Question

The ratio of intensities of rays emitted from two different coherent Sourees is \(\alpha .\) For the interference pattern by them, $\left[\left(I_{\max }+I_{\min }\right) /\left(I_{\max }-I_{\min }\right)\right]$ will be equal to (A) \(\\{(1+\sqrt{\alpha}) / 2 \alpha\\}\) (B) \(\\{(1+\alpha) / 2 \alpha\\}\) (C) \(\\{(1+\sqrt{\alpha}) / 2\\}\) (D) \(\\{(1+\alpha) /(2 \sqrt{\alpha})\\}\)

Short answer

The short answer for the given question is: (D) \(\dfrac{(1+\alpha)}{(2 \sqrt{\alpha})}\)

Step-by-step solution

Find the relationship between I1 and I2 using given intensity ratio

Given that the ratio of intensities of rays emitted from two different coherent sources is: \(\alpha = \dfrac{I_1}{I_2}\) From the given ratio we can express \(I_1\) in terms of \(\alpha\), and \(I_2\): \(I_1 = \alpha I_2\) Next, we will find the intensities for the maximum and minimum interference patterns.

Find Imax and Imin

The maximum intensity \(I_{\max }\) occurs when the phase difference \(\theta = 0\). Hence, the formula for maximum interference pattern intensity is: \(I_{\max } = I_1 + I_2 + 2\sqrt{I_1 I_2}\) Similarly, the minimum intensity \(I_{\min }\) occurs when the phase difference \(\theta = \pi\). Hence, the formula for minimum interference pattern intensity is: \(I_{\min } = I_1 + I_2 - 2\sqrt{I_1 I_2}\)

Substitute I1 in terms of I2

Replace \(I_1\) with \(\alpha I_2\) in the equations for \(I_{\max }\) and \(I_{\min }\): \(I_{\max } = \alpha I_2 + I_2 + 2\sqrt{\alpha I_2^2}\) \(I_{\min } = \alpha I_2 + I_2 - 2\sqrt{\alpha I_2^2}\)

Calculate the expression for the given ratio

Now, we need to find the expression for \(\dfrac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}\) Substitute the expressions obtained for \(I_{\max }\) and \(I_{\min }\) from the previous step: \(\dfrac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }} = \dfrac{(\alpha I_2 + I_2 + 2\sqrt{\alpha I_2^2}) + (\alpha I_2 + I_2 - 2\sqrt{\alpha I_2^2})}{(\alpha I_2 + I_2 + 2\sqrt{\alpha I_2^2}) - (\alpha I_2 + I_2 - 2\sqrt{\alpha I_2^2})}\) Simplify the expression: \(\dfrac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }} = \dfrac{2(\alpha I_2 + I_2)}{4\sqrt{\alpha I_2^2}}\) For this expression, notice how \(I_2\) cancels out in both numerator and denominator: \(\dfrac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }} = \dfrac{2(\alpha + 1)}{4\sqrt{\alpha}}\) Finally, simplify to get the required expression: \(\dfrac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }} = \dfrac{(1+\alpha)}{2\sqrt{\alpha}}\) Now, compare the expression with the given options to find the correct answer. The correct answer is: (D) \(\dfrac{(1+\alpha)}{(2 \sqrt{\alpha})}\)