Question

In a young's double slit experiment d = 1mm, $\lambda =6000A^0$ and D - 1m (where d, $\lambda$ D have usual meaning) . Each of slit individually produces the same intensity on the screen. The minimum distance between two points on the screen having 75% intensity of the maximum intensity is $\frac{k}{10}$ mm. The value of K is ___

Answer 1

$Imax=(\sqrt{I_0}+\sqrt{I_0}{)}^2$
$I=\frac{3}{4}{I}_{max}=3I_o=I_0+I_0+2\sqrt{I_0}\sqrt{I_0}cos\theta$
$\therefore \theta =\frac{\pi }{3}\Rightarrow \Delta x=\frac{yd}{D}=\frac{\lambda }{6}$
$\therefore y=\frac{D\lambda }{2}=0.1mm$
$\therefore$ minimum distance on either sides of central maximum = 2 $\times$ 0.1 = 0.2 mm $=\frac{2}{10}$ mm