The smallest angle of incidence necessary for total internal reflection (TIR) to occur at the second boundary of a block with a refractive index of 1.52 is determined by the critical angle, which must be calculated using Snell's Law. The discussion clarifies that an angle of 90 degrees cannot be the solution, as it does not satisfy the conditions for TIR. Instead, the angle of incidence at the first boundary must be less than the critical angle to ensure that the light ray is internally reflected at the second boundary. The confusion arises from misinterpretation of the problem's requirements regarding the angles involved.
PREREQUISITESStudents and professionals in physics, particularly those focusing on optics, as well as educators seeking to clarify concepts related to total internal reflection and light behavior at interfaces.
Have you done even the minimal amount of research, like looking at the TIR wiki?pkc111 said:
What do you mean by "produce 90", and what does it have to do with the question?pkc111 said:
see reasoning posted 1 hr ago; 90 degrees refracted angle is the condition for TIR to beginA.T. said:
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.Motore said:
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.Motore 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.pkc111 said:
If you read the posts you would understand.sophiecentaur said:
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.pkc111 said:
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.)sophiecentaur said:
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.pkc111 said:
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.pkc111 said:
