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 $\phi$ is given by $\frac{I_{0}}{P}\left(5 + 4 cos \phi\right)$ . 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 $1:4$
$I_{m a x}=I_{0}=I_{1}+I_{2}+2\sqrt{I_{1} \times I_{2}}=I+4I+2\times 2I$
$I_{0}=9I\Rightarrow I=\frac{I_{0}}{9}$
Intensity at any point:
$I^{'}=I_{1}+I_{2}+2\sqrt{I_{1} \times I_{2}}cos\phi$
$I^{'}=I+4I+2\sqrt{I \times 4 I}cos\phi$
$I^{'}=5I+4Icos\phi \, ;I^{'}=I\left[5 + 4 cos \phi\right]$
$I^{'}=\frac{I_{0}}{9}\left[5 + 4 cos \phi\right]=\frac{I_{0}}{P}\left[5 + 4 cos \phi\right]$
$P=9$