Interference FAQ:
1. What is a phase shift, or phase difference?
Consider two identical waves. If the waves don’t exactly line up (for instance, crest aligned with
crest in a sine wave), then there is a phase shift, or phase difference, between them.
Phase shift can be measured in radians, degrees, or fractions of a wavelength. A 360˚ phase shift
(same as 2π radians) is a shift of one wavelength. Since this is the repeat distance of the
wave, the waves are still aligned. A 180˚ phase shift (same as π radians) is a shift of one-half
wavelength. Now the crest of one wave lines up with the trough of the other.
2. What does phase difference have to do with interference?
If the crests of the waves line up (phase shift of 0, λ, 2λ, etc) the resulting wave is bigger, giving
constructive interference. If a crest lines up with a trough (phase shift of λ/2, 3λ/2, etc), there
is cancellation (destructive interference)
3. Why does a difference in path length cause a phase shift between the waves
traveling those paths?
If two identical waves travel different distances to reach the same point, one will have
progressed through more (repeat distances) wavelengths than the other. If the waves started
out in phase, they will no longer be in phase at their destination. They will be out of phase by
the (fractional) number of wavelengths that fit in the path length difference.
4. I don’t get thin-film interference.
That’s not a question.
5. When is there reflection from an interface between two materials?
There is a reflection whenever the indices of refraction differ. It doesn’t matter which index is
bigger.
6. When is there a phase shift due to reflection at an interface?
There is a ½ wavelength phase shift when light is reflected at an interface with a medium that
has a higher index of refraction. There is no phase shift otherwise.
7. In thin-film interference, your equations on the slides are different than the equations
in the book.
In thin-film interference, as in any interference problem, there are at least two paths that light
rays follow. In a single-layer film, those paths are a reflection from the top surface, and a
reflection from the bottom surface. For perpendicular incidence, the path-length difference
between these is 2t, where t is the film thickness.
If this were the only contribution to the phase difference, then constructive interference would
occur when 2t=mλ, were m is an integer. But there is also a ½ wavelength phase shift when
light is reflected from a medium with a higher index of refraction. This could happen at one,
both, or neither of the interfaces in a single layer film. If it happens at neither or both, then it
doesn’t contribute to the phase difference between the paths, and constructive interference is
still given by 2t=mλ. If it happens at only one of the interfaces, constructive interference is
given by 2t=(m+ ½)λ. This is the difference in the equations.
8. In thin film interference, part of the path is in a material with n=n1 and part of the path
is in a different material, with n=n2. Since the wavelength is different in each
medium, how do I know whether to use λ/n1 or λ/n2?
It’s true that the light traverses both materials, but usually the extra distance traveled by one of
the waves (this is the path length difference) is only in one of the materials. You need to use
the index for this material, since that is what tells you where the crest will be after traversing
the extra path length.
9. What is the difference between interference and diffraction?
Not really very much. Interference with lots of sources is called diffraction. For instance, a
diffraction grating has many, many slits that act as sources of light.
10. Why is single-slit diffraction called diffraction? – there is only one slit.
From Huygen’s principle, every point in space inside the slit acts as a source of spherical waves.
So you could call this infinite-slit interference, hence diffraction.
11. Why are the angles for destructive interference in diffraction from a single slit of
width a given by asin " min = m# ( m = ±1, ± 2, ± 3, etc ). Isn’t this the equation for
constructive interference from two slits of separation a?
From Huygen’s principle, every point in space inside the slit acts as a source of spherical waves.
So you could call this infinite-slit
interference, hence diffraction. If there were only two
!
!
sources, at either end of the slit, then it would be the same as two-slit interference with slit
separation a, and the angles of constructive interference would be given by the above
relation, with the addition of m=0. But there are many other sources distributed between the
two slits. All these combine to give the condition above for destructive interference for a
single slit of width a.