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Anti-reflection Coating

A linearly polarized plane wave of wavelength lembda, is travelling in air and ia incident normally on a plane sheet of glass. Show that the ratio of the reflected to the incident electric field amplitude is


where n(air) and n(glass) are the refractive indices of air and glass respectively.

The glass sheet is now coated on the incident side with a material with a refractive index n(coat) = squareroot((n(air)*n(glass))) and a thickness d = lembda/4. Prove that there is now, in principle, no reflected power.

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Solution Preview

Please refer to the attachment.

We consider the general case of inclined incidence from the first principles and there from derive the expression for the special case of normal incidence.

Incident rays Reflected rays
1 2
E θ1 θ1

ε1 C E
A θ2 B
ε2 D

Transmitted rays

The fig shows a plane electromagnetic wave incident at the boundary between two perfectly dielectric (nonmagnetic) mediums (permittivities ε1 and ε2). Part of the wave is reflected back into the first medium and part is transmitted into the second. The fig. shows two rays of the wave

We know that speed of an EM wave in any medium is determined only by the permittivity and
permeability of the medium and is given by : v = 1/√με ......(1)

a) In the above fig., in the time interval ray 2 travels the distance CB, the transmitted part of ray 1 travels distance AD and reflected part distance AE. Hence,

CB/AD = v1/v2 where v1 and v2 are the velocities in media 1 and 2 respectively.

CB = AB cos(90O - θ1) = AB sinθ1

AD = AB sinθ2

Hence, CB/AD = v1/v2 = sinθ1/sinθ2 ........(2)
___ ____
From (1) we have : v1 = 1/√μ1ε1 and v2 = 1/√μ2ε2 .......(3)

Permeability of nonmagnetic media is same as ...

Solution Summary

Step by step solution provided. The expert examines anti-reflection coating. The refractive indices of air and glass respectively are determined.