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 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

The correct option is A $\frac{3}{4}$

$\varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}I=I_1+I_2+2\sqrt{I_1{I}_2}cos\varphi I^{\prime }=I+I+2Icos\frac{\pi }{3}=3Iand{I}_0=I+I+2Icos{0}^{\circ }=4I\therefore \frac{I^{\prime }}{I_0}=\frac{3}{4}$