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Physics 1CLecture 22B
"Somehow light is particle and wave. The experimenter makes the choice. You get what you interrogate for. And you want to know if I'm a wave or a particle."
--Tom Stoppard
Quiz 2 InfoIt will be a Scantron test that covers Chapters 14 and 21.
A list of equations, constants, and conversions will be provided on the quiz.
You are to write the version of your test on the Scantron form.
You will use the Quiz Code Number that you were assigned at the first Quiz.
You are expected to abide by UC Policy on Integrity of Scholarship.
From Last TimeA light ray travels from medium 1 to medium 3 as shown in the figure below. What can we say about the relationship between the index of refraction for medium 1 (n1) and the index of refraction for medium 3 (n3)?
Use Snell’s Law twice:
Equation the two n2:30o
Refraction of LightLast time we calculated the angle of refraction going from air to glass.
What if instead of going from air to glass with an incident angle of 30.0o, it went from glass to air with the same incident angle:
€
n1 sinθ1 = n2 sinθ2
€
sinθ2 =n1n2sinθ1
€
θ2 = sin−1 n1n2sinθ1
⎛
⎝ ⎜
⎞
⎠ ⎟
€
θ2 = sin−1 1.521.0003
sin30°⎛
⎝ ⎜
⎞
⎠ ⎟
€
θ2 = sin−1 0.760( ) = 49.4°
Here the angle increased (slow to fast).
Refraction of LightWhat if instead having an incident angle of 30.0o
(from glass to air) it had an incident angle of 60.0o (still glass to air):
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n1 sinθ1 = n2 sinθ2
€
sinθ2 =n1n2sinθ1
€
θ2 = sin−1 n1n2sinθ1
⎛
⎝ ⎜
⎞
⎠ ⎟
Here the angle doesn’t exist. The angle was so great that it refracted the incident ray back to the first medium (reflected?).
€
θ2 = sin−1 1.521.0003
sin60°⎛
⎝ ⎜
⎞
⎠ ⎟
€
θ2 = sin−1 1.32( ) = ?????
Total Internal ReflectionThis is known as Total Internal Reflection (TIR).
It can only occur if you move from a slow medium to a fast medium such that the refracted ray is bent away from the normal compared to the incident ray.Here you can see light ray 5 undergoing total internal reflection.
This means that at angle that light ray 5 hits at, no light enters the second medium.
Total Internal ReflectionWe define the critical angle as a particular angle of incidence that will result in an angle of refraction of 90o.For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary (TIR).
This ray will obey the Law of Reflection at the surface boundary.
Total Internal ReflectionExampleA light ray originally in water is incident on air at an angle of 45o with respect to the normal. Will the light ray be refracted into air or be reflected back into the water?
AnswerThe diagram is given but it is up to you determine if it is greater than the critical angle.
airH20
45o
Total Internal ReflectionWhat is the critical angle between light moving from water to air?
Use Snell’s Law:
Since 45o is less than the critical angle it will not be totally internally reflected, the light ray will enter the air region.
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n1 sinθ1 = n2 sinθ2
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n1 sinθc = n2 sin90°
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sinθc =n2n1
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θc = sin−1 n2n1
⎛
⎝ ⎜
⎞
⎠ ⎟ = sin−1
1.00031.33
⎛
⎝ ⎜
⎞
⎠ ⎟
€
θc = sin−1 0.752( ) = 48.8°
Answer
MiragesMirages occur because of TIR.
Hot air is less dense than cool air.
The less dense hot air has a small index of refraction.
The more dense cool air has a higher index of refraction.
TIR happens for light in cool air (slow) off of hot air (fast).
MikeEva
PrismsExampleAn incident ray in air is headed straight towards an equilateral plastic prism (n=1.50). The ray is parallel to the bottom of the prism. Use Snell’s Law to find the angle (with respect to the normal) that the light ray exits the prism on the right.
AnswerThe diagram is given but it is up to you to draw the normal(s) and path of the ray.
PrismsAnswer
Start with:
Thus, the incident angle is 30o from air to plastic. Using Snell’s Law we find:
60o
60o
30o
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n1 sinθ1 = n2 sinθ2
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θ2 = sin−1 n1n2sinθ1
⎛
⎝ ⎜
⎞
⎠ ⎟
€
θ2 = sin−1 11.5sin30°
⎛
⎝ ⎜
⎞
⎠ ⎟ = sin−1 0.333( )
€
θ2 =19.47°
θ2
PrismsAnswerNow, we need to look at the second boundary.We can now examine the small top triangle created by the ray in the prism.The bottom left angle on this triangle will be:
This means that the bottom right angle of the triangle will be:
60o
49.47o70.53o
θ2
€
90°−19.47° = 70.53°
€
180°− 70.53°− 60° = 49.47°
PrismsAnswerIs the 49.47o, the angle we will use in Snell’s Law?No, it is not with respect to the normal.We draw the normal and find:
At the second boundary, will this ray be refracted or totally internally reflected?
Check by calculating the critical angle.
60o
49.47o
θ3
€
90°− 49.47° = 40.53°€
θ3 = 40.53°
€
sinθc =n1n2
€
θc = sin−1 11.5⎛
⎝ ⎜
⎞
⎠ ⎟ = sin−1 0.667( )
€
θc = 41.81°
refracted, barely
PrismsAnswer
Now we need to calculate the outgoing ray with Snell’s Law again:
θ3
€
θ3 = 40.53°
θ4
€
n2 sinθ3 = n1 sinθ4
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θ4 = sin−1 n2n1sinθ3
⎛
⎝ ⎜
⎞
⎠ ⎟
€
θ4 = sin−1 1.51sin40.53°
⎛
⎝ ⎜
⎞
⎠ ⎟ = sin−1 0.975( )
€
θ4 = 77.10° This is the outgoing angle with respect to the normal
PrismsAnswerOverall, this incident ray will be pushed downward compared to its original direction.Don’t forget your geometry when dealing with prisms and Snell’s Law.
θ1+θ2=180o
θ1+θ2=90o
θ1+θ2+θ3
=180o
Clicker Question 22B-1In the prism example we just performed, a light ray was deflected downward when it moved through a prism when the pointy side was up. If you inverted the prism (so the pointy side was down) how would the direction of the incident ray change after passing completely through the front and back sides of the prism?A) The light ray would still be deflected downward.
B) The light ray would now be deflected upward.
C) The light ray would not be deflected (it would move in the same direction that it had originally).
Clicker Question 22B-2If nwater=1.50 and nglass=1.33, then total internal reflection at an interface between this glass and water:A) occurs whenever the light goes from glass to water.
B) occurs whenever the light goes from water to glass.
C) may occur when the light goes from glass to water.
D) may occur when the light goes from water to glass.
E) can never occur at this interface.
For Next Time (FNT)
Start reading Chapter 23
Finish working on the homework for Chapter 21
