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Snell's Law

Snell's Law tells us how light bends when it moves out of one medium and into another.  In particular, it tells us how a ray of light in air bends as it enters a glass lens.

Figure 1 -- Light Entering Glass:

Light entering glass
Whenever light passes from one medium to another with a different refractive index, its path bends.  If the refractive index of the medium it's leaving is N0 and that of the medium it's entering is N1, and if the angle from perpendicular with which it hits the interface is θ0 and that which it enters the new medium is θ1, then Snell's Law says that

1)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/f2/eqe_temp_image_RhpbfO.png


On the remainder of this page we'll  present a simple derivation of this law.

Change in Wavelength with Velocity

Frequency can't vary as radiation passes through an interface -- the same number of wave crests must "come out" as "went in".  So, since velocity divided by frequency is wavelength, the wavelength must change at the interface.  If we use λ for wavelength, ν for frequency, N for refractive index, and V for the velocity of light in the medium, then within a single medium we have:

2)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/f2/eqe_temp_image_AeFF5s.png


And at the interface between two media we have:

3)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/f2/eqe_temp_image_e_LMHh.png

Bending of the Waves At the Interface

We will now assume that a beam of light travels as a sequence of parallel, flat wave crests, and that the direction of travel is always perpendicular to each wave crest.  The distance between the crests is the wavelength, and it changes as the beam enters a new medium, as a result of which each crest must bend at the interface.  Using this assumption, and looking at figure 2, we can determine how much the wave crests must bend, and consequently how much the beam's direction must change.

Figure 2 -- Snell's Law:Snell's law, by simple trigonometry

The segment labeled "r" in figure 2 is the hypotenuse of two right triangles in the diagram, labeled T0 and T1.  The red side of triangle T0 has length λ0, and the red side of triangle T1 has length λ1.  Since they're right triangles, we can see that

4)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/f2/eqe_temp_image_m3xMiO.png

Dividing the first of these into the second, we have

5)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/f2/eqe_temp_image_Zi56_f.png

Plugging equation (3) into equation (5) we obtain the result we wanted, which is

6)    file:///media/disk/home/slawrence/website/physics_insights/physics/formulas/f2/eqe_temp_image_WAZTcc.png





Page created on 1/4/2009