Q148 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Physics
The ratio of maximum and minimum intensities in an interference pattern is $36: 1$. The ratio of the amplitude of the two interfering waves will be
- A$5: 7$
- B$7: 4$
- C$4: 7$
- Correct answerD$7: 5$
Solution
$\frac{I_{\max }}{I_{\min }}=\frac{36}{1}$
We know, $I_{\max }=\left(a+a_2\right)^2$
$$
\begin{array}{rlrl}
I_{\min } & =\left(a_1-a_2\right)^2 \\
& & \frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2} & =\frac{36}{1} \Rightarrow \frac{a_1+a_2}{a_1-a_2}=\frac{6}{1} \\
\Rightarrow & & a_1+a_2 & =6 a_1-6 a_2 \\
\Rightarrow & & 5 a_1 & =7 a_2
\end{array}
$$
$\therefore$ The ratio of amplitude, $\frac{a_1}{a_2}=\frac{7}{5}$
Hence, the correct answer is option: 'D'.