Question

Find the fringe width of the fringe pattern.

• A. $0.05cm$
• B. $0.25cm$
• C. $0.01cm$
• D. $0.1cm$

Answer 1

The correct option is A $0.05cm$

$S^{\prime }$ is the virtual point source such that light from $S$ and $S^{\prime }$ interfere on the the screen, then this situation resembles with the Young's double slit experiment.

Let d be the distance between the two point source.

$d=2SP=2mm$

$D=2mv=3\times {10}^{8}m{s}^{-1}\nu =6\times {10}^{14}Hz$

using $v=\nu \lambda$

On solving, $\lambda =500nm$

fringe width, $\beta =\frac{\lambda D}{d}$

This gives $\beta =.05cm$