As shown in the figure, in Young's double slit experiment, a thin plate of thickness t = 10 μm and refractive index μ = 1 . 2 is inserted infront of slit S 1 . The experiment is conducted in air

Question

As shown in the figure, in Young's double slit experiment, a thin plate of thickness $t = 10 μm$ and refractive index $μ = 1.2$ is inserted infront of slit $S_{1}$ . The experiment is conducted in air ( $μ = 1$ ) and uses a monochromatic light of wavelength $λ = 500 nm$ . Due to the insertion of the plate, central maxima is shifted by a distance of $x β_{0}$ . $β_{0}$ is the fringe-width before the insertion of the plate. The value of the $x$ is ______.

TestHub · Accepted Answer

Fringe shift due to slab in Young's double slit experiment is given by $∆ x = \frac{D (μ - 1) t}{d}$

Since, fringe width $β_{0} = \frac{D}{d} λ$

Then, $∆ x = \frac{β_{0} (μ - 1) t}{λ}$

Putting the values, we have

$= \frac{10 \times 10^{- 6} (1.2 - 1)}{5 \times 10^{- 7}} β_{0}$

$= \frac{10^{- 5} \times 0.2}{5 \times 10^{- 7}} β_{0} = 4 β_{0}$

Hence, the value of $x = 4$ .

Physics - Wave Optics Question with Solution | TestHub

Doubts & Discussion