Question

The maximum intensity of fringes in Young's Experiment is $I$. If one of the slits is closed, then intensity at that place becomes $I_0$. Then relation between $I$ and $I_0$ is

Answer 1

Step 1: Formula used

Suppose slit widths are equal, so they produce waves of equal intensity, say $I\text{'}$.

Resultant intensity at any point will be,$I_R=4I\text{'}{\cos }^2\theta$

$I\text{'}\to$Intensity of waves produced by the slit

$I_R\to$Resultant intensity at any point

$\theta \to$Phase difference between the waves at the observation point

Step 2: Conditions

For maximum intensity, $\cos \theta =1$

$\because I_{max}=4I\text{'}=I\to \left(1\right)$

If one of the slits is closed, the resultant intensity at the same point will be,

$I\text{'}=I_0\to \left(2\right)$

Step 3: Compare the equations $\left(1\right)$ and $\left(2\right)$

Comparing the equations, we get,

$I=4I_0$

Thus, the relation between $I$ and $I_0$ is $I=4I_0$.