Question

In Young's double slit experiment, the band width of the fringes observed is \( \beta \), when light of wavelength \( \lambda \) is used. With same experimental set up, to double the band width of the fringes, the wavelength of light required is

• A. \( \lambda \)
• B. \( \frac{\lambda}{2} \)
• C. \( 2\lambda \)
• D. \( \frac{\lambda}{4} \)
• E. \( \frac{\lambda}{8} \)

Hint:

Fringe width directly proportional to wavelength.

Answer 1

The Correct Option is C

Concept: \[ \beta = \frac{\lambda D}{d} \]

Step 1: Relation.
\[ \beta \propto \lambda \]

Step 2: To double \( \beta \).
\[ \lambda \rightarrow 2\lambda \]