Snell's Law and Dispersion of Light
Theory:
The index of refraction of light is the ratio of the velocity of light in a vacuum to the velocity of
light in a medium. The index of refraction is also defined by Snell’s law of a refracting surface.
When light passes from one medium to another, the angle of incidence ϕ1 is related to the angle of
refraction ϕ2 by the equation
nl sinϕ1 = n2 sinϕ2
(1)
where nl and n2 are the indices of refraction for medium 1 and 2, respectfully. The index of
refraction of air is approximately one. Therefore,for light traveling from air into a medium of index
of refration n, the ratio of the Sin ϕ1 / Sin ϕ2 should be the index of refraction n.
Dispersion of light is the separation of light into its component colors by a refracting surface.
The separation of colors is due to the fact that the index of refraction of a medium changes with
wavelength of the light. The smaller wavelength light (violet) has a higher index of refraction and
therefore is bent more than longer wavelength light (red).
In part 1, the index of refraction of a cylindrical lens will be determined and Snell’s law will be
verified. In part 2, light travels through a prism. The light is first refracted at point D and then
travels across the prism to point E where it is refracted again (Fig. 1). The angles are labeled on the
diagram.
Fig. 1
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For the refraction at D and E, Snell’s law gives
sin i1 = n sin r1
(2)
and n sin r2 = sin i2
(3)
where n is the index of refraction of the prism and the index of refraction of air is set to 1. δl and
δ2 are the angle the light is deviated from its path at surfaces 1 and 2. δ is then the total angle of
deviation and is given by
δ = δ1 + δ2
(4) but
δ1 = il - rl
(5) and
so
δ = (il +i2) - (rl + r2)
(7)
δ2 = i2 - r2
(6)
In triangle DEF, rl + r2+ β = 180o (8) In trapezoid AEFD, since angles at E and D are right angles,
α + β = 180°
(9)
Therefore Eqn (8) becomes rl + r2 = α
(10)
and Eqn (7) becomes δ = il + i2 - α (11)
If we choose a special case, where the angle of deviation is at a minimum (δ0). It can be shown with
the use of calculus that il = i2 and rl = r2. Eqn. (11) becomes
δ0 = 2i − α or
i = (α + δ0) / 2
and Eqn. 10 becomes r = α / 2 (12)
(14)
Therefore Eqn. (2) becomes
n = sin(½(α+δ0))
/sin(α/2)
(15)
Apparatus:
1. Cylindrical lens
2. Ray table
3. Light Source
4. Optical Bench
5. Spectrometer
6. Prism
7. Mercury Light Source
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Part 1
1. Mount the ray table base and the light source on the optics bench as shown in Fig. 2. Put the ray
table on the base with the Cartesian grid (mm scale) facing up. Turn the ray table so the 0o line points
to the light source. Put the light source on its bracket so that the multiple slits are facing the ray
table. Adjust the slit mask on the front of the light source so the light source projects one ray of light
across the middle of the top surface of the ray table.
Fig. 3
Fig. 2
2. Put the cylindrical lens on the ray table so the flat surface of the lens faces the light source and
the edge of the lens in on the 90o line with the lens exactly centered on the 0o line (Fig. 3). Without
distrubing the alignment of the cylindrical lens, rotate the ray table and set the angle of incidence to
the values listed in data table. Enter the corresponding angle of refraction in the table and calculate
the Sin(ϕ1) and Sin(ϕ2). Plot Sin(ϕ1) versus Sin(ϕ2). The slope of the line is the index of refraction
of the lens.
Part 2
1. A spectrometer consists of a collimator, a prism, a telescope and a dial for measuring angles (see
Fig. 4). The function of the collimator is to produce a slit of parallel light rays. The telescope will
bring the parallel rays of light into focus.
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2. Without the prism in place, look through the telescope and see if the slit and the cross hairs are
in focus. Normally, they should be. However, to focus the telescope, the eyepiece is moved in or
out and to focus the collimator the slit near the source is moved in or out. The slit opening can be
adjusted by a threaded screw on the slit opening. Line up the cross hairs on the center of the slit and
set the dial on the base of the spectrometer to read zero degrees.
3. Turn the prism table so that the path of the beam of light through the prism is approximately
symmetrical. Move the telescope into the path of the refracted beam and adjust its position so that
the image of the slit is roughly in the center of the field of view. Slowly turn the prism table in such
a direction as to diminish the angle of deviation, until the deviation becomes a minimum. The test
for this position is that any further motion of the prism table in the same direction will reverse the
direction of motion of the slit image. In locating this position of minimum deviation, if the image
moves out of the field of the telescope, the telescope must be moved so as to follow it. When the
position of the prism for minimum deviation has been found, and the prism table set as near to it as
possible, adjust the position of the telescope until the image of the slit falls on the intersection of the
cross hairs. Record the setting of the telescope to the nearest tenth of a degree.
4. Repeat procedure 3 for a different color line and record the setting in the data section. Calculate
the index of refraction for each color by using Eqn. 15.
5. Plot the index of refraction versus wavelength. Adjust the scale so that the variation can be seen.
Data:
Part 1
Incidence Angle (φ1)
Refracted Angle (φ1)
Sin (φ1)
10˚
20˚
30˚
40˚
50˚
60˚
70˚
80˚
Index of Refraction from the graph __________
Known Index of Refraction __________
Percent Error __________
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Sin (φ2)
Part 2
α = _______
Wavelength of brightest
Mercury lines
Color
Minimum angle of deviation
Index of refraction
5790
5461
4358
4047
Questions:
1. Dispersion of light is common in nature. Give an example of dispersion and explain why it
occurs.
2. A student claims that dispersion is caused by the glass modifying light as it travels through it.
White light is modified to produce red light by a certain thickness of glass. Likewise, violet light is
produced by a greater thickness of glass. Dispersion in a prism is caused by the fact that violet light
travels further through the glass than red light. Design a simple experiment to disprove the student’s
claim.
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