Question

In $YDSE$ , the slits have different widths. As a result, the amplitude of waves from two slits is $A$ and $2A$ respectively. If $I_0$ be the maximum intensity of the interference pattern then the intensity of the pattern at a point where the phase difference between waves is $\varphi$ is given by $\frac{I_0}{P}\left(5+4cos\varphi \right)$ . Where P is in integer. Find the value of P?

• A. $9$
• B. $3$
• C. $6$
• D. $1$

Answer 1

Answer: (A)

As the amplitudes are $A$ and $2A$ then the ratio of intensities is $1:4$
$I_{max}=I_0=I_1+I_2+2\sqrt{I_1\times I_2}=I+4I+2\times 2I$
$I_0=9I\Rightarrow I=\frac{I_0}{9}$
Intensity at any point:
$I^{{}^{\prime }}=I_1+I_2+2\sqrt{I_1\times I_2}cos\varphi$
$I^{{}^{\prime }}=I+4I+2\sqrt{I\times 4I}cos\varphi$
$I^{{}^{\prime }}=5I+4Icos\varphi ;I^{{}^{\prime }}=I\left[5+4cos\varphi \right]$
$I^{{}^{\prime }}=\frac{I_0}{9}\left[5+4cos\varphi \right]=\frac{I_0}{P}\left[5+4cos\varphi \right]$
$P=9$