MCQ
The ratio of maximum to minimum intensity due to superposition of two waves is $\frac{{49}}{9}$ Then the ratio of the intensity of component waves is .
- ✓$6.25$
- B$0.64$
- C$\frac{{4}}{49}$
- D$\frac{{9}}{49}$
✓
Answer
Correct option: A.
$6.25$
a
$\frac{I_{\max }}{I_{\min }}=\frac{(a+b)^{2}}{(a-b)^{2}}$
$\frac{I_{\max }}{I_{\min }}=\frac{(a+b)^{2}}{(a-b)^{2}}$
$=\frac{49}{9}$
where $a, b$ are the amplitudes $\frac{a+b}{a-b}=\frac{7}{3}$
$3 a+3 b=7 a-7 b$
or, $4 a=10 b$
or, $\frac{a}{b}=\frac{10}{4}=\frac{5}{2}$
also we know, $\frac{I_{1}}{I_{2}}=\frac{a^{2}}{b^{2}}=\left(\frac{a}{b}\right)^{2}=\frac{25}{4}$
So the intensity ratio of component waves $6.25$
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