Question

In Young's double-slit experiment, the distance between slits is $d=0.25cm$ and the distance of the screen $D=120cm$ from the slits. If the wavelength of light used is $\lambda =6000\overset{^\circ }{A}$ and $I_{0}$ is the intensity of central maximum, then the minimum distance of the point from the centre, where the intensity is $\frac{I_{0}}{2}$ is $k\times 10^{- 6}\text{m}$ . What is the value of $k$ ?

Answer 1

$I=\frac{I_{0}}{2}=I_{0}cos^{2}\frac{\phi}{2}$
$\Rightarrow cos\frac{\phi}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{\phi}{2}=\frac{\pi }{4}$
$\Rightarrow \phi=\frac{\pi }{2}$
$\Rightarrow \frac{2 \pi }{\lambda }\left(\frac{y}{D}\right)d=\frac{\pi }{2}$
$\Rightarrow y=\frac{\lambda D}{4 d}=\frac{\left(6000 \times \left(10\right)^{- 10}\right) \times 1 .2}{4 \times \left(0 .25\right) \times \left(10\right)^{- 2}}$
$=72\times 10^{- 6}m$