Question

Light of wavelength $520mm$ passing through a double slit, produces interference pattern of relative intensity versus angular position $\theta$ as shown in the figure. Find the separation d between the slits.

Answer 1

$I_{Net}=4I_o{\cos }^2\left(\frac{\varphi }{2}\right)$

where, $\varphi \to$Phase difference

$\frac{I_{Net}}{4I_o}={\cos }^2\left(\frac{\varphi }{2}\right)$ & $\varphi =(\frac{2\pi }{\lambda }\times \Delta u)$

$I_{relative}={\cos }^2\left(\frac{\pi d\theta }{\lambda }\right)$ & $\varphi =\frac{2\pi }{\lambda }\times d\times \tan \theta$

where, $\theta \to$Angle of deflection

For $\theta ={0.75}^o=\frac{3^o}{4}=\frac{3}{4}\times \frac{\pi }{180}rad$

$\frac{\varphi }{2}=\frac{\pi d}{\lambda }\times \theta$ ($\theta \approx \tan \theta$ for small $\theta$)

$\theta =\frac{\pi }{240}$

$0={\cos }^2(\frac{\pi d}{520}\times \frac{\pi }{240})$

$\frac{\pi }{2}=\frac{{\pi }^2\times d}{520\times {10}^{-9}\times 240}$

$d=\frac{520\times {10}^{-19}\times 240}{2\pi }$

$d=\frac{52\times 24\times {10}^{-17}}{6.28}$

$d=198.72\times {10}^{-4}mm$

$d=1.98\times {10}^{-2}mm$