Question

The maximum intensity of fringes in Youngs experiment is $I$. If one of the slit is closed then the intensity at that place becomes $I_{0}$ . Which of the following relation is true?

• A. $I=I_{0}$
• B. $I=2I_{0}$
• C. $I=4I_{0}$
• D. There is no relation between $I$ and $I_{0}$

Answer 1

Answer: (C)

$I=4I_{0}$

Answer 2

Suppose slit width's are equal so they produces waves of equal intensity say I'. Resultant intensity at any point $I_{R}=4 I' \cos ^{2} \phi$ where $\phi$ is the phase difference between the waves at the point of observation. For maximum intensity, $\phi=0^{\circ}$ $\Rightarrow I_{\max }=4 I'=I$ ...(i) If one of slit is closed, resultant intensity at the same point will be $I'$ only ie, $I'=I_{0}$...(ii) Comparing Eqs. (i) and (ii) we get $I=4 I_{0}$