Question

In YDSE experiment D is distance of screen from slit, d is seperation between slits and $\lambda$ is wavelength of light used. If distance of nearest point to the central maxima where intensity is same as that due to single slit is $\frac{nD\lambda }{md}$ then n + m is

Answer 1

$I_R=4I_{0}co{s}^2\frac{\varphi }{2}$ here $\varphi$ is phase difference

For $I_R=I_0\Rightarrow I_0=4I_{0}co{s}^2\frac{\varphi }{2}\Rightarrow \varphi ={120}^{\circ }=\frac{2\pi }{3}$

By $\varphi =\frac{2\pi }{\lambda }\Delta x\Rightarrow \frac{2\pi }{3}=\frac{2\pi }{\lambda }\Delta x\Rightarrow \Delta x=\frac{\lambda }{3}$

By $\Delta x=dsin\theta =dtan\theta =\frac{dy}{D}$

$\frac{\lambda }{3}=\frac{dy}{D}\Rightarrow y=\frac{D\lambda }{3d}$
$\Rightarrow n=1,m=3\Rightarrow n+m=4$