In a Young's double slit experiment set up, the two slits are kept $0.4\text{ mm}$ apart and screen is placed at $1\text{ m}$ from slits. If a thin transparent sheet of thickness $20 \, \mu\text{m}$ is introduced in front of one of the slits then center bright fringe shifts by $20\text{ mm}$ on the screen. The refractive index of transparent sheet is given by $\frac{\alpha}{10}$, where $\alpha$ is _________

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In a Young's double slit experiment set up, the two slits are kept $0.4\text{ mm}$ apart and screen is placed at $1\text{ m}$ from slits. If a thin transparent sheet of thickness $20 \, \mu\text{m}$ is introduced in front of one of the slits then center bright fringe shifts by $20\text{ mm}$ on the screen. The refractive index of transparent sheet is given by $\frac{\alpha}{10}$, where $\alpha$ is __________.

Young's Double Slit Fringe Shift Analysis

The problem involves calculating the refractive index of a transparent sheet placed in a Young's double-slit experiment, given the fringe shift.

Given Data

Slit separation, $d = 0.4\text{ mm} = 0.4 \times 10^{-3}\text{ m}$
Screen distance, $D = 1\text{ m}$
Sheet thickness, $t = 20 \, \mu\text{m} = 20 \times 10^{-6}\text{ m}$
Fringe shift, $\Delta y = 20\text{ mm} = 20 \times 10^{-3}\text{ m}$
Refractive index, $\mu = \frac{\alpha}{10}$

Fringe Shift Formula

When a thin transparent sheet of thickness $t$ and refractive index $\mu$ is introduced in front of one slit, the central bright fringe shifts. The formula for this shift $\Delta y$ is:

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

Calculation of Refractive Index

Substitute the given values into the fringe shift formula: $ 20 \times 10^{-3}\text{ m} = \frac{1\text{ m}}{0.4 \times 10^{-3}\text{ m}} (\mu - 1) (20 \times 10^{-6}\text{ m}) $
Simplify the equation: $ 20 \times 10^{-3} = (2.5 \times 10^3) (\mu - 1) (20 \times 10^{-6}) $ $ 20 \times 10^{-3} = (50 \times 10^{-3}) (\mu - 1) $
Solve for $(\mu - 1)$: $ \mu - 1 = \frac{20 \times 10^{-3}}{50 \times 10^{-3}} = \frac{20}{50} = 0.4 $
Solve for the refractive index $\mu$: $ \mu = 1 + 0.4 = 1.4 $

Determining Alpha

The refractive index is given as $\mu = \frac{\alpha}{10}$. Equating this with the calculated value:

$ 1.4 = \frac{\alpha}{10} $

Solve for $\alpha$:

$ \alpha = 1.4 \times 10 = 14 $

Therefore, the value of $\alpha$ is 14.

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