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Unformatted text preview: 35 .4 . 1 Apparent depth A coin under water a ears to be:
2 farther from the surface than it really is.
3 as dee as it reall is. Extra: Find the ratio of the apparent depth to the real
depth for the case when the observer is directly above the
coin, i.e. when the incident angle is approximately 0°? Explanation: From the sketch below one sees that the
apparent depth CB is less than the real depth CA. Extra: In the small angle approximation eleC/h’,
8220th. Snell’s law gives 6,:n02, or hfh’:n. 35 .5 .1 Color Dispersion Violet Red In a glass medium, the index of refraction of a violet lay
is higher than that for a red ray. Consider red and violet
rays through a glass prism. Let the unprimed angles to be
for the red, and the primed ones for the violet. Choose 6’ 6’ 0’
0’ 9’ Extra: Compare 04 with 9’4. Which ray gives a more
overall bending toward the baseline? Explanation: nviolet>nrem so 9’2<02. In other words, upon entrance into the prism, the violet ray bends more
toward the base line. Inspection on the geometric relations: 02+03+B=180°, 9’2 +B’3 +3=1800 leads to
0’3>03. Extra: Since 0’3>83, then 0’4>04. So as the violet ray exits the prism, once again it bends more toward the baseline. The violet ray gives more bending both at the
entrance point and at the exit point. 35 .4.2 Light passing through a slab Consider case A. A light ray passes through a slab with
index of refraction n2=1.5, which is submerged in a liquid with index of refraction n1: n3=1.2. Compare 93 with
01:300.
Extra: Consider case B, where n2 is changed to n’2:2.0. Assuming the incident angle 9’1 is the same as 91 of case
A, compare the new angle 0’3 with 03 of case A. Which
one is larger, 0’3 or 93? Explanation: Based on the Snell’s law and the set up,
nlsin 01=n2sin 02=n3sin 03. Since n3=n1, so 03:61.
Extra: Notice that the explanation just given is
independent of n2. So long as 9’1: 01, we have 0’3: 0’1: 81: 93. 35 .7 . 1 Total reflection Consider the incidence of a light my from a denser
medium to a lighter medium. At 02: 9:, where 0c is the
critical angle, 01:900. Determine 02 range, where there is
total reﬂection. Extra: Show that for the media of air—water, where the
index refraction of water is 1.33, 06: 48.80. Explanation: The Snell’s law says that nlsinelznzsinez.
At the critical angle, sinec=(nI/n2)sin90°=(n1fn2). If
02>80, sin81:(n2:'n1)sinezzsinezr’sinec > 1. So 81 cannot
be realized, i.e. there is no refracted ray in medium 1.
Thus in the region 8c <02<90°the ray in medium 2 is totally reﬂected.
Extra: For the present case, sinec=1f1.33. This gives 8c: 48.8”. 35 .8 . 1 Fermat’s Principle Compare the time for light to travel from A to B along the
path ACB, where the angle of incident equals to the angle of relfection, i.e. 91:9,, to that along AC’B. Choose one: tAC’B > tACB tAC’B = tACB [AC’B < tACB Hint: Consider two lines connected to image A’. Inspect
the relationships between AC’ & A’C’, and AC & A’C. ExplanationThe distance from A’ to B, via C’ is farther
than that via C, i.e. A’C’B > A’CB. By inspection, one
sees that AC’=A’C’, and AC = A’C. So AC’B > ACB. In
other words, tAC.B > tACB. Ans:1. 35.8.2 Least time A boy is at the lodge L on the beach. He must rescue a
drowning girl in the water at P. Among the three paths
illustrated which path is closest to the least time? Choose one: 2 L—>b+P 3 L—>c+P — All of the above Explanation: The boy runs faster than he swims. It
minimizes the time required when the boy covers more
distance on the beach than in the water. The least time
path should satisfy Snell’s law. From the sketch, among
the three paths illustrated, the path which passes through c
should be closest to the least time path. ...
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 Fall '08
 Turner
 Physics, Snell's Law, Total internal reflection, Geometrical optics, Snell’s law

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