Question

Consider an YDSE that has different slits width, as a result, amplitudes of waves from two slits are $A$ and $2A$, respectively. If $I_0$ be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is $\varphi$ is :

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

Answer 1

The correct option is C $\frac{I_0}{9}[5+4\cos \varphi ]$

As amplitudes are $A$ and $2A$, so intensities would be in the ratio $1:4$, let us say $I$ and $4I$
$I_{max}=I_0=I+4I+2\sqrt{4i^2}=9I$ $\Rightarrow$ $I=I_{0}9$
Intensity at any point, $I^{{}^{\prime }}=I+4I+2\sqrt{4I^2}\cos \varphi$
$I^{\prime }=5I+4I\cos \varphi =\frac{I_0}{9}(5+4\cos \varphi )$