Question

Assume 100 pm X-ray beam is passed through YDSE. Interference pattern is observed on a photographic plate placed 40 cm away from the slits. What should be the separation between the slits so that the separation between two successive maxima is 0.1 mm.

• A. $4\mu m$
• B. $0.4\mu m$
• C. $4nm$
• D. $40\mu m$

Answer 1

Answer: (B)

$0.4\mu m$

In YDSE, the separation between two successive maxima is given by:
$\beta =\frac{\lambda D}{d}$ or $d=\frac{\lambda D}{\beta }$, where, d is the separation between the slits.
So, $d=\frac{{10}^{-10}\times 0.4}{0.1\times {10}^{-3}}=4\times {10}^{-7}m$