Question

In Young's Double Slit Experiment intensity at a point is (1/4) of the maximum intensity. Angular position of the point is:

• A. $si{n}^{-1}\left(\frac{2\lambda }{d}\right)$
  
• B. $si{n}^{-1}\left(\frac{\lambda }{2d}\right)$
• C. $si{n}^{-1}\left(\frac{\lambda }{3d}\right)$
• D. $si{n}^{-1}\left(\frac{\lambda }{d}\right)$

Answer 1

The correct option is C $si{n}^{-1}\left(\frac{\lambda }{3d}\right)$

Interesting at a point $\left(I\right)$ where phase difference is $\delta bxYDSE$ in given by :

$I=Imax{\cos }^2\delta /2$ [$Imax$: maximum intensity ]

at the point where $I=\frac{Imax}{4}$, we get :

$\frac{Imax}{4}=Imax{\cos }^2\delta /2$

$\Rightarrow {\cos }^2\delta /2=\frac{1}{4}\Rightarrow \cos \delta /2=\pm \frac{\delta }{2}=\frac{\pi }{3}$ or $\frac{2\pi }{3}$

$\Rightarrow \delta =2\frac{\pi }{3}$ or $\frac{2\pi }{3}$, well take $\frac{2\pi }{3}$ only.

Thus path difference $\left(△x\right)$ at that point in,

$△x=\frac{\lambda }{2\pi }\delta =\frac{2\lambda }{2\pi }.\frac{\pi }{3}.\Rightarrow △x=\frac{\lambda }{B}$.

Also, $△x=d\sin \theta$ ($\theta :$ angular position )

$\Rightarrow \sin \theta =\frac{\lambda }{3d}\Rightarrow \theta ={\sin }^{-1}\left(\frac{\lambda }{3d}\right)$