Calculating Angle Between Rays PQ & RS

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Homework Help Overview

The discussion revolves around calculating the angle between two rays, PQ and RS, with an additional ray TQ striking a surface at point Q. The problem involves concepts related to optics, specifically refraction and reflection, with a refractive index of 1.50 provided.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss various attempts to calculate the angle, with some expressing uncertainty about their results and questioning the clarity of the problem's diagram. There are inquiries about the concept of deviation and its relation to the angles involved.

Discussion Status

Some participants have provided guidance on the relationship between incident and reflected angles, while others are seeking clarification on the concept of deviation. The discussion is ongoing, with multiple interpretations and approaches being explored.

Contextual Notes

There are mentions of unclear diagrams and the need for better visual representation to facilitate understanding. Additionally, some participants note the challenge of self-teaching and the limitations of their resources regarding the concept of deviation.

Taylor_1989
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Homework Statement


Calculate the angle between the ray PQ and the ray RS

Anouther ray TQ also strikes the surface at Q. The refractive index of the glass is 1.50


Homework Equations



c= 1/ refractive index

The Attempt at a Solution


The answer on my sheet say it is 20 degree, I have had to different attempts which have give me 20.9 and 21.6 neither which i think is right. I cal the 1st on by: 1/1.50=41.8/2 which gave 20.9 and the second by 85-41.8/2 which gave 21.6. I took the refractive index from a question further down, which I have put in, but I don't need help with that question. As this is in my book and taken from a past paper, I don't know that If i looked at the exam paper, I could just measure the angle. But as it says calculate I assume, I have to use the equations
 

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hi, your diagram is not clear. there is no point P in the diagram. its difficult to understand the problem with the given diagram. Taylor, if your problem can't be understood by people here, there is less chance that people will help you. always post questions in clear manner
so that even a baby can understand it...
 
IssacNewton said:
hi, your diagram is not clear. there is no point P in the diagram. its difficult to understand the problem with the given diagram. Taylor, if your problem can't be understood by people here, there is less chance that people will help you. always post questions in clear manner
so that even a baby can understand it...

Thanks for the response, I wonder why I haven't has a response, I will try and draw a more clear diagram.
 
Right I have managed to, get a copy of the exam paper, question. I have attached, so I hope this helps people ans my question.

Many thanks.
 

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Right, I think I a seen where I have gone wrong, after searching on the internet for about 4hrs "bad times". The question is about something called deviation? As I understand deviation is the difference between the sin i and sin r (not refractive index). So I am right in working out the question this way: 90 degree - 85 degree= 5 degree, so that is the angle of deviation for 1 half? Then I did 85 * 2= 170 degree and 5*2= 10, 170-10= 160 so the 180-160=20.

If possible could someone verify this, and give me a better explanation on deviation if there is one as deviation is not included much in my books.
 
I don't know about deviation, but I did get the right answer. And I'm not sure how you got the answer, so I'll just try to help from the beginning.

Firstly, there is an incident ray PQ, and reflected ray QR. And it gives you the incident angle (which is measured from the normal N). So what can you say about the angle of the reflected ray? (again measured from the normal N)
 
BruceW said:
I don't know about deviation, but I did get the right answer. And I'm not sure how you got the answer, so I'll just try to help from the beginning.

Firstly, there is an incident ray PQ, and reflected ray QR. And it gives you the incident angle (which is measured from the normal N). So what can you say about the angle of the reflected ray? (again measured from the normal N)

I can sort of see you coming from, but 100% could you please expand, or show the exact calculation, so I can work back from how you worked it out. Also what are you counting as the normal? The N and M on the diagram.

I do apologize if this is a bit of a stupid question, but I am self taught, so I only understand what I have read.
 
taylor, angle of incidence is same as angle of reflection. that is one of the laws of geometric optics.. so can you see which one is angle on incidence and which one is angle reflection at point Q

Edit: http://en.wikipedia.org/wiki/Angle_of_incidence
 
  • #10
angle of incidence is PQN, since QN is normal at point Q. now what's angle of reflection ?
 
  • #11
IssacNewton said:
angle of incidence is PQN, since QN is normal at point Q. now what's angle of reflection ?

Yer my bad, I was looking at the question thinking it was a lot harder than it looks. I completely forgot about the incident = reflection. Rookie mistake, least I learned from this mistake, to read the question more carefully.

Thanks for the help, probs would have been stuck on this for a long time, until I realized where I was going wrong.
 
  • #12
so what's the answer you got ?
 

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