Question

In Young's doubles slit experiment, the angular position of a point on the central maxima whose intensity is one fourth of maximum intensity, then

• A.
  ${\sin }^{-1}\left(\frac{\lambda }{d}\right)$
• B.
  ${\sin }^{-1}\left(\frac{\lambda }{2d}\right)$
• C.
  
  ${\sin }^{-1}\left(\frac{\lambda }{3d}\right)$
  
• D.
  ${\sin }^{-1}\left(\frac{\lambda }{4d}\right)$

Answer 1

The correct option is C


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

In Young's double slit experiment the resultant intensity at a point having phase difference ${}^{\prime }{\varphi }^{\prime }$ is,

$I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)....\left(1\right)$

For, $I_{max};\cos \frac{\varphi }{2}=1$

$\Rightarrow I_{max}=4I_0$

According to question, $I=\frac{I_{max}}{4}=\frac{4I_0}{4}$

Then (1) becomes,

$\frac{4I_0}{4}=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

$\Rightarrow \frac{1}{2}=\cos \left(\frac{\varphi }{2}\right)$

$\Rightarrow \frac{\varphi }{2}={\cos }^{-1}\left(\frac{1}{2}\right)$

$\Rightarrow \frac{\varphi }{2}=\frac{\pi }{3}$

$\therefore \varphi =\frac{2\pi }{3}$

Hence, the corresponding path difference is,

$y=\frac{\lambda }{2\pi }\times \varphi$

$\Rightarrow y=\frac{\lambda }{2\pi }\times \frac{2\pi }{3}=\frac{\lambda }{3}$

For the bright fringe,

$y=d\sin \theta$

$\Rightarrow \sin \theta =\frac{y}{d}=\frac{\lambda }{3d}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.