Question

In Young's double slit experiment using a source of wavelength \(\lambda\) interference bands are observed on a screen. If the separation between the slits alone is halved in this experiment, then the angular separation \(\omega\) of the fringes on the screen becomes

• A. \(2\omega\)
• B. \(\sqrt{2}\omega\)
• C. \(\frac{\omega}{2}\)
• D. \(\frac{\omega}{\sqrt{2}}\)
• E. \(\frac{\omega}{4}\)

Hint:

\(\omega = \frac{\lambda}{d}\). Smaller slit separation gives wider fringes.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Angular fringe width \(\omega = \frac{\lambda}{d}\), where \(d\) is slit separation.

Step 2: Detailed Explanation:
\(\omega \propto \frac{1}{d}\). If \(d\) is halved (\(d' = \frac{d}{2}\)), then \(\omega' = \frac{\lambda}{d/2} = \frac{2\lambda}{d} = 2\omega\)

Step 3: Final Answer:
Angular separation becomes \(2\omega\).