interference physics 202 professor lee carkner lecture 24
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Interference
Physics 202Professor Lee
CarknerLecture 24
Interference
Due to a phase difference between the incoming waves the amplitude of the resultant wave can be larger or smaller than the original
Light can experience such interference as well
The interference of light demonstrates the wave nature of light
Speed of Light Why does light bend when entering a new
medium?
n = c/v Consider a wavefront half in air and half in
glass
Note that:
Light also changes wavelength in a new medium (v = f
Frequency stays the same
Phase Change
If light travels through a medium of length L with index of refraction n, the number of wavelengths in that medium is:
N = L/new = Ln/
The difference in wavelengths between two different paths is:
N2 - N1 = (L/)(n2-n1)
Phase
We can represent phase in different ways
Phase differences are seen as brightness variations
Destructive interference ( = ½) produces a dark spot
Intermediate interferences produces intermediate brightness
Diffraction
When a planar wavefront passes through a slit the wavefront flares out
Diffraction can be produced by any sharp edge e.g. a circular aperture or a thin edge In order to see light it has to pass through a
circular aperture (e.g. your eye), so diffraction effects all images
Diffraction
Basic Interference
In Young’s experiment sunlight passes through a double slit with a screen on the other side
The two resultant wavefronts overlap to produce constructive and destructive interference
The interference patterns appear on the screen and show bright and dark maxima and minima, or fringes
Interference Patterns
The two rays travel different distances and so will be in or out of phase depending on if the difference is a multiple of 1 or an odd multiple of 0.5 wavelengths L = 1, L = ½,
What is the path length difference (L) for a given set-up?
Path Length Difference Consider a double slit
system where d is the distance between the slits and is the angle between the normal and the point on the screen we are interested in
The path length difference for rays emerging from the two slits is
L = d sin This is strictly true only when
the distance to the screen D is much larger than d
Maxima and Minima
d sin = m For minima (dark spots) the path
length must be equal to a odd multiple of half wavelengths:
d sin = (m+½)
Location of Fringes
For example: m=1max = sin-1 (/d)
min = sin-1 (1.5/d) Zeroth order maxima is straight in front of the
slits and order numbers increase to each side
Interference Patterns
You can also find the location of maxima in terms of the linear distance from the center of the interference pattern (y):
For small angles (or large D) tan = sin
where m is the order number The same equation holds for for minima if
you replace m with (m+½)
Wavelength of Light
If you measure d and (or y and D) you can solve the above equations for the wavelength of light
Young found 570 nm, fairly close to the true value of 555 nm
Next Time
Read: 35.6-35.8 Homework: Ch 35, P: 8, 21, 35, 36