Question

Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity.

Answer 1

Let $y_1$ and $y_2$ be the displacement of two waves having same amplitude $a$ and phase difference $\varphi$ between them. Then,

$y_1=a\sin \omega t$

$y_2=a\sin (\omega t+\varphi )$

The resultant displacement, $y=y_1+y_2$

$y=a\sin \omega t+a\sin (\omega t+\varphi )=a\sin \omega t+a\sin \omega t\cos \varphi +a\cos \omega t\sin \varphi =a\sin \omega t(1+cos\varphi )+\cos \omega t\left(a\sin \varphi \right)$

Define $R\cos \theta =a(1+\cos \varphi )$ and $R\sin \theta =a\sin \varphi$

$y=R\sin \omega t\cos \theta +R\cos \omega t\sin \theta$

$y=R\sin (\omega t+\theta )$

Here R is resultant amplitude at some point P. By using the above equations ($R\cos \theta =a(1+\cos \varphi )$ and $R\sin \theta =a\sin \varphi$), we can easily find out $R^2$

$R^2=4a^2{\cos }^2\frac{\varphi }{2}$

Now, for maximum intensity

${\cos }^2\frac{\varphi }{2}=1$

$\varphi =2n\pi$ where n =0,1,2....

$\varphi =0,2\pi ,4\pi ,6\pi ....$

Therefore, $I_{max}=4a^2$

For minimum intensity

${\cos }^2\frac{\varphi }{2}=0$

$\varphi =(2n+1)\pi$ where n=0,1,2,3....

$\varphi =\pi ,3\pi ,5\pi ....$

Therefore, $I_{min}=0$