Question

The two slits in Young's interference experiment have widths in the ratio n:1. The ratio of the intensities of the maxima and minima in the interference pattern is

• A. $\frac{\sqrt{n}+1}{\sqrt{n}}$
  
• B. $\frac{\sqrt{n}}{\sqrt{n}-1}$
  
• C. ${\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)}^2$
  
• D. ${\left(\frac{n+1}{n}\right)}^2$

Answer 1

The correct option is C ${\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)}^2$


The intensity of light emerging from a slit is proportional to its width. Since the amplitude is proportional to the square-root of the intensity, we have
$\frac{A_1}{A_2}-\sqrt{\frac{I_1}{I_2}}=\sqrt{\frac{n}{1}}=\sqrt{n}$
The ratio of intensities is given by,
$\frac{I_1}{I_2}=(\frac{A_1+A_2}{A_1-A_2}{)}^2$ = ${\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)}^2$