Question

In a Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{6}(\lambda$ being the wavelength of light used) is $I$. If $I_{0}$ denotes the maximum intensity, $\frac{I}{I_{0}}$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

In Young's double slit experiment intensity at a point is given by
$I=I_{0} \cos ^{2}\left(\frac{\phi}{2}\right)$
or $\frac{I}{I_{0}}=\cos ^{2}\left(\frac{\phi}{2}\right)\,\,\,\,\,\,\dots(i)$
Phase difference, $\phi=\frac{2 \pi}{\lambda} \times$ path difference
$\therefore \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6} $
or $ \phi=\frac{\pi}{3}\,\,\,\,\,\dots(ii)$
Substitute eqn. (ii) in eqn. (i), we get
$\frac{I}{I_{0}}=\cos ^{2}\left(\frac{\pi}{6}\right)$
or $\frac{I}{I_{0}}=\frac{3}{4}$