Question

In a Youngs 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, $\frac{I}{I_0}$ is equal to

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

Answer 1

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

For a path difference of $\varphi$, the intensity at point is $I=I_{0}co{s}^2\frac{\varphi }{2}$

where $I_0$ is the maximum intensity

Here $\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}$

Thus $\frac{I}{I_0}=co{s}^2\left(\frac{\pi }{6}\right)$

$=\frac{3}{4}$