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Question
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?
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Solution
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,
\[I_2 = \frac{I_1 \times 64}{100}\]
\[ \Rightarrow \frac{I_2}{I_1} = \frac{64}{100} = \frac{16}{25}\]
\[\text{And }\frac{I_2}{I_1} = \frac{{a_2}^2}{{a_1}^2}\]
\[ \Rightarrow \frac{a_2}{a_1} = \frac{4}{5}\]
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.
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