Question

In 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 $I_0$ denotes the maximum intensity, then $\frac{I}{I_0}$ is equal to

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

Answer 1

The correct option is C

$\frac{3}{4}$

For maximum intensity $\varphi =0$,

$I_0=a^2+a^2+2a\times acos\varphi =4a^2$

When $x=\frac{\lambda }{6}$

$\varphi =\frac{2\pi }{\lambda }x=\frac{2\pi }{\lambda }\frac{\lambda }{6}=\frac{\pi }{3}={60}^{\circ }$

Then

$I=a^2+a^2+2a\times acos{60}^{\circ }=3a^2$

$\therefore \frac{I}{I_0}=\frac{3a^2}{4a^2}=\frac{3}{4}$