Question

Two sources give interference pattern which is observed on a screen, D distance apart from the sources. The fringe width is 2b. If the distance D is now doubled, then the fringe width will become –

• A. b
• B. $\frac{\beta}{2}$
• C. 4b
• D. Remains same

Answer 1

Answer: (C)

4b

Answer 2

Fringe width = $\frac{D\lambda}{d}$

\ fringe width µD.

If D is now doubled ; then fringe width also doubles