Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Statement I: The maximum intensity in interference pattern is four times the intensity due to each slit.
Statement II: Intensity is directly proportional to square of amplitude.

Wave Optics

Solution:

Let $a$ be the amplitude of waves from each slit. Intensity is directly proportional to square of amplitude i.e., $I \propto a^{2}$. Due to each slit $I_{1} \propto a^{2}$
When destructive interference occurs, then resultant amplitude $=a-a=0$
$\therefore $ Minimum intensity, $I_{\min }=0$
When constructive interference occurs, the resultant amplitude $=a+a=2 a$
$\therefore $ Maximum intensity, $I_{\max } \propto(2 a)^{2}$
Hence, $I_{\max } \propto 4 a^{2}$
or $I_{\max }=4$ times the intensity due to each slit.