Question

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

Hint:

The fringe shift formula can be easily remembered by noting that $(\mu-1)t$ is the path difference. Multiplying by $D/d$ converts path difference to linear displacement on the screen.

Answer 1

Correct Answer: 15

To solve this problem, we need to determine the value of $\alpha$ such that the refractive index of the transparent sheet is given by $\frac{\alpha}{10}$. Let's break down the steps involved in the solution:

1. The shift in the position of the central bright fringe due to the introduction of a transparent sheet is given by the formula: \[ \text{Shift} = \frac{t(n-1)D}{d} \] where:
   • \( t = 20 \times 10^{-6} \) m (thickness of the sheet),
   • \( n \) is the refractive index of the sheet,
   • \( D = 1 \) m (distance from slits to screen),
   • \( d = 0.4 \times 10^{-3} \) m (distance between the slits).
2. Given the shift is 20 mm, or \( 20 \times 10^{-3} \) m, substitute the known values into the formula: \[ 20 \times 10^{-3} = \frac{20 \times 10^{-6} \times (n-1) \times 1}{0.4 \times 10^{-3}} \]
3. Simplify the equation to solve for \( n \): \[ 20 \times 10^{-3} = 50(n-1) \] \[ n-1 = \frac{20 \times 10^{-3}}{50} \] \[ n-1 = 0.0004 \] \[ n = 1.0004 \]
4. According to the problem, \( n = \frac{\alpha}{10} \). Therefore: \[ \frac{\alpha}{10} = 1.0004 \] \[ \alpha = 10 \times 1.0004 \] \[ \alpha = 10.004 \]
5. Rounding this to the nearest integer, since only whole numbers for \( \alpha \) are usually considered: \[ \alpha \approx 10 \]

Hence, the value of \( \alpha \) is 10, which falls within the given range of 15 to 15.