Question

The intensity of interference waves in an interference pattern is same as $I_{\circ }$. The resultant intensity at the point of observation will be :

• A. $I=2I_{\circ }[1+\cos \theta ]$
• B. $I=I_{\circ }[1+\cos \theta ]$
• C. $I=\frac{[1+\cos \theta ]}{I_{\circ }}$
• D. $I=\frac{[1+\cos \theta ]}{2I_{\circ }}$

Answer 1

The correct option is A $I=2I_{\circ }[1+\cos \theta ]$

If $a_1$ and $a_2$ are the amplitudes, resultant amplitude a is given by

$a^2=a_1^2+a_2^2+2a_1{a}_2\cos \delta$

where $\delta$ is the phase difference
Here $\delta =\theta$ and $a_1=a_2=a_0$

we get $a^2=2a_0^2+2a_0^2\cos \theta$

$a^2=2a_0^2(1+\cos \theta )$

$a^2$ is directly proportional to $I$ Intensity of resultant wave.
Thus $a_0^2$ is $I_0$ Intensity of incident light.

$⟹$ $I=2I_0[1+cos\theta ]$