Question

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

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

Answer 1

Answer: (C)

Amplitude $\propto$ Width of slit
$\Rightarrow A_2=3A_1$
$\frac{I_{\max }}{I_{\min }}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{|\sqrt{I_1}-\sqrt{I_2}|}\right)}^2$
$\because$ Intensity $I\propto A^2$
$\Rightarrow \frac{I_{\max }}{I_{\min }}={\left(\frac{A_1+A_2}{|A_1-A_2|}\right)}^2$
$={\left(\frac{A_1+3A_1}{|A_1-3A_1|}\right)}^2$
$={\left(\frac{4A_1}{2A_1}\right)}^2=4:1$