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 $\varphi .$

• A. $I_0{\cos }^2(\varphi /2)$
• B. $\frac{I_0}{3}{\sin }^2(\varphi /2)$
• C. $\frac{I_0}{9}(5+4\cos \varphi )$
• D. $\frac{I_0}{9}(5+8\cos \varphi )$

Answer 1

Answer: (C)

Resultant intensity when two waves with phase difference $\varphi$ interfere is
$I=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}$ cos$\varphi$
As, intensity $I\propto A^2$ or $I=k{A}^2,$
where $k=a$ constant and A is amplitude.
So, $I=A_1^2+A_2^2+2A_1{A}_2\cos \varphi$
Here, $A_1=A$ and $A_2=2A$
$\Rightarrow I=A^2+{\left(2A\right)}^2+2\left(A\right)\left(2A\right)\cos \varphi$
$=A^2\left(5+4\cos \varphi \right){\dots }_{\left(i\right)}$
Intensity is maximum $\left(I_o\right),$ when $\cos \varphi =1$
$\Rightarrow I_o=A^2\left(5+4\times 1\right)=9A^2$
$\Rightarrow A^2=I_o/9\dots \left(ii\right)$
So, resultant intensity,
$I=A^2\left(5+4\cos \varphi \right)$
[From Eqs. (i) and (ii)]
$\Rightarrow I=\frac{I_o}{9}\left(5+4\cos \varphi \right)$