Question

In Young's double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is

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

Answer 1

The Correct Option is C

$\hspace25mm I=I_{max}cos^2 \big(\frac{\phi}{2}\big)$
$\therefore \hspace20mm \frac{I_{max}}{4}=I_{max}cos^2 \frac{\phi}{2}$
$\hspace20mm cos \frac{\phi}{2}=\frac{1}{2}$
or $\hspace20mm \frac{\phi}{2}=\frac{\pi}{3}$
$\therefore \hspace20mm \phi =\frac{2 \pi}{3}=\big(\frac{2 \pi}{\lambda}\big)\Delta x \hspace20mm$ ...(i)
where $\hspace10mm \Delta x=d \, sin \, \theta$
Substituting in E (i), we get
$\hspace25mm sin \, \theta=\frac{\lambda}{3d}$
or $\hspace 25mm \theta=sin^{-1}\big(\frac{\lambda}{3d}\big)$
$\therefore$ Correct answer is (c).