Question

In a Young's Double Slit Experiment, if the distance between the slits is halved and the distance to the screen is doubled, what happens to the fringe width?

• A. Doubled
• B. Halved
• C. Quadrupled
• D. Unchanged

Hint:

Fringe width depends on both the screen distance and the slit separation together, do not work out the effect of each change separately and then add them, combine the two ratios by dividing one by the other. Also note that here the slit separation is being reduced, not increased, which pushes the fringe width up even further.

Answer 1

The Correct Option is C

Topic - Wave Optics (Interference):
This question focuses on Young's Double Slit Experiment (YDSE) and how the geometric parameters of the setup influence the interference pattern.
Step 1: Understanding the Question:
We need to find the ratio of the new fringe width to the old fringe width based on changes in slit separation (\(d\)) and screen distance (\(D\)).
Step 2: Key Formula or Approach:
The fringe width \(\beta\) is given by: \[ \beta = \frac{\lambda D}{d} \] Where:
\(\lambda\) = Wavelength of light.
\(D\) = Distance between slits and screen.
\(d\) = Distance between the two slits.
Step 3: Detailed Solution:
1. Let the initial parameters be \(D\) and \(d\). Initial fringe width \(\beta = \frac{\lambda D}{d}\).
2. New screen distance \(D' = 2D\).
3. New slit separation \(d' = \frac{d}{2}\).
4. Substitute these into the formula for new fringe width \(\beta'\):
\[ \beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{(d/2)} \] \[ \beta' = \frac{2 \lambda D}{d/2} = 4 \left( \frac{\lambda D}{d} \right) \] \[ \beta' = 4\beta \] Step 4: Final Answer:
The fringe width becomes quadrupled (4 times).