Unformatted text preview: , ±2, ...
For minima, we want destructive
r2 - r1 = (m+½)λ, with m = 0, ±1, ±2, ...
For water, it is pretty easy to keep
the two sources in phase - you drop
two small stones at the same time.
For radio, with MHz - GHz frequencies,
this can be done electronically. But for
visible light, with f ≈ 1015 Hz, this is
How do we get around this?
Sunday, February 24, 2013 In-phase light sources - Old Technique
How do we get
We take a
divide it into
What do we see
the two inphase light
sources? Diffraction - more about
this in two lectures.
Sunday, February 24, 2013 Two in-phase light sources. In-phase light sources Standard geometrical approximations for analysis:
0) variations in path length over width of band or slit << wavelength
1) the two slits are narrow compared to their separation
2) The distance to the screen is large compared to the slit separation.
Sunday, February 24, 2013 Geometry and Algebra Since the path difference is dsinθ, and we get maxima when
the difference is an integral number of wavelengths, when
mλ = dsinθ or sinθ = mλ/d.
The vertical positions of the bands on the screen become
y = Rtanθ ≈ Rsinθ = mRλ/d.
Sunday, February 24, 2013 In-phase light sources To the left: a picture of
alternating light & dark
maxima at y ≈ mRλ/d.
minima at y ≈ (m+½)Rλ/d. Sunday, Fe...
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This document was uploaded on 02/18/2014 for the course PHYS 228 at Rutgers.
- Spring '12