Question

If a transparent slab of refractive index $\mu$=1.5 and thickness t=$2.5\times {10}^{-5}$ is inserted in front of one of the slits of Young's double slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5 mm and that between slits and screen is 100 cm.

• A. 5 cm
• B. 2.5 cm
• C. 0.25 cm
• D. 0.1 cm

Answer 1

The correct option is B 2.5 cm

As, Path difference $\Rightarrow$

$△x=\frac{yd}{D}$

so, $y=\frac{△xD}{d}$

$y=\frac{t(\mu -1)D}{d}$

$\Rightarrow \frac{2.5\times {10}^{-5}(1.5-1)\left(\frac{100}{100}\right)}{\frac{0.5}{1000}}$

$y\Rightarrow 0.025m$

$\Rightarrow 2.5cm$