Two coherent sources of intensity ratio α interfere. In...

Let the intensities of the two coherent sources be $I_1$ and $I_2$.

Given the intensity ratio $\dfrac{I_1}{I_2} = \alpha$.

The maximum intensity in the interference pattern is given by:

I_\max = (\sqrt{I_1} + \sqrt{I_2})^2 = I_1 + I_2 + 2\sqrt{I_1 I_2}

The minimum intensity is given by:

I_\min = (\sqrt{I_1} - \sqrt{I_2})^2 = I_1 + I_2 - 2\sqrt{I_1 I_2}

Calculating the difference and sum of the maximum and minimum intensities:

I_\max - I_\min = 4\sqrt{I_1 I_2}

I_\max + I_\min = 2(I_1 + I_2)

Taking the ratio of these two quantities:

\dfrac{I_\max-I_\min}{I_\max+I_\min} = \dfrac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \dfrac{2\sqrt{I_1 I_2}}{I_1 + I_2}

Dividing the numerator and the denominator by $I_2$:

\dfrac{I_\max-I_\min}{I_\max+I_\min} = \dfrac{2\sqrt{\dfrac{I_1}{I_2}}}{\dfrac{I_1}{I_2} + 1}

Substituting $\dfrac{I_1}{I_2} = \alpha$:

\dfrac{I_\max-I_\min}{I_\max+I_\min} = \dfrac{2\sqrt{\alpha}}{\alpha + 1}

Answer: $\dfrac{2\sqrt{\alpha}}{1+\alpha}$