Interference Physics 202 Professor Lee Carkner Lecture 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 ( = ) 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 Youngs 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 = , 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+) Location of Fringes For example: m=1 max = 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+) 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
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