Question

Two coherent sources of intensity ratio ‘$\alpha$’ interfere. In interference pattern $\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=$

• A. $\frac{2\alpha }{1+\alpha }$
• B. $\frac{2\sqrt{\alpha }}{1+\alpha }$
• C. $\frac{2\alpha }{1+\sqrt{\alpha }}$
• D. $\frac{1+\alpha }{2\alpha }$

Answer 1

Answer: (B)

$\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{{\left(a_1+a_2\right)}^2-{\left(a_1-a_2\right)}^2}{{\left(a_1+a_2\right)}^2+{\left(a_1-a_2\right)}^2}$
$[\because I_{\max }={\left(a_1+a_2\right)}^2,I_{\min }={\left(a_1-a_2\right)}^2$
where $a=$ amplitude $]$
$=\frac{4a_1{a}_2}{2\left(a_1^2+a_2^2\right)}=\frac{2a_1{a}_2}{a_1^2+a_2^2}$
Now, dividing the numerator and denominator by $a_1{a}_2,$ we get
$\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{2}{\left[\frac{a_1}{a_2}+\frac{a_2}{a_1}\right]}=\frac{2}{\left[\sqrt{\alpha }+\frac{1}{\sqrt{\alpha }}\right]}=\frac{2\sqrt{\alpha }}{(\alpha +1)}$