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)

In Young's double slit experiment, intensity at any point on the screen is given by
$I=I_1+I_2+2\sqrt{I_1{I}_2}cos\varphi \dots \left(i\right)$
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 Eq. $\left(i\right)$]
$I=I_0+I_0+2\sqrt{I_0{I}_0}cos\varphi$
$I=2I_0+2I_{0}cos\varphi$
$=2I_0\left(1+cos\varphi \right)=2I_0\cdot 2co{s}^2\frac{\varphi }{2}$
$I=4I_{0}co{s}^2\frac{\varphi }{2}\dots \left(ii\right)$
When path difference is $\lambda$, then phase difference $\varphi =2\pi$
$\therefore$ From Eq. $\left(ii\right)$,
$I=4I_{0}co{s}^2\frac{2\pi }{2}$
$I=4I_0=k$ (given) $\dots \left(iii\right)$
Again when path difference is $\frac{\lambda }{3}$, then phase difference $\varphi =\frac{2\pi }{\lambda }\cdot \frac{\lambda }{3}=\frac{2\pi }{3}$
$\therefore$ From Eq. $\left(ii\right)$, $I=4I_{0}co{s}^2\left(\frac{2\pi /3}{2}\right)=4I_{0}co{s}^2\frac{\pi }{3}$
$=4I_0\cdot \frac{1}{4}=k\cdot \frac{1}{4}=\frac{k}{4}$