Question

In a Young's double slit experiment the intensity of the resultant wave at a point P on the screen is I where the path difference between the waves from coherent sources $S_1$ and $S_2$ is $\lambda$. Then the intensity of the resultant wave at a point where the path difference is $\lambda /4$ is given by.

• A. $I/\sqrt{2}$
• B. $2I$
• C. $4I$
• D. $I/2$

Answer 1

The correct option is D $I/2$

Given,

The intensity of the resultant wave at a point P on the screen is $I$ when the path difference is $\lambda$

Here we have to find the intensity of the resultant wave at a point where the path difference is $\lambda /4$.

We have Intensity, $I=I$ ${\cos }^2\frac{\varphi }{2}$ ............... (1)

Where $\varphi$ is phase difference

Also, Phase difference, $\varphi$ $=\frac{2\pi }{\lambda }$ X path difference .................(2)

Given, path difference is $\frac{\lambda }{4}$

So, pahse difference, $\varphi$ = $\frac{2\pi }{\lambda }$ X $\frac{\lambda }{4}$

That is, $\varphi$ = $\frac{\pi }{2}$

Substituting the value of $\varphi$ in equation (1) we get,

Intensity, $I$ = $I$ ${\cos }^2\frac{\pi }{4}$