Question

consider the situation of t eprevious problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on teh screen?

Answer 1

Given that the mirror reflects 64% of energy (intensity) of light.
so,$\frac{I_1}{I_2}=0.64=\frac{16}{25}\Rightarrow \frac{r_1}{r_2}=\frac{4}{5}So,\frac{I_{max}}{I_{min}}=\frac{(r_1+r_2{)}^2}{(r_1-r_2{)}^2}=\frac{(4+5{)}^2}{(4-5{)}^2}=81:1$