Question

Two coherent sources of intensity ratio $ \beta $ interfere. Then the value of $ (I_{\max }-I_{\min })/(I_{\max }+I_{\min }) $ is:

• A. $\frac{1+\beta}{\sqrt{\beta}}$
• B. $\sqrt{\left(\frac{1+\beta}{\beta}\right)}$
• C. $\frac{1+\beta}{2\beta}$
• D. $\frac{2 \sqrt{\beta}}{1+\beta}$

Answer 1

The Correct Option is D

We know $I \propto a^{2}$ or $a \propto \sqrt{I} $ $\therefore \frac{a_{1}}{a_{2}}=\sqrt{\left(\frac{I_{1}}{I_{2}}\right)}$ So, $\frac{I_{\max }}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}$ $=\frac{\left(a_{1}+a_{2}\right)}{\left(a_{2}-a_{1}\right)^{2}}=\frac{(1+\sqrt{\beta})^{2}}{(1-\sqrt{\beta})^{2}}$ Applying componendo and dividendo $\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}=\frac{(1+\sqrt{\beta})^{2}+(1-\sqrt{\beta})^{2}}{(1+\sqrt{\beta})^{2}-(1-\sqrt{\beta})^{2}} $ or $=\frac{2+2 \beta}{4 \sqrt{\beta}} $ or $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{2 \sqrt{\beta}}{1+\beta}$