The figure below depicts two wave disturbances with common wavelength 1.5 meters moving out in phase from two points 10 meters apart.

• The waves strike a surface 30 meters away from the line of their origin, as indicated below.
• We wish to find the difference in the distances traveled by the two waves to the indicated point, which lies a distance 2 meters above the perpendicular by sector of the segment connecting the points from which the waves originate.
• We wish to also determine by how many wavelengths the wave originating from the lower point will lag behind the wave originating from the upper point.

The further wave will therefore lag the closer by a distance of .65 m, which we see is .43 of a wavelength.

• The figure depicts how when the closer wave is a peak, the further wave will be near a 'valley', and when the ways combined they will nearly cancel one another out.
• The graph that the bottom right of the figure depicts how the two ways are nearly 180 degrees out of phase.

In fact the waves are .43 of a wavelength, or .43 * 360 degrees, out of phase.

The figure below depicts a plane wave approaching a barrier with two slits.

• The slits are separated by distance d.
• If the direction of the wave is perpendicular to the segment connecting the slits, the incoming wave will drive new waves out from the slits in perfect phase.
• We wish to determine how these waves will interact when they arrive at a distant point located at a position at an angle `theta with respect to the central axis of the figure.

We see that the extra distance traveled by the wave from the lower slit will be the short side of the indicated triangle.

• By simple trigonometry this distance is d sin(`theta).
• We observe that if this extra distance is equal to 1 full wavelength, then the waves will arrive at the distant point in phase and will therefore reinforce one another.
• The same is true if the extra distance equal to 2, 3, 4, ..., n full wavelengths.
• We therefore say that the condition for positive interference of the waves is that d sin(`theta) = n `lambda, where `lambda is the wavelength.

Video Clip #1

If the extra distanc is equal to 1/2 wavelength, then the peaks of one wave will meet the valleys of the other and the ways will cancel each other out.

• The same is true for distances of 3/2, 5/2, 7/2, ..., n-1/2 wavelengths.
• We might therefore say that the condition for destructive interference is that d sin(`theta) = (n - 1/2) `lambda.

In the figure below we consider this situation when the distance from the slits to a screen is R.

• We look at a point on the screen located a distance y < < R from the central axis of the slits, as depicted below.
• Since y < < R, the angle `theta turns out to be very close to y / R (as measured in radians) so that to a good approximation we have sin(`theta) = y / R.
• The path difference corresponding to this position y is therefore (d / R) * y.
• The quantity d / R is just the ratio of slit separation to screen distance; for small slit separation and large screen distance this is a very small number.

In class we observed separation approximately 1 cm between bright spots, with a distance R of approximately 600 cm and and estimated distance of .01 cm between a series of fairly equally spaced threads converging in a V.

• From this day we estimate that the wavelength of the light was 1.6 * 10^-7 cm.
• This is on the right order of magnitude, but the actual wavelength of the laser used is about 6.7 * 10^-7 cm.
• If our estimate of the separation of the threads (which was the hardest thing to measure) was too low by a factor of 4, then the wavelength estimate would be very nearly correct.
• Alternatively, the observed separation and known wavelength of the light tell us that the thread separation must have been about .04 cm at the point for the light was transmitted.