Question:
In a YDSE experiment, a sheet of thickness \( t \) and \( \mu = 1.56 \) is introduced at a slit.
The central maxima shifts to the position of the \( 7^{\text{th}} \) maxima.
The wavelength of light is 480 nm. If \( t = x \, \mu\text{m} \), find the value of \( x \).
In a YDSE experiment, a sheet of thickness \( t \) and \( \mu = 1.56 \) is introduced at a slit.
The central maxima shifts to the position of the \( 7^{\text{th}} \) maxima.
The wavelength of light is 480 nm. If \( t = x \, \mu\text{m} \), find the value of \( x \).
Updated On: Apr 9, 2026
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Correct Answer: 6
Solution and Explanation
Step 1: Understanding the Concept:
Introducing a transparent sheet in front of one slit in Young's Double Slit Experiment (YDSE) introduces an additional optical path length, causing the entire fringe pattern to shift.
Step 2: Key Formula or Approach:
Path difference introduced by the sheet: \( \Delta p = (\mu - 1)t \). If the central maxima shifts to the \( n^{th} \) maxima position: \[ (\mu - 1)t = n\lambda \]
Step 3: Detailed Explanation:
Given: \( \mu = 1.56 \), \( n = 7 \), and \( \lambda = 480 \, \text{nm} = 0.48 \, \mu m \). \[ (1.56 - 1)t = 7 \times 0.48 \] \[ 0.56 \times t = 3.36 \] \[ t = \frac{3.36}{0.56} \] \[ t = 6 \, \mu m \] Thus, \( x = 6 \).
Step 4: Final Answer:
The value of \( x \) is 6.
Introducing a transparent sheet in front of one slit in Young's Double Slit Experiment (YDSE) introduces an additional optical path length, causing the entire fringe pattern to shift.
Step 2: Key Formula or Approach:
Path difference introduced by the sheet: \( \Delta p = (\mu - 1)t \). If the central maxima shifts to the \( n^{th} \) maxima position: \[ (\mu - 1)t = n\lambda \]
Step 3: Detailed Explanation:
Given: \( \mu = 1.56 \), \( n = 7 \), and \( \lambda = 480 \, \text{nm} = 0.48 \, \mu m \). \[ (1.56 - 1)t = 7 \times 0.48 \] \[ 0.56 \times t = 3.36 \] \[ t = \frac{3.36}{0.56} \] \[ t = 6 \, \mu m \] Thus, \( x = 6 \).
Step 4: Final Answer:
The value of \( x \) is 6.
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