Question

The maximum intensity of fringes in Young's experiment is $I$. If one of the slits is closed, the intensity at the same point becomes $I_0$. Then relation between $I$ & $I_0$.

• A. $I=I_0$
• B. $I=2I_0$
• C. $I=4I_0$
• D. These is no relation

Answer 1

The correct option is C $I=4I_0$

In Young's experiment, the resultant intensity at a point is given by,
$I_R=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$
Where, $I_0$ is the intensity of each source.
For maximum intensity, $\varphi =2n\pi$, so that, $\cos \varphi =1$

$\therefore I_{max}=4I_0=I......\text{(given)}$

If one of the slit is closed, the intensity at that point will be equal to, $I_0$.

Hence, $\left(C\right)$ is the correct answer.