 # Interference Physics 202 Professor Lee Carkner Lecture 24.

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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: N 2 - N 1 = (L/ )(n 2 -n 1 )

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=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+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