Question

The two coherent sources of intensity ratio \(\beta\) produce interference. The fringe visibility will be (A) \(2 \beta\) (B) \((\beta / 2)\) (C) \(\\{\sqrt{\beta} /(1+\beta)\\}\) (D) \(\\{(2 \sqrt{\beta}) /(1+\beta)\\}\)

Short answer

The correct answer for the fringe visibility is option (D): \(\\{(2 \sqrt{\beta}) /(1+\beta)\\}\).

Step-by-step solution

Calculate the intensities of the two sources

Let the intensities of the two sources be I₁ and I₂, with an intensity ratio β: \[ \beta = \frac{I_2}{I_1} \] To calculate fringe visibility, we need to find the individual intensities.

Find the formula for fringe visibility in terms of intensities

Fringe visibility (V) is given by the formula: \[ V = \frac{I_{max} - I_{min}}{I_{max} + I_{min}} \] where Iₘₐₓ is the maximum intensity, and Iₘᵢₙ is the minimum intensity. We can express the maximum and minimum intensities as: \[ I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} \] \[ I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} \] Now let's plug these expressions into the formula for V.

Calculate fringe visibility (V) in terms of I₁ and I₂

Substitute the expressions for Iₘₐₓ and Iₘᵢₙ into the formula for V: \[ V = \frac{(I_1 + I_2 + 2\sqrt{I_1 I_2}) - (I_1 + I_2 - 2\sqrt{I_1 I_2})}{(I_1 + I_2 + 2\sqrt{I_1 I_2}) + (I_1 + I_2 - 2\sqrt{I_1 I_2})} \] Simplify the expression: \[ V = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} \]

Express fringe visibility (V) in terms of β

Substitute the ratio β back into the formula for V: \[ V = \frac{4\sqrt{I_1 I_1 \beta}}{2(I_1 + I_1\beta)} \] Cancel out I₁: \[ V = \frac{4\sqrt{\beta}}{2(1 + \beta)} \] Now we have fringe visibility in terms of the intensity ratio β: \[ V = \frac{2\sqrt{\beta}}{1 +\beta} \] Let's compare our result with the given options.

Match the fringe visibility with the correct option

Our calculated fringe visibility V is: \[ V = \frac{2\sqrt{\beta}}{1 +\beta} \] This expression matches with option (D): (D) \(\\{(2 \sqrt{\beta}) /(1+\beta)\\}\) Therefore, the correct answer is option (D).