Question

In a Young's double slit experiment, $d=1\, mm , \lambda=6000\,\mathring{A}$ and $D=1\, m$ (where $d, \lambda$ and $D$ have usual meaning). Each of slit individually produces same intensity on the screen. The minimum distance between two points on the screen having $75 \%$ intensity of the maximum intensity is in mm.

Answer 1

The intensity at distance $x$-from central order bright on screen is
$I=I_{0} \cos ^{2} \frac{\pi x}{\beta}$
where $I_{0}=$ maximum intensity $\beta=$ fringe width
If $I=\frac{3}{4} I_{0} \Rightarrow \cos \frac{\pi x}{\beta}=\frac{\sqrt{3}}{2}=\frac{\pi}{6} \cos$
$\therefore x=\frac{\beta}{6}$
Hence the required distance $=2 \times \frac{\beta}{6}=\frac{\beta}{3}=0.20\, mm$