Question

The width of one of the two slits in a Young's double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit width. The ratio of the maximum to the minimum intensity in interference pattern will be

• A. $\frac{1}{a}$
• B. $\frac{9}{1}$
• C. $\frac{2}{1}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\frac{9}{1}$

Answer 2

$A_{\max}$ and $A_{\min}$.

Also $\frac{I_{\max}}{I_{\min} = \left( \frac{A_{\max}}{A_{\min}}^{2} = \left( \frac{3A}{A} \right)^{2} = \frac{9}{1} \right)}$