Calculate Thickness of Mica Sheet in Young Double Slit Experiment

In Young's Double Slit Experiment, when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the light rays, an additional optical path length of $(\mu - 1)t$ is introduced.

As a result, the entire fringe pattern shifts. The shift in the path difference at the position of the central maximum due to the insertion of the mica sheet is equal to: $\Delta x = (\mu - 1)t$

According to the question, the central fringe shifts to the position previously occupied by the $7^{\text{th}}$ bright fringe (with both slits uncovered). The optical path difference corresponding to the $7^{\text{th}}$ bright fringe is $7\lambda$.

Equating the optical path differences, we get: $(\mu - 1)t = 7\lambda$

Substitute the given values into the equation:

• $\mu = 1.56$
• $\lambda = 450\text{ nm} = 450 \times 10^{-9}\text{ m}$

$(1.56 - 1)t = 7 \times 450 \times 10^{-9}\text{ m}$

$0.56 \times t = 3150 \times 10^{-9}\text{ m}$

$t = \frac{3150}{0.56} \times 10^{-9}\text{ m}$

$t = 5625 \times 10^{-9}\text{ m}$

Comparing with the given expression for thickness, $t = \alpha \times 10^{-9}\text{ m}$, we get: $\alpha = 5625$