Question

In Young's double slit experiment the ratio of the slit widths is 1 :4. The ratio of maximum and minimum intensities in the interference pattern will be

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

Answer 1

The correct option is C 9 :1

$\begin{array}{c}\frac{w_1}{w_2}=\frac{1_1}{1_2}=\frac{A^2}{B^2}\\ \frac{4}{1}=\frac{A^2}{B^2}\\ \frac{A}{B}=\frac{2}{1}=A=2B\\ \frac{1max}{1min}=\frac{{\left(A+B\right)}^2}{{\left(A-B\right)}^2}={\left(\frac{2B+B}{2B-B}\right)}^2\\ ={\left(\frac{3B}{B}\right)}^2=\frac{9}{1}\end{array}$