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
$I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$
or $\frac{I}{I_0}={\cos }^2\left(\frac{\varphi }{2}\right)\dots (i)$
Phase difference, $\varphi =\frac{2\pi }{\lambda }\times$ path difference
$\therefore \varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}$
or $\varphi =\frac{\pi }{3}\dots (ii)$
Substitute eqn. (ii) in eqn. (i), we get
$\frac{I}{I_0}={\cos }^2\left(\frac{\pi }{6}\right)$
or $\frac{I}{I_0}=\frac{3}{4}$