The intensity ratio of the two interfering beams of light is $β$. The value of $\left[\frac{I_{max}-I_{min}}{I_{max}+I_{min}}\right]$ is

$\frac{2\sqrt{β}}{(1+β)}$

The correct answer is Option (2) → $\frac{2\sqrt{β}}{(1+β)}$

For two interfering beams with intensity ratio $\beta = \frac{I_{1}}{I_{2}}$:

$I_{max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^2 = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}$

$I_{min} = (\sqrt{I_{1}} - \sqrt{I_{2}})^2 = I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}}$

Thus,

$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{4\sqrt{I_{1}I_{2}}}{2(I_{1} + I_{2})}$

$= \frac{2\sqrt{I_{1}I_{2}}}{I_{1} + I_{2}}$

Using $\beta = \frac{I_{1}}{I_{2}} \; \Rightarrow \; I_{1} = \beta I_{2}$:

$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{2\sqrt{\beta I_{2} \cdot I_{2}}}{\beta I_{2} + I_{2}}$

$= \frac{2\sqrt{\beta}}{\beta + 1}$

Final Answer: $\;\; \frac{2\sqrt{\beta}}{1 + \beta}$