## total internal reflection

1. The problem statement, all variables and given/known data

Light is incident normally on the short face of a 30-60-90 (degree) prism. A drop of liquid is placed on the hypotenuse of the prism.

If the index of the prism is 1.68, find the maximum index that the liquid may have if the light is to be totally reflected.
2. Relevant equations

sin ik = n1/n2

3. The attempt at a solution

sin 60 = n1/1.68

n1= 1.45

it said that my answer is close enough only differ in significant figure. however, they didn't accept my answer. anyone how to solve this problem?? thanks
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i was doing this problem in mastering physic..
and it turned out

"Very close. Check the rounding and number of significant figures in your final answer."

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## total internal reflection

Since the calculation gives 1.45492..., it is possible that Mastering Physics made a rounding error and thinks it should be 1.46.

 Quote by Redbelly98 Since the calculation gives 1.45492..., it is possible that Mastering Physics made a rounding error and thinks it should be 1.46.
OMG..!!
thanks..
it works..

mastering physics sometimes is really killing me with its rounding thingies.. :(

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Good catch, RB!

 Quote by marpple mastering physics sometimes is really killing me with its rounding thingies.. :(
You will probably not be surprised to learn that you are not the first to complain about problems with mastering physics.
 Mentor Blog Entries: 10 It was just a hunch. If sin(60°)=0.8660... is rounded to 0.87, one gets 1.4616→1.46

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 Quote by Redbelly98 If sin(60°)=0.8660... is rounded to 0.87, one gets 1.4616→1.46
That's where I was going to go. I wondered if the 60° was meant to be "exact" or really to only 2 sig figs (or 1 ). But if that were true, then the answer should have been rounded to 1.5, which won't fly with Mastering Physics...
 alright... another questions.. still regarding to internal reflection.. :) A light ray in air strikes the right-angle prism shown in the figure (Intro 1 figure) angle B=32.0. This ray consists of two different wavelengths. When it emerges at face AB, it has been split into two different rays that diverge from each other by 8.50 degree . Find the index of refraction of the prism for each of the two wavelengths. so far,, what i did is... for wavelength one... n1 sin theta1 = n2 sin theta2 1 . sin 90 = n2 . sin 12 n2 = sin 90 / sin 12 = 4.809 for wavelength two... n1 sin theta1 = n2 sin theta2 1 . sin 90 = n2 sin (12+8.5) n2 = sin 90/ sin 20.5 = 2.85 anyone knows where the mistakes are...?? thanks

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 Quote by marpple alright... another questions.. still regarding to internal reflection..
This problem has nothing to do with total internal reflection, just plain old refraction. The first refraction is trivial--the rays pass straight through. You only have to worry about the second surface, where it goes from glass to air. What's the angle of incidence at that surface? What are the angles of refraction?

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 Quote by Doc Al That's where I was going to go. I wondered if the 60° was meant to be "exact" or really to only 2 sig figs (or 1 ). But if that were true, then the answer should have been rounded to 1.5, which won't fly with Mastering Physics...
Yes, they (Mastering Physics) have botched this one. It's not uncommon in physics problems to give nice round figures and assume they're "exact", as they seem to have done with the prism angles here.

A commercial prism will typically have better than 0.05 degrees (and sometimes better than 0.01 degrees) tolerance. Not that introductory physics students should be expected to know that!

 Quote by Doc Al This problem has nothing to do with total internal reflection, just plain old refraction. The first refraction is trivial--the rays pass straight through. You only have to worry about the second surface, where it goes from glass to air. What's the angle of incidence at that surface? What are the angles of refraction?
ahh..i thought it was total internal reflection since the ray coming in 90 degree,, but.. for total internal reflection 90 degree should be for the angle of refraction.

so, we dont need to consider from air to glass?
 Quote by Redbelly98 Yes, they (Mastering Physics) have botched this one. It's not uncommon in physics problems to give nice round figures and assume they're "exact", as they seem to have done with the prism angles here. A commercial prism will typically have better than 0.05 degrees (and sometimes better than 0.01 degrees) tolerance. Not that introductory physics students should be expected to know that!
yess...yess...
blame mastering physics

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 Quote by marpple so, we dont need to consider from air to glass?
Nothing to consider. The ray passes straight through (as shown in the diagram) since the angle of incidence is 0 degrees.

All the action takes place at the glass to air surface.

 Quote by Doc Al have to worry about the second surface, where it goes from glass to air. What's the angle of incidence at that surface? What are the angles of refraction?
so, for the first wavelength
the angle of incidence is 32 degree, and the angle of refraction is 12 ??

for the second wavelength
the angle of incidence is 32 degree, and the angle of refraction is 20.5 ??

am i on the right track??

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 Quote by marpple so, for the first wavelength the angle of incidence is 32 degree, and the angle of refraction is 12 ?? for the second wavelength the angle of incidence is 32 degree, and the angle of refraction is 20.5 ??
Your angles are off. Remember that the angles are measured from the normal to the surface, so you have to figure them out.

 Quote by Doc Al Your angles are off. Remember that the angles are measured from the normal to the surface, so you have to figure them out.
which one is not correct??
the angle of the incidence or the angle of refraction??

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 Quote by marpple which one is not correct?? the angle of the incidence or the angle of refraction??
None of them are correct.

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