Two light waves amplitudes in the ratio 3 : 1 produce interference. The ratio of the maximum to minimum intensity is

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Let the amplitudes of the two light waves be $A_{1}$ and $A_{2}$ .

Given $\frac{A _{1}}{A _{2}} = \frac{3}{1}$

The maximum intensity $I_{ma x}$ is proportional to $(A_{1} + A_{2})^{2}$ and the minimum intensity $I_{min}$ is proportional to $(A_{1} - A_{2})^{2}$ .

$\frac{I _{ma x}}{I _{min}} = \frac{( A _{1} + A _{2} ) ^{2}}{( A _{1} - A _{2} ) ^{2}}$

Substituting the ratio of amplitudes:

$\frac{I _{ma x}}{I _{min}} = \frac{( 3 + 1 ) ^{2}}{( 3 - 1 ) ^{2}} = \frac{16}{4} = \frac{4}{1}$

The ratio of maximum to minimum intensity is $4 : 1$ .

Answer: $4 : 1$

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