Question

A long horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1 m away from the slit. If the mirror reflects only $64\%$ of the light falling on it, the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen is:

• A. 8 : 1
• B. 3 : 1
• C. 81 : 1
• D. 9 : 1

Answer 1

The correct option is C 81 : 1

Intensity of direct ray $=I_0=k{A}_0^2$
Intensity of reflected ray $=\frac{64}{100}{I}_0=k{\left(\frac{8A_0}{10}\right)}^2$
$\frac{I_{max}}{I_{min}}=\frac{(A_0+0.8A_0{)}^2}{(A_0-0.8A_0{)}^2}={\left(\frac{1.8}{0.2}\right)}^2=\frac{81}{1}$.