Question

In Young’s double slit experiment, if the separation between the slits is halved, and the distance between the slits and the screen is doubled, then the fringe width compared to the unchanged one will be

• A. Unchanged
• B. Halved
• C. Doubled
• D. Quadrupled
• E. Fringes will disappear

Hint:

Fringe width $\propto \frac{D}{d}$, so increasing $D$ and decreasing $d$ increases spacing.

Answer 1

The Correct Option is D

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

Step 1: Apply changes.
\[ D \rightarrow 2D,\quad d \rightarrow \frac{d}{2} \]

Step 2: New fringe width.
\[ \beta' = \frac{\lambda (2D)}{d/2} = \frac{2\lambda D}{d/2} \] \[ = 4 \cdot \frac{\lambda D}{d} = 4\beta \]

Step 3: Conclusion.
\[ \beta' = 4\beta \]