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Light - Angle of Deviation...

by futb0l
Tags: angle, deviation, light
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futb0l
#1
Jul5-04, 12:27 AM
P: n/a
I am having trouble understanding this question, can somebody here please explain it to me... should be quite simple as this is only high school stuff.

Calculate the angle of deviation at a glass-air interface for an angle of incidence of 65degrees and refractive index of glass of 1.55 .

edit: ooops, sorry i posted in the wrong forum... suposed to be in homework help.
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AKG
#2
Jul5-04, 12:48 AM
Sci Advisor
HW Helper
P: 2,586
I believe it's called Snell's Law:

[tex]\frac{\sin \theta _1}{n_1} = \frac{\sin \theta _2}{n_2}[/tex]

[itex]\theta _1[/tex] is the angle of incidence, [itex]n_1[/itex] is the index of refraction of the first medium (i.e. if it's going from water to plastic, water is the first medium), [itex]\theta _2[/itex] is the angle of deviation (wait, I don't remember it being called "angle of deviation," I thought it was angle of refraction or something, oh well...) and you can guess what [itex]n _2[/itex] is. I believe you have 3 of these values, so you can find the fourth. And don't take my word, look up this law and make sure I have it right (I can't imagine it being any different, but you never know).

EDIT: Oh, and in case you needed, the third value that isn't explicitly given to you is the index of refraction of air, which I'm pretty sure is 1, but look that up too.
futb0l
#3
Jul5-04, 01:35 AM
P: n/a
i know how to calculate the angle of refraction, but i am still not sure on how to calculate the 'angle of deviation' which is quite confusing. btw, Snell's Law is suposed to be...

[tex]n_1 \sin \theta _1 = n_2 \sin \theta_2[/tex]

kuenmao
#4
Jul5-04, 05:21 AM
P: 76
Light - Angle of Deviation...

futb0l got the equation right.
I believe that angle of deviation is NOT the angle of refraction...rather, the angle formed by the refracted ray, and the incident ray. In simpler terms, it means "how big an angle the light ray is bent". My method of finding this angle would be to subtract the angle of refraction from the angle of incidence. Why? Draw a figure yourself.
futb0l
#5
Jul5-04, 05:24 AM
P: n/a
allright thanks.
futb0l
#6
Jul5-04, 05:47 AM
P: n/a
I tried using your method [kuenmao] but when I did it I have an answer of 29 degrees. The answer sheet says it's 19 degrees, maybe I didn't do the question correctly or something.
Jamez
#7
Jul6-04, 08:24 AM
P: 19
wait, wait, wait... From Glass(1.55) to Air (1) at 65 degrees?
thats gonna total internally refect isn't it? i just calculated the critical angle of glass to air, it's 40.17 degrees.
futb0l
#8
Jul6-04, 07:57 PM
P: n/a
yeah, i thought about that too... and knowing that it will internally reflect, the angle of deviation should be 50 degrees... because 90 - 65 = 25 and then you times it by 2...

however, the answer sheet says it's 19... which I think is wrong.
hotsugababe
#9
Nov10-10, 08:09 PM
P: 1
Hey, i dont know how long ago this questions from but
the answer 19 is right!
This is how you do it:
the formula as you said is
[tex]n_1 \sin \theta _1 = n_2 \sin \theta_2[/tex]

what you do is
[tex]n_1(1) \sin ( 65 = n_2(1.55) \sin \theta_2[/tex]
and then you get
Sin(65) divided by 1.55 = refractive angle
the refractive angle is 35
Your mistake was you just subtracted 35 from 65 and got 30
you should have subtracted:
SIN (65) - SIN(35)
And then gotten the inverse sin
which would have given you
19.4 which is 19 degrees
:D yay!


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