Question

Two coherent sources of intensity ratio ‘$α$’ 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)

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