In a double-slit interference pattern, it is the amplitudes of the light waves f

Question

In a double-slit interference pattern, it is the amplitudes of the light waves from each slit that add, not the light intensities. Usually, in the analysis of interference patterns, the radiation intensity (or amplitude) is assumed to be exactly the same for both slits. Because of inaccurate optical alignment, however, the illumination is often not the same for the two slits. If the intensity of light going through one slit is 0.010 mW/mm2 while the intensity going through the second slit is 0.030 mW/mm2, what is the intensity at an interference maximum and at an interference minimum?

Explanation / Answer

For this type of problem we will use the basic expressions of much interference

Et= E1 +E2

the electric field intensity is the squared

I = Et * Et

I = (E1 +E2 ) . (E1+E2)

I = E1 E1 + E2 E2 + 2 E1. E2

I = I1 +I2 + 2 I12

El termino I12 es la parte responsable de la interferencia

Data

I1 = 0.010 mW/mm²

I2 = 0.03 mW/mm²

I = E2   E = sqrt I

I12 =2 sqrt ( I1 I2) Cos

I12 = sqrt(0.01 0.03) Cos

I12 = 0.0173 Cos

We have a maximum for angles = 0, 2

Cos 0 = 1

I =I1 +I2 +2 I12

I = 0.01 + 0.03 + 2 0.0173

I = 0.0746 mW /mm²

minimum = , 3

I = 0.01 + 0.03 - 2 0.0173

I = 0.0054 mW/mm²

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