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

Answer: (C)

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

Answer 2

\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).