Question

In young’s double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is

• A. 2 : 1
• B. 4 : 1
• C. 9 : 1
• D. 8 : 1

Answer 1

Answer: (C)

If $W_1$ and $W_2$ are widths of two slits, then
$\frac{I_{1}}{I_{2}}=\frac{W_{1}}{W_{2}}=4$
Also, $\frac{I_{1}}{I_{2}}=\frac{A^{2}_{1}}{A^{2}_{2}}\,\therefore \frac{A^{2}_{1}}{A^{2}_{2}}=4$ or $\frac{A_{1}}{A_{2}}=2$
$\frac{I_{max}}{I_{min}}=\left(\frac{A_{1}+A_{2}}{A_{1}-A_{2}}\right)^{2}=\frac{\left(\frac{A_{1}}{A_{2}}+1\right)^{2}}{\left(\frac{A_{1}}{A_{2}}-1\right)^{2}}=\frac{\left(2+1\right)^{2}}{\left(2-1\right)^{2}}=\frac{9}{1}$