Question:medium

In a Young's Double Slit Experiment (YDSE), the separation between the two coherent slits is halved, and the perpendicular distance between the plane of the slits and the viewing screen is doubled. If the initial fringe width was \( \beta \), what will the new modified fringe width \( \beta' \) be?

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Fringe width is directly proportional to screen distance (\( D \)) and inversely proportional to slit separation (\( d \)). Halving a value in the denominator always effectively doubles the overall result.
Updated On: May 30, 2026
  • \( \beta \)
  • \( 2\beta \)
  • \( 4\beta \)
  • \( \frac{\beta}{4} \)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Young's Double Slit Experiment is a classic demonstration of the interference of light.
When coherent light waves from two narrow slits overlap, they create an interference pattern on a screen consisting of alternating bright and dark bands.
The "fringe width" (\(\beta\)) is the distance between any two consecutive bright fringes or two consecutive dark fringes.
The width of these fringes is determined by the geometric parameters of the setup and the wavelength of the light source used.
Step 2: Key Formula or Approach:
The formula for the fringe width in a standard YDSE setup is:
\[ \beta = \frac{\lambda D}{d} \]
Where:
\(\lambda\) = wavelength of the light used.
\(D\) = distance between the slits and the screen.
\(d\) = separation distance between the two slits.
Step 3: Detailed Explanation:
Let the initial fringe width be represented by \(\beta = \frac{\lambda D}{d}\).
According to the changes described in the problem:
1. The new slit separation is halved: \(d' = \frac{d}{2}\).
2. The new screen distance is doubled: \(D' = 2D\).
3. The wavelength of the source remains unchanged: \(\lambda' = \lambda\).
Now, let's substitute these modified experimental variables into the fringe width equation to find the new value \(\beta'\):
\[ \beta' = \frac{\lambda' D'}{d'} \]
\[ \beta' = \frac{\lambda \cdot (2D)}{(d/2)} \]
To simplify this expression, we rearrange the fraction by moving the divisor in the denominator (\(2\)) to the numerator:
\[ \beta' = 2 \cdot 2 \cdot \left( \frac{\lambda D}{d} \right) \]
\[ \beta' = 4 \left( \frac{\lambda D}{d} \right) \]
Since the original fringe width was \(\beta = \frac{\lambda D}{d}\), we substitute it back:
\[ \beta' = 4\beta \]
The fringe width increases fourfold because it is directly proportional to the screen distance (which doubled) and inversely proportional to the slit separation (which was halved).
Step 4: Final Answer:
The new modified fringe width \(\beta'\) is \(4\beta\).
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