The description of interference is established, and the explanation of Michelsons' experiment is given, and tested.

1. Light interference
Learning object 8
- Zeanna Janmohamed
figure 1.1 young’s double slit experiment
http://www.microscopyu.com/articles/polarized/images/interferencefigure5.jpg

2. 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

3. 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/sp
ectrum.gif

4. 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/l
asers/images/nero1.jpg

5. 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

6. 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

7. 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 medium
http://fc03.deviantart.net/fs70/i/2010/347/f/4
/colourfull_light_waves_by_originaliamme-
d34see0.jpg

8. 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 interometer
http://hyperphysics.phy-
astr.gsu.edu/hbase/phyopt/phopic/miche
l.jpg

9. 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

10. 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/jcwy
ant/JoseDiaz/AnimatedGifs/Mich
elson.gif

11. 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 light
http://teklaplus.pl/wp-
content/uploads/2013/03/30578
3_89211.jpg

12. 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.