\right)\times 1.2=\left(\right.2.5-1\left.\right)t \\ t=\frac{0 . 5 \times 1 . 2}{1 . 5}=0.4µm$