Question

In Young's double slit experiment using monochromatic light of wavelength $\lambda$ . the intensity of light at a point on the screen where path difference is $\lambda$ , is $k$ units. The intensity of light at a point, where path difference is $\lambda /3$ is

• A. $k/2$
• B. $k/3$
• C. $k/4$
• D. $2k/3$

Answer 1

Answer: (C)

$k/4$

Answer 2

In Young's double slit experiment, intensity at any point on the screen is given by
$I=I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}cos\, \phi \ldots\left(i\right)$
where, $I_{1}$ and $I_{2}$ are intensity of two sources
But both sources are identical, hence
$I_{1}=I_{2}=I_{0}$ $[\therefore$ from E $\left(i\right)$]
$I=I_{0}+I_{0}+2\sqrt{I_{0}I_{0}}cos \phi$
$I=2I_{0}+2I_{0}\,cos\, \phi$
$=2I_{0} \left(1+cos\,\phi\right)=2I_{0}\cdot2cos^{2} \frac{\phi}{2}$
$I=4I_{0} cos^{2} \frac{\phi}{2} \ldots\left(ii\right)$
When path difference is $\lambda$, then phase difference $\phi=2\pi$
$\therefore$ From E $\left(ii\right)$,
$I=4I_{0} \, cos^{2} \frac{2\pi}{2}$
$I=4I_{0}=k$ (given) $\ldots\left(iii\right)$
Again when path difference is $\frac{\lambda}{3}$, then phase difference $\phi=\frac{2\pi}{\lambda}\cdot\frac{\lambda}{3}=\frac{2\pi}{3}$
$\therefore$ From E $\left(ii\right)$, $I=4I_{0} cos^{2} \left(\frac{2\pi /3}{2}\right)=4I_{0} cos^{2} \frac{\pi}{3}$
$=4I_{0} \cdot\frac{1}{4}=k\cdot\frac{1}{4}=\frac{k}{4}$