Question

The equation of two light waves are $y_1=3\cos \omega t$ and $y_2=4\cos \omega t$. The ratio of maximum to minimum intensity produced by the superposition of these waves will be,

• A. $36$
• B. $49$
• C. $64$
• D. $81$

Answer 1

The correct option is B $49$

Given:
$y_1=3\cos \omega t$
$y_2=4\cos \omega t$
Amplitude of first wave, $A_1=3$
Amplitude of second wave, $A_2=4$

The relation between amplitude and intensity is,
$I\propto A^2$

$\Rightarrow \frac{I_{max}}{I_{min}}={\left(\frac{A_{max}}{A_{min}}\right)}^2={\left(\frac{A_1+A_2}{A_1-A_2}\right)}^2$

$\therefore \frac{I_{max}}{I_{min}}=\frac{(3+4{)}^2}{(3-4{)}^2}=\frac{49}{1}$

Hence, $\left(B\right)$ is the correct answer.