A beam of light of a single vacuum wavelength is divided by a beam splitter into two…

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Wave propagation and light in media · Interference and diffraction

Optics and Photonics, Ch Harmonic waves, phasors and group velocity

Problem

A beam of light of a single vacuum wavelength is divided by a beam splitter into two beams which are brought together again on a detector. At the detector the two fields are parallel, so they may be added as scalars, and they are written

E_1\cos\omega t \qquad\text{and}\qquad E_2\cos(\omega t + \delta),

with

E_1

and

E_2

constant amplitudes and

\delta

a fixed phase difference. The detector cannot follow the field itself, only its square averaged over a cycle, and the irradiance of a beam is proportional to that average:

I = K\langle E^2\rangle

with

K

a positive constant that need not be evaluated. Absorption in the beam splitter is ignored. ((a)) Starting from the cycle average of the square of the total field, show that the irradiance on the detector is

I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta

, where

I_1

and

I_2

are the irradiances the two beams give alone. You may use

\langle\cos^2\omega t\rangle = \tfrac12

. ((b)) With beam 2 blocked the detector reads

2.0\ \mathrm{W\,m^{-2}}

; with beam 1 blocked it reads

8.0\ \mathrm{W\,m^{-2}}

. Find the irradiance when both beams are present and

\delta = \pi/3

Related theory

Interference and diffraction

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