Question

In Young’s double slit experiment, the amplitudes of the two waves incident on the two slits are $A$ and $2A.$ If $I_0$ is the maximum intensity, then the intensity at a spot on the screen, where the phase difference between the two interfering waves is $\phi.$

• A. $I_{0} \cos ^{2}(\phi / 2)$
• B. $\frac{I_{0}}{3} \sin ^{2}(\phi / 2)$
• C. $\frac{I_{0}}{9}(5+4 \cos \phi)$
• D. $\frac{I_{0}}{9}(5+8 \cos \phi)$

Answer 1

Answer: (C)

Resultant intensity when two waves with phase difference $\phi$ interfere is
$I=I_{1}+I_{2}+2\sqrt{I_{1}}\sqrt{I_{2}}$ cos$\phi$
As, intensity $I \propto A^{2}$ or $I = kA^{2},$
where $k= a$ constant and A is amplitude.
So, $I=A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2} \cos\phi$
Here, $A_{1}=A$ and $A_{2}=2A$
$\Rightarrow I=A^{2}+\left(2A\right)^{2}+2\left(A\right)\left(2A\right)\cos \phi$
$=A^{2}\left(5+4 \cos\phi\right) \ldots_{\left(i\right)}$
Intensity is maximum $\left(I_{o}\right),$ when $\cos \phi= 1$
$\Rightarrow I_{o}=A^{2}\left(5+4\times1\right)=9A^{2}$
$\Rightarrow A^{2}=I_{o}/ 9\ldots \left(ii\right)$
So, resultant intensity,
$I=A^{2}\left(5+4 \cos\phi\right)$
[From Eqs. (i) and (ii)]
$\Rightarrow I=\frac{I_{o}}{9}\left(5+4\cos\phi\right) $