Question

In Young's double-slit experiment, the separation between the slits is $d=0.25cm$ and the distance of the screen from the slits is $D=100cm$ . 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

Answer: 7

$\text{Path diff.}=\frac{\text{xd}}{\text{D}}$
$\Rightarrow \text{Path diff.}=\frac{4\times 10^{-5}\times \text{0.25}\times 10^{-2}}{1}$
$\text{Path diff.}=1\times 10^{-7}$
$\text{Phase diff.}=\frac{\text{path diff.}}{\lambda }\times 2\pi$
$\Rightarrow \text{Phase diff.}=\frac{1\times 10^{-7}}{6\times 10^{-7}}\times 2\pi$
$\text{Phase diff.}=\frac{2\pi }{6}\Rightarrow \text{Phase diff.}=\frac{\pi }{3}\Rightarrow \varphi =60^{{}^{\circ }}$
${\text{I}}_{\text{R}}={\text{I}}_1+{\text{I}}_2+2\sqrt{{\text{I}}_1{\text{I}}_2}cos60^{{}^{\circ }}\Rightarrow {\text{I}}_{\text{R}}={\text{I}}_1+{\text{I}}_2+2\text{I}\times \frac{1}{2}$
${\text{I}}_{\text{R}}=3\text{I}\Rightarrow {\text{I}}_{\text{R}}=\frac{3{\text{I}}_0}{4}$
$\Rightarrow p=3,q=4\Rightarrow p+q=7$