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.
• B.
• C.
• D.

Answer 1

The correct option is A

The intensity at a point on the screen where the path difference is $\Delta$ is given by

$I=I_{0}co{s}^2(\frac{\pi }{\lambda }\times \Delta x)I=I_{0}co{s}^2(\frac{\pi }{\lambda }\times \frac{\lambda }{6})\Rightarrow \frac{I}{I_0}=\frac{3}{4}$