Question

Consider the situation of the previous 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 the screen?

Answer 1

Given:
The mirror reflects 64% of the energy or intensity of light.
Let intensity of source = I₁.
And intensity of light after reflection from the mirror = I₂.
Let a₁ and a₂ be corresponding amplitudes of intensity I₁ and I₂.
According to the question,
$$\begin{gathered} I_2=\frac{I_1\times 64}{100} \\ \Rightarrow \frac{I_2}{I_1}=\frac{64}{100}=\frac{16}{25} \\ \mathrm{And}\frac{I_2}{I_1}=\frac{{a_2}^2}{{a_1}^2} \\ \Rightarrow \frac{a_2}{a_1}=\frac{4}{5} \end{gathered}$$
$$\begin{gathered} \mathrm{We}\mathrm{know}\text{that}\frac{I_{\max }}{I_{\min }}=\frac{{\left(a_1+a_2\right)}^2}{{\left(a_1-a_2\right)}^2} \\ =\frac{{\left(5+4\right)}^2}{{\left(5-4\right)}^2} \\ {I}_{\max }:I_{\min }=81:1 \end{gathered}$$

Hence, the required ratio is 81 : 1.