Question

In an interference experiment, the ratio of amplitudes of coherent waves is $\frac{a_1}{a_2}=\frac{1}{3}$. The ratio of maximum and minimum intensities of fringes will be

• A. $4$
• B. $2$
• C. $9$
• D. $18$

Answer 1

The correct option is A $4$

Given $\frac{a_1}{a_2}=\frac{1}{3}$
Ratio of intensities, $\frac{I_1}{I_2}={\left(\frac{a_1}{a_2}\right)}^2=\frac{1}{9}$
Now, $\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2={\left(\frac{1+3}{1-3}\right)}^2=4$.