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 the light used) is $I$. If $I_o$ denotes the maximum intensity, then $\left(I/I_o\right)$ 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)

$\frac{3}{4}$

phase difference at path difference $\lambda /6$
is $\varphi =\frac{2\pi }{6}=\frac{\pi }{3}$
and $I=I_{0}co{s}^2\left(\frac{\varphi }{2}\right)$
$=I_{0}co{s}^2\left(\frac{\pi }{6}\right)$
$=I_0\left(\frac{\sqrt{3}}{2}{)}^2\right(\because cos\frac{\pi }{6})=\frac{\sqrt{3}}{2}$
$=I_0\frac{3}{4}$
So, $\frac{I}{I_0}=\frac{3}{4}$