Question

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(=1.5), the central fringe shifts by 0.2 cm. The value of t is_________ cm.

• A. \(8 \times 10^{-4}\)
• B. \(6.0 \times 10^{-3}\)
• C. \(5.0 \times 10^{-3}\)
• D. \(5.6 \times 10^{-4}\)

Hint:

The fringe shift `S` is often expressed in terms of fringe width \( \beta = \frac{\lambda D}{d} \). The phase difference introduced is \( \Delta \phi = \frac{2\pi}{\lambda}(n-1)t \), and the shift is \( S = \frac{\beta}{2\pi} \Delta\phi \). This leads to the same formula \( S = \frac{D}{d}(n-1)t \). Notice that the shift is independent of the wavelength of light used.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a Young's Double Slit Experiment (YDSE) problem. Introducing a transparent sheet in the path of one of the slits introduces an additional optical path length. This causes the entire interference pattern, including the central maximum, to shift on the screen. We need to find the thickness of the sheet given the shift.
Step 2: Key Formula or Approach:
When a transparent sheet of thickness `t` and refractive index `n` is introduced, it creates an additional optical path difference of \( \Delta x = (n-1)t \).
This path difference causes a shift `S` in the fringe pattern on the screen, given by:
\[ S = \frac{D}{d} \Delta x = \frac{D}{d}(n-1)t \] where `D` is the distance to the screen and `d` is the slit separation. We can rearrange this formula to solve for `t`.
Step 3: Detailed Explanation:
Given values:
Slit separation, \(d = 0.1\) cm.
Screen distance, \(D = 50\) cm.
Refractive index, \(n = 1.5\).
Fringe shift, \(S = 0.2\) cm.
We use the formula for fringe shift and solve for the thickness `t`.
\[ S = \frac{D}{d}(n-1)t \] \[ t = \frac{S \cdot d}{D(n-1)} \] Substitute the given values into the equation. It's convenient to keep all units in cm.
\[ t = \frac{(0.2 \, \text{cm}) \cdot (0.1 \, \text{cm})}{(50 \, \text{cm})(1.5 - 1)} \] \[ t = \frac{0.02}{50 \times 0.5} \] \[ t = \frac{0.02}{25} \, \text{cm} \] To simplify the fraction:
\[ t = \frac{2 \times 10^{-2}}{25} = \frac{2}{25} \times 10^{-2} = 0.08 \times 10^{-2} \, \text{cm} \] \[ t = 8 \times 10^{-4} \, \text{cm} \] Step 4: Final Answer:
The thickness of the transparent sheet is \(8 \times 10^{-4}\) cm. This corresponds to option (A).