Question

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of \( 7^{\text{th}} \) bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm, the thickness of mica sheet is \( \alpha \times 10^{-9} \text{ m} \). The value of \( \alpha \) is _______.

Answer 1

Correct Answer: 5625

Step 1: Understanding the Concept:
Introducing a transparent sheet of thickness \( t \) in one arm of a Young's Double Slit Experiment (YDSE) adds an extra optical path length. This shifts the central maxima to where the geometric path difference compensates for the optical delay.

Step 2: Key Formula or Approach:
Shift in path difference: \( \Delta p = (\mu - 1)t \).
If it shifts to the \( n^{\text{th}} \) maxima, then \( (\mu - 1)t = n\lambda \).

Step 3: Detailed Explanation:
Given: \( \mu = 1.56 \), \( n = 7 \), \( \lambda = 450 \text{ nm} = 450 \times 10^{-9} \text{ m} \).
We set up the equality for shift to the 7th bright fringe:
\[ (1.56 - 1)t = 7 \times 450 \times 10^{-9} \]
\[ 0.56t = 3150 \times 10^{-9} \]
\[ t = \frac{3150}{0.56} \times 10^{-9} = 5625 \times 10^{-9} \text{ m} \]
Comparing this with \( \alpha \times 10^{-9} \), we get \( \alpha = 5625 \).

Step 4: Final Answer:
The value of \( \alpha \) is 5625.