Question

The intensity ratio $\frac{I_1}{I_2}$ of the two interfering sources in Young's experiment is 4. The ratio $\frac{I_{max}}{I_{min}}$ is:

• A. 4 : 1
• B. 2 : 1
• C. 3 : 1
• D. 9 : 1

Answer 1

The correct option is D 9 : 1

$I_1=I,I_2=4I$ $\therefore$ $I_{max}=(\sqrt{1}+{\sqrt{4I}}^2),I_{min}=(\sqrt{4I-\sqrt{I}}{)}^2$
$\frac{I_{max}}{I_{min}}=\frac{1+4I+2\sqrt{1\times 4I}}{4I+1-2\sqrt{4I\times I}}=\frac{9}{1}$