Two beams of light of intensity $I_{1}$ and $I_{2}$ interfere to give an interference pattern. If the ratio of maximum intensity to that of minimum intensity is $\frac{25}{9}$ , then ratio of $\frac{I_{1}}{I_{2}}$ is

\frac{5}{3}

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4

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\frac{81}{625}

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Solution

The correct option is B 16
$\frac{I_{m a x}}{I_{m i n}} = \frac{25}{9} o r {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = \frac{25}{9}$
where a denotes amplitude.
$\frac{a_{1} + a_{2}}{a_{1} - a_{2}} = \frac{5}{3}$ or $5 a_{1} - 5 a_{2} = 3 a_{1} + 3 a_{2}$
or, $5 a_{1} - 5 a_{2} = 3 a_{1} + 3 a_{2}$ or $2 a_{1} = 8 a_{2}$
or, $\frac{a_{1}}{a_{2}}$ = 4
or ${(\frac{a_{1}}{a_{2}})}^{2}$ = 16 = $\frac{I_{1}}{I_{2}}$ .

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