Question

In Young's double-slit experiment, the separation between the slits is $d=0.25 \, cm$ and the distance of the screen from the slits is $D=100 \, cm$ . When the wavelength of light used in the experiment is $\lambda =6000 \, \mathring{A} $ , the intensity at a distance $x=4\times 10^{- 5} \, m$ from the central maximum is $\frac{p}{q}I_{0}$ , where $I_{0}$ is the intensity of central maxima and $p$ and $q$ are the smallest integers. What is the value of $p+q$ ?

Answer 1

$\text{Path diff.}=\frac{\textit{xd}}{\textit{D}}$
$\Rightarrow \text{Path diff.}=\frac{4 \times 10^{- 5} \times \text{0.25} \times 10^{- 2}}{1}$
$\text{Path diff.} = 1 \times 1 0^{- 7}$
$\text{Phase diff.}=\frac{\text{path diff.}}{\lambda }\times 2\pi $
$\Rightarrow \text{Phase diff.}=\frac{1 \times 10^{- 7}}{6 \times 10^{- 7}}\times 2\pi $
$\text{Phase diff.}=\frac{2 \pi }{6}\Rightarrow \text{Phase diff.}=\frac{\pi }{3}\Rightarrow \phi=60^{^\circ }$
$\textit{I}_{\text{R}}=\textit{I}_{1}+\textit{I}_{2}+2\sqrt{\textit{I}_{1} \textit{I}_{2}}cos60^{^\circ }\Rightarrow \textit{I}_{\text{R}}=\textit{I}_{1}+\textit{I}_{2}+2\textit{I}\times \frac{1}{2}$
$\textit{I}_{\text{R}}=3\textit{I}\Rightarrow \textit{I}_{\text{R}}=\frac{3 \textit{I}_{0}}{4}$
$\Rightarrow p=3,q=4\Rightarrow p+q=7$