Question

A long horizontal slit is place $1mm$ above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen $1m$ 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 $81:1$

Before reflection, intensity of light is $I_o$(say) and after reflection it becomes ${0.64I}_o$

$I_{max}=I_o+{0.64I}_o+2\sqrt{I_o\times {0.64I}_o}=3.24I_o$

$I_{min}=I_o+{0.64I}_o-2\sqrt{I_o\times {0.64I}_o}=0.04I_o$

$\frac{I_{max}}{I_{min}}=\frac{324}{4}=81:1$