Question

Consider an YDSE that has different slit widths, as a result, amplitude 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 for phase difference $\varphi$ 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 amplitude of the waves are $A$ and $2A$ and,

We know that $I\propto A^2$

So intensities of the waves would be in the ratio $1:4$, let us say $I$ and $4I$.

The intensity of the resultant wave is given by,

$I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

$I_R\to I_{max},$ when $\cos \varphi =1$,

Given that,

$I_{max}=I_0=I+4I+2\sqrt{4I^2}=9I$

$\Rightarrow I=\frac{I_0}{9}$

Intensity at a point where phase difference is $\varphi$ any point,

$I^{{}^{\prime }}=I+4I+2\sqrt{4I^2}\cos \varphi$

$I{}^{\prime }=5I+4I\cos \varphi =\frac{I_0}{9}(5+4\cos \varphi )$

Hence, option $\left(\text{C}\right)$ is correct.