Light interference

Learning object 8 - Zeanna Janmohamed

figure 1.1 young’s double slit experiment http://www.microscopyu.com/articles/polarized/images/interferencefigure5.jpg

interference

• Two waves will interfere constructively when they are in phase and have the same frequency and destructively when they are out of phase

• Electromagnetic waves do the same thing! – Waves do not have the same precision though – EMR waves do not have the same phase

difference to completely reinforce or cancel one another

Different kinds of interference• We need to consider the different kinds of

wavelengths • Need to also consider the relative phase of the

waves – Light bulbs have different wavelengths in different

phases and this is why they produce white light

Figure 1.1 electromagnetic spectrum http://www.yorku.ca/eye/spectrum.gif

lasers• Lasers produce monochromatic light that produces

one bright beam of exactly one wavelength (hence being monochromatic)

• The waves are also coherent and therefore are all in phase with each other

Figure 1.3 lasers http://www.laserfest.org/lasers/images/nero1.jpg

More complications with light interference

1. Planes of the electric and magnetic fields are important when adding EMR waves

2. Addition of two waves is straightforward though

- Wave intensities are I and the phase difference is ∆ϕ- When intensities are equal and the phase difference is 0

therefore the total intensity is 4 times the intensity of one alone

- When intensities are not equal the resultant intensity is 0 because the values cancel

Check and reflect

• What would be the resulting intensity if the initial waves are equal and the phase difference is 2pi?

• What would be the resulting intensity if the initial waves are opposite and the phase difference is 3pi?

Answers: a. cos(2pi) = 1, therefore it would be 4x b. cos(3pi) = 0 therefore it would be 0x

Phase difference

• Sometimes it is hard to think of phase differences, therefore we try and think of them in terms of wavelengths – If the wave is in a medium we must use the

wavelength of the EMR wave in that medium

Figure 1.4 light in a mediumhttp://fc03.deviantart.net/fs70/i/2010/347/f/4/colourfull_light_waves_by_originaliamme-d34see0.jpg

Michelson interferometer

• A beam of light is split with part going one direction and another another direction both of different lengths (this causes a phase difference)

• Then combine the light once more • This produces an interference pattern • A glass compensator plate makes the total distance

each beam travels in glass equal

Figure 1.5 michelson interometerhttp://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/phopic/michel.jpg

Michelson interferometer calculations

• If we gradually adjust the movable mirror from bright to dim (zero path difference to path difference of half wavelength) – Note that change in path difference d is twice the

distance the mirror moves • Formula is derived: mλ=2d

• Where m = any integer » Lambda = wavelength » D = distance

Interferometer patterns

• When the path difference is 0 = bright disk formed at the detector

• 0.5 wavelength = dark spot – Interference patterns continues in that manner

Figure 1.6 Michelson interferometer http://fp.optics.arizona.edu/jcwyant/JoseDiaz/AnimatedGifs/Michelson.gif

Check and reflect• A Michelson interferometer uses a laser beam of red

light. The mirror must be adjusted to give maximum bright output, what are the intervals of movement that the mirror needs to be adjusted by to produce light. What would occur if this point is missed?

• Hint: red light is between 620 and 750 nm, you may use either, but the explanation must be appropriate

Figure 1.7 red lighthttp://teklaplus.pl/wp-content/uploads/2013/03/305783_89211.jpg

Check and reflect answers

Following the formula mλ=2d, the path difference must be increased by one wavelength in order to achieve the next maximum. If the wavelength in question is 750nm (far end of red), then the distance it must be moved is in intervals of 0.5(750) or 375nm.

If you miss this value you will come across path difference that is not 0 and will have a dim spot achieved.