Total internal reflection in prism

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Homework Help Overview

The discussion revolves around the concept of total internal reflection in a prism, specifically a 30-60-90 degree prism. The original poster presents a problem involving the index of refraction of a liquid placed on the hypotenuse of the prism and seeks to determine the maximum index for total reflection. Additionally, there is a related question about light rays in a right-angle prism and their behavior at different wavelengths.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the index of refraction and the significance of rounding in the context of Mastering Physics. There are attempts to clarify the angles of incidence and refraction for different wavelengths and to identify potential errors in reasoning.

Discussion Status

The conversation includes various interpretations of the problems, with some participants providing guidance on the significance of angles and the implications of rounding errors. There is an ongoing exploration of the correct application of Snell's law and the conditions for total internal reflection.

Contextual Notes

Participants note the challenges posed by the Mastering Physics platform regarding rounding and significant figures, as well as the potential for misinterpretation of angles in relation to the normal. There is also mention of the assumption that certain angles in the problem are exact, which may not align with typical precision in physics problems.

marpple
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Homework Statement



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.

Homework Equations



sin ik = n1/n2

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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Your answer looks fine to me.
 
Doc Al said:
Your answer looks fine to me.

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."

any clues about using mastering physics regarding to rounding the answer?
 
Since the calculation gives 1.45492..., it is possible that Mastering Physics made a rounding error and thinks it should be 1.46.
 
Redbelly98 said:
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.. :(
 
Good catch, RB!

marpple said:
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.
 
It was just a hunch.

If sin(60°)=0.8660... is rounded to 0.87, one gets 1.4616→1.46
 
Redbelly98 said:
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 :wink:). 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.. :)

YF-33-42.jpg


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
 
  • #10
marpple said:
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?
 
  • #11
Doc Al said:
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 :wink:). 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!
 
  • #12
Doc Al said:
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. :biggrin:

so, we don't need to consider from air to glass?
Redbelly98 said:
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 :wink:
 
  • #13
marpple said:
so, we don't 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.
 
  • #14
Doc Al said:
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??
 
  • #15
marpple said:
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.
 
  • #16
Doc Al said:
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??
 
  • #17
marpple said:
which one is not correct??
the angle of the incidence or the angle of refraction??
None of them are correct. :frown:
 
  • #18
Doc Al said:
None of them are correct. :frown:

LOL
:biggrin:

ic,,is this because i took the normal wrongly..??
 
  • #19
marpple said:
ic,,is this because i took the normal wrongly..??
Beats me, since you didn't show your normal on the diagram. Looks like you took the angle of incidence measured from parallel to the surface instead of normal, and the angles of refraction measured from the horizontal line. :rolleyes:
 
  • #20
Doc Al said:
Beats me, since you didn't show your normal on the diagram. Looks like you took the angle of incidence measured from parallel to the surface instead of normal, and the angles of refraction measured from the horizontal line. :rolleyes:

LOL...
how did you know?? :shy:

ic,, i'll work on it :)
 
  • #21
Doc Al said:
What's the angle of incidence at that surface? What are the angles of refraction?

well, let me try again :biggrin:

for the first wavelength
the angle of incidence is 58 degree, and the angle of refraction is 70 ??

for the second wavelength
the angle of incidence is 58 degree, and the angle of refraction is 78.5 ??
 
  • #22
marpple said:
for the first wavelength
the angle of incidence is 58 degree, and the angle of refraction is 70 ??

for the second wavelength
the angle of incidence is 58 degree, and the angle of refraction is 78.5 ??
Now you're cooking. :cool: :approve:
 
  • #23
Doc Al said:
Now you're cooking. :cool: :approve:

noooo...
it's still wrong :cry:
 
  • #24
What values did you get?
 
  • #25
Doc Al said:
What values did you get?

Find the index of refraction of the prism for each of the two wavelengths.

n1 = 0.902 and n2 = 0.865
 
  • #26
marpple said:
n1 = 0.902 and n2 = 0.865
Show how you got those values. (Don't confuse the two indices of refraction when applying Snell's law.)
 
  • #27
Doc Al said:
Show how you got those values. (Don't confuse the two indices of refraction when applying Snell's law.)

for the first wavelength..
na sin 58 = n1 sin 70
n1 = (sin 58 . 1 ) / sin 70 = 0.902

for the scond wavelength...
na sin 58 = n2 sin 78.5
n2 = (sin 58 . 1 ) / sin 78.5 = 0.865


actually, I'm not so sure about which one is the first wave and which one is the second.
so, here i assumed, one with 12 degree angle to the horizontal line as the first wave
 
  • #28
marpple said:
for the first wavelength..
na sin 58 = n1 sin 70
n1 = (sin 58 . 1 ) / sin 70 = 0.902

for the scond wavelength...
na sin 58 = n2 sin 78.5
n2 = (sin 58 . 1 ) / sin 78.5 = 0.865
You have things backwards. It should be:
n_{prism}\sin\theta_i = n_{air}\sin\theta_r

actually, I'm not so sure about which one is the first wave and which one is the second.
so, here i assumed, one with 12 degree angle to the horizontal line as the first wave
It doesn't matter which is first or second, since Snell's law is symmetric:

n_1\sin\theta_1 = n_2\sin\theta_2

What does matter is that you have the index of refraction matched with the correct angle. :wink:

In this case, the ray is going from prism to air, so the prism angle is the angle of incidence.
 
  • #29
Doc Al said:
You have things backwards. It should be:
n_{prism}\sin\theta_i = n_{air}\sin\theta_r


It doesn't matter which is first or second, since Snell's law is symmetric:

n_1\sin\theta_1 = n_2\sin\theta_2

What does matter is that you have the index of refraction matched with the correct angle. :wink:

In this case, the ray is going from prism to air, so the prism angle is the angle of incidence.

omg..i'm still thinking that it goes from air to prism.. LOL
 

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