Problem : If there were no atmosphere, what color would be sky appear?
If there were no atmosphere, there would be no atoms to scatter any of the light from the sun downwards towards us and the sky would appear black (as it appears on the moon!).Problem : Use a dimensional argument to show that the proportion of the electric field of a light ray scattered by an atomic oscillator is proportional to λ ^{-2} . Let E _{i} and E _{s} be the incident and scattered amplitudes, respectively, with E _{s} corresponding to a distance d from the scatterer. Assume E _{s}âàùE _{I} , E _{s}âàù1/r and also that the scattered amplitude is proportional to the volume of the scatterer.
The problem basically gives us that E _{s}âàù = , where V is the volume of the scatterer and K incorporates anything (including constants) that we might have left out. Clearly the quantity VK/r must be dimensionless, so K must have units of (length) ^{-2} . The only other factor that the scattering might reasonably depend on is the wavelength. So K = f (λ) . λ has units of length so Kâàùλ ^{-2} so E _{i}/E _{s}âàùλ ^{-2} .Problem : Derive an expression for the time taken by light to travel through a substance consisting of m layers of material each with thickness d _{i} and index of refraction n _{i} .
The time taken to travel through each layer is given by t _{i} = d _{i}/v _{i} so the total time is a sum:
t = d _{i}/v _{i} |
t = n _{i} d _{i} |
Problem : Propose a simple argument to show that for reflection from a planar surface, Fermat's principle demands that the incident and reflected rays share a common plane with the normal to the surface.
Take two points P and Q above the surface. The light moves from P , reflects off the surface and ends up at Q . A normal to the surface and a line joining P and Q define a plane called the plane of incidence. Choose some point R on the surface, but also in the plane of incidence (these define a line L ); we will show a ray must reflect at some such point. Choose another point R' also on the surface, but not in the plane of incidence. For any such point it must be true that we can draw a line from R' to a point R perpendicular to L . The distance PR' is then given by which is by definition greater than PR , unless R' lies on L (we have defined it not to, though). Similarly the distance R'Q > RQ since R'Q = . Thus for any point on the surface but not on the plane of incidence we can find a shorter path of the reflecting ray through a point in the plane of incidence. Thus, by Fermat's principle, it should take this shortest path (since we are dealing with a single medium, shortest distance corresponds to shortest time). Since R , P and Q are all in the plane of incidence, PR and RQ must lie in the plane of incidence, which also contains the normal to the surface (we defined it this way).Problem : Derive the law of reflection θ _{i} = θ _{r} using Fermat's principle.
Using the result of the last problem we will assume P the point from which the light comes, Q the point to which the light goes, and R , the point from which the light reflects are all coplanar. Once again, since we are considering a single medium only the path of least time will be the path of least distance. Let the x- axis correspond to the surface. Without loss of generality let both P and Q be a distance y _{0} above the surface, with P on the y -axis and Q a distance x _{0} away (at (x _{0}, y _{0}) ). The point R can be anywhere along the x -axis. Call the coordinate of R (x _{1}, 0) . The distance PR is given by . The distance from RQ is given by . Thus the total distance PRQ is:
D = + |
fracdLdx _{1} = x _{1}(y _{0} ^{2} + x _{1} ^{2})^{-1/2} + (x _{1} - x _{0})(y _{0} ^{2} + (x _{0} - x _{1})^{2})^{-1/2} = 0 |
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