Question

Two slits in Young's experiment have widths in the ratio $1 : 25$. The ratio of intensity at the maxima and minima in the interference pattern, $ \frac{I_{\max}}{I_{\min}}$ is

• A. $\frac{49}{121}$
• B. $\frac{4}{9}$
• C. $\frac{9}{4}$
• D. $\frac{121}{49}$

Answer 1

Answer: (C)

As, intensity $I \propto$ width of slit W
Also, intensity $I \propto$ square of amplitude $A$
$\therefore \frac{I_{1}}{I_{2}}=\frac{W_{1}}{W_{2}}=\frac{A_{1}{ }^{2}}{A_{2}{ }^{2}}$
But $\frac{W_{1}}{W_{2}}=\frac{1}{25} $ (given)
$\therefore \frac{A_{1}{ }^{2}}{A_{2}{ }^{2}}=\frac{1}{25} \text { or } \frac{A_{1}}{A_{2}}=\sqrt{\frac{1}{25}}=\frac{1}{5}$
$\therefore \frac{I_{\max }}{I_{\min }}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}}=\frac{\left(\frac{A_{1}}{A_{2}}+1\right)^{2}}{\left(\frac{A_{1}}{A_{2}}-1\right)^{2}}$
$=\frac{\left(\frac{1}{5}+1\right)^{2}}{\left(\frac{1}{5}-1\right)^{2}}=\frac{\left(\frac{6}{5}\right)^{2}}{\left(-\frac{4}{5}\right)^{2}}=\frac{36}{16}=\frac{9}{4}$