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 $\lambda =6000\stackrel{{}^{\circ }}{A}$ 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

Answer: 72

$I=\frac{I_0}{2}=I_{0}co{s}^2\frac{\varphi }{2}$
$\Rightarrow cos\frac{\varphi }{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{\varphi }{2}=\frac{\pi }{4}$
$\Rightarrow \varphi =\frac{\pi }{2}$
$\Rightarrow \frac{2\pi }{\lambda }\left(\frac{y}{D}\right)d=\frac{\pi }{2}$
$\Rightarrow y=\frac{\lambda D}{4d}=\frac{\left(6000\times {\left(10\right)}^{-10}\right)\times 1.2}{4\times \left(0.25\right)\times {\left(10\right)}^{-2}}$
$=72\times 10^{-6}m$