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
$\Rightarrow A _{2}=3 A _{1}$
$\frac{ I _{\max }}{ I _{\min }}=\left(\frac{\sqrt{ I _{1}}+\sqrt{ I _{2}}}{\left|\sqrt{ I _{1}}-\sqrt{ I _{2}}\right|}\right)^{2}$
$\because$ Intensity $I \propto A ^{2}$
$\Rightarrow \frac{ I _{\max }}{ I _{\min }}=\left(\frac{ A _{1}+ A _{2}}{\left| A _{1}- A _{2}\right|}\right)^{2}$
$=\left(\frac{ A _{1}+3 A _{1}}{\left| A _{1}-3 A _{1}\right|}\right)^{2}$
$=\left(\frac{4 A _{1}}{2 A _{1}}\right)^{2}=4: 1$