# What is the smallest angle of incidence necessary so that TIR occurs?

• I
• pkc111
In summary: I don't see how that helps either.The angle of incidence is always the same, so the angle of refraction must also be the same. That is the principle of total internal reflection.
pkc111
TL;DR Summary
The ray diagram below shows a ray of incidence entering a block with a refractive index of 1.52. Determine the smallest angle of incidence necessary so that the refracted ray will totally internally reflect off the second boundary.

So I figure the answer is 90 degrees...but I don't feel that is what they are after?

pkc111 said:
Summary: The ray diagram below shows a ray of incidence entering a block with a refractive index of 1.52. Determine the smallest angle of incidence necessary so that the refracted ray will totally internally reflect off the second boundary.

View attachment 250004

So I figure the answer is 90 degrees...but I don't feel that is what they are after?
Have you done even the minimal amount of research, like looking at the TIR wiki?

sophiecentaur
Absolutely

Motore
My reasoning is that the angle of the refracted ray from the top surface is parallel to the incident angle in the bottom surface. So for the top one to be 90 degrees the bottom one has to be also.

Motore
Hints: Total internal reflection critical angle -> Snell's law -> smallest incidence angle necessary

90 is the smallest to produce 90..I don't see your point sorry?

Have you looked at the equation for total internal reflection critical angle? If so, did you compute the critical angle? If it comes out 90 degrees, then you have made a mistake.

pkc111 said:
90 is the smallest to produce 90..
What do you mean by "produce 90", and what does it have to do with the question?

A.T. said:
What do you mean by "produce 90", and what does it have to do with the question?
see reasoning posted 1 hr ago; 90 degrees refracted angle is the condition for TIR to begin

see my working... theta 1 = theta 3, so for theta 3 to equal 90, theta 1 must also...right?

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But that is not what the problem is asking of you. You need to find the smallest Θ1 that on the second boundry the ray doesn't exist the block (i.e. is totaly internaly reflected). So Θ2 is essentialy the total internal reflection critical angle. Again, compute the total internal reflection critical angle and by snell's law you will find the smallest Θ1.

case in point...how do you get TIR to occur by varying the bottom incident beam angle? It doesn't look possible to me?

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Motore said:
But that is not what the problem is asking of you. You need to find the smallest Θ1 that on the second boundry the ray doesn't exist the block (i.e. is totaly internaly reflected). So Θ2 is essentialy the total internal reflection critical angle. Again, compute the total internal reflection critical angle and by snell's law you will find the smallest Θ1.
No that is exactly what is being asked...ie what minimum value of theta 1 produces a total internal reflection at the top boundary? ie for what value of theta 1 is theta 3 = 90 degrees.

No. Read the problem again. It says total internal reflection of the second boundry. You cannot have total internal reflection outside the block because air has a lower index of refraction than the block. The light enters the block at the angle we are seeking so it will totaly internaly reflect inside the block and never leave it.
Again, look at the Wiki page of total internal relfection.

Motore said:
The light enters the block at the angle we are seeking so it will totaly internaly reflect inside the block and never leave it.
Exactly...and I am saying there is no angle for which this is true..you just have to look at the geometry to see that (2 parallel lines ie air/block incident ray parallel to block/air refracted ray). You don't need any Snells law calculations to see that.

The question does not demand that the light ray never leave the block. It requires that the ray which is refracted upon entry at the lower face is then totally reflected at the upper face. There is no requirement for total internal reflection of the reflected ray as it strikes the lower face on the rebound.

Edit: However, I see the point. There is no such angle. Symmetry demands that the ray is deflected toward the upper face by enough that TIR can never occur.

Thank you jbriggs444

pkc111 said:
90 is the smallest to produce 90..I don't see your point sorry?
That doesn't make sense. Angles in optics are relative to the normal. You would have found that out from any of the hundreds of links about refraction in general.

sophiecentaur said:
That doesn't make sense. Angles in optics are relative to the normal. You would have found that out from any of the hundreds of links about refraction in general.
If you read the posts you would understand.

pkc111 said:
If you read the posts you would understand.
I only need to read that you talk of a minimum angle of 90. That makes no sense, whatever the previous posts say. More than 90 will take you round the other side of the interface. This is a very basic question that you will find in any A level level optics book and you need to use the standard conventions for solving a problem like this.
As for multiple internal reflections, if the sides of the block are parallel, the beam will not escape when the angle of incidence is great enough to establish TIR once. (See optical fibre)

PS a proper diagram from the OP would allow us all to know what he means. This is one of the most common difficulties when PF is asked a seemingly basic question.

What is your issue Sophie centaur? The question was resolved 24 hrs ago by someone who actually read the posts.

I don't think I could recommend this site to anyone...Way too much attitude

sophiecentaur said:
That doesn't make sense.
I agree that the OP isn't communicating terribly clearly. Hint to @pkc111: if everyone is saying you aren't making sense, you are more likely to get help if you rephrase what you are asking. (Edit: in fact you did - fifteen posts in - possibly earlier but the working in the photo is illegible.)

That said, I believe (as @jbriggs444 noted) the point is that the symmetry of the situation is such that the angle of incidence at the second surface is equal to the angle of refraction at the second surface. Thus it's impossible to get total internal reflection in this case, because the angle of incidence at the second surface would need to exceed the critical angle and it cannot because the angle of refraction at the first surface is less than or equal to the critical angle. The only way to get total internal reflection is to inject light into one of the side faces.

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pkc111 said:
I don't think I could recommend this site to anyone...Way too much attitude
The least I can do is to give you a reply. After sitting and thinking about it, I find that you must be right = so , apologies.
The light can't get in through the first face if it is to be TIR'd at the next face. I guess that makes sense; usually, to launch a TIR's beam, you would go in via the end of the block.
I guess the questioner was probing deeper than one might have thought at first.

@Ibix got there first.

Wow..so much drama over a question that was very straight forward...And answered so simply and easily by a poster over 24 hrs ago. The question was complete from the beginning. Some posters should think about how much they are really here to help or just to throw your weight around

weirdoguy and sophiecentaur
pkc111 said:
The question was complete from the beginning.
No, sorry, that is not true. I too was confused by your original post (OP), since you did not label any angles and did not specify what you meant by 90 degrees. TIR problems are covered a lot on the Internet, so even if you did not have a better figure from the question you were trying to work on, you can use Google Images to find very similar TIR problem figures, and link to them or paste them into your question.

And many posters were trying to help you by guessing about what you were asking, and trying their best to provide the guidance that they thought you might need. Making helpers guess what you are asking about wastes a lot of folks' time, which degrades the discussion forum process. We strive to have a high signal-to-noise ratio in posts and replies here (which is one of the big attractions of the PF), so when we have numerous posts trying to guess what the OP was asking, that is not good.

Anyway, it sounds like you found the reply that you wanted part-way through the thread, so I'll go ahead and tie off the thread now.

sophiecentaur, Motore and Nugatory

## 1. What is total internal reflection (TIR)?

Total internal reflection is a phenomenon that occurs when a light ray travels from a denser medium to a less dense medium at an angle of incidence greater than the critical angle. This results in the light being completely reflected back into the denser medium, instead of being refracted.

## 2. How is the angle of incidence related to TIR?

The angle of incidence is the angle at which a light ray enters a medium. The angle of incidence must be greater than the critical angle for TIR to occur. If the angle of incidence is less than the critical angle, the light ray will be refracted instead of being reflected.

## 3. What is the critical angle?

The critical angle is the smallest angle of incidence at which TIR occurs. It is determined by the refractive indices of the two media that the light ray is traveling through. The critical angle can be calculated using Snell's law: sin(critical angle) = n2/n1, where n1 is the refractive index of the denser medium and n2 is the refractive index of the less dense medium.

## 4. Is the critical angle the same for all materials?

No, the critical angle varies depending on the refractive indices of the two materials. Different materials have different refractive indices, so the critical angle will be different for each combination of materials.

## 5. How does the angle of incidence affect the amount of light reflected during TIR?

The angle of incidence has a direct impact on the amount of light reflected during TIR. The closer the angle of incidence is to the critical angle, the more light will be reflected. As the angle of incidence increases beyond the critical angle, the amount of reflected light decreases until eventually, all of the light is refracted instead of being reflected.

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