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Question

(A) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.Let the amplitudes be $a_1$ and $a_2$

Vedclass · Accepted Answer

$9:1$

Vedclass · Answer

(A) The intensity $I$ of a wave is directly proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.Let the amplitudes be $a_1$ and $a_2$. Given $\frac{a_1}{a_2} = \frac{2}{1}$.The maximum intensity $I_{\max}$ is proportional to $(a_1 + a_2)^2$ and the minimum intensity $I_{\min}$ is proportional to $(a_1 - a_2)^2$.Therefore,the ratio of maximum to minimum intensity is given by:$\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \left( \frac{\frac{a_1}{a_2} + 1}{\frac{a_1}{a_2} - 1} \right)^2$.Substituting the given ratio $\frac{a_1}{a_2} = 2$:$\frac{I_{\max}}{I_{\min}} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \left( \frac{3}{1} \right)^2 = \frac{9}{1}$.Thus,the ratio is $9:1$.

If the ratio of the amplitudes of two waves is $2:1$,then the ratio of their maximum and minimum intensity is:

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