Question

In a Young's double slit experiment, $d=1mm,\lambda =6000\mathring{A}$ and $D=1m$ (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

Answer: 0.20

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 $=2\times \frac{\beta }{6}=\frac{\beta }{3}=0.20mm$