Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# 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 - Physics

Sum

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.

Concept: Interference
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 17 Light Waves
Q 23 | Page 382
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