Two coherent sources have an intensity ratio | vedclass

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(B) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = \frac{100}{1}$.Let the amplitudes be $a_1$ and $a_2$. Since $I \propt

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$121 : 81$

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(B) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = \frac{100}{1}$.Let the amplitudes be $a_1$ and $a_2$. Since $I \propto a^2$,we have $\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{100}{1}} = \frac{10}{1}$.The ratio of maximum to minimum intensity is given by the formula:$\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.Substituting the values,we get:$\frac{I_{\max}}{I_{\min}} = \left( \frac{10 + 1}{10 - 1} \right)^2 = \left( \frac{11}{9} \right)^2 = \frac{121}{81}$.Thus,the ratio is $121 : 81$.

Two coherent sources have an intensity ratio of $100 : 1$ and are used to obtain the phenomenon of interference. What is the ratio of the maximum to the minimum intensity?

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