A long narrow horizontal slit is paced 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 1.0 m away from the slit. 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?

#### Solution

Given:-

The mirror reflects 64% of the energy or intensity of light.

Let intensity of source = I_{1}.

And intensity of light after reflection from the mirror = I_{2}.

Let a_{1} and a_{2} be corresponding amplitudes of intensity I_{1} and I_{2}.

According to the question,

\[I_2 = \frac{I_1 \times 64}{100}\]

\[ \Rightarrow \frac{I_2}{I_1} = \frac{64}{100} = \frac{16}{25}\]

\[\text{And }\frac{I_2}{I_1} = \frac{{a_2}^2}{{a_1}^2}\]

\[ \Rightarrow \frac{a_2}{a_1} = \frac{4}{5}\]

We know that \[\frac{I_\max}{I_\min} = \frac{\left( a_1 + a_2 \right)^2}{\left( a_1 - a_2 \right)^2}\]

\[ = \frac{\left( 5 + 4 \right)^2}{\left( 5 - 4 \right)^2}\]

\[ I_\max : I_\min = 81 : 1\]

Hence, the required ratio is 81 : 1.