Question

An interference pattern is observed by Young's double slit experiment. If now the separation between coherent source is halved and the distance of screen from coherent sources is doubled, then now fringe width

• A. becomes double
• B. becomes one-fourth
• C. remains same
• D. becomes four times

Answer 1

Answer: (D)

Fringe width $(W)=\frac{D \lambda}{d}$
Now, distance between the coherent sources is halved
i.e., $d'=d / 2$ and distance of coherent source and screen is doubled
i.e., $D'=2 D$
$\therefore W'=\frac{D' \lambda}{d'}=\frac{2 D \lambda}{(d / 2)}=\frac{4 D \lambda}{d}$
$\Rightarrow \frac{W'}{W}=4$
$\Rightarrow W'=4\, W$ (becomes four times)