Calculating Angle Between Rays PQ & RS

  • Thread starter Taylor_1989
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In summary, to calculate the angle between the ray PQ and the ray RS, you need to use the law of reflection which states that the angle of incidence is equal to the angle of reflection. You also need to take into account the normal at point Q, which is the line QN. Once this is understood, the angle between PQ and RS can be easily calculated using the given incident angle and the law of reflection.
  • #1
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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  • #2
*Another
 
  • #3
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...
 
  • #4
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.
 
  • #5
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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  • #6
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.
 
  • #7
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)
 
  • #8
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.
 
  • #9
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 ?
 

1. How do I calculate the angle between two rays?

To calculate the angle between two rays, you will need to use the formula: θ = cos^-1[(PQ * RS) / (|PQ| * |RS|)]. This formula uses the dot product of the two vectors PQ and RS, as well as their magnitudes.

2. What is the difference between a ray and a vector?

A ray is a line that extends infinitely in one direction, while a vector has both magnitude and direction. The angle between two rays can be calculated using the dot product of their corresponding vectors.

3. Is it possible for the angle between two rays to be negative?

No, the angle between two rays is always positive. If the result of the calculation is negative, it means that one of the vectors is pointing in the opposite direction of the other, resulting in an angle greater than 180 degrees.

4. Can the angle between two rays be greater than 180 degrees?

Yes, the angle between two rays can be greater than 180 degrees. This occurs when the dot product of the two vectors is negative, indicating that the rays are pointing in opposite directions.

5. How can I use the angle between two rays in real-world scenarios?

The angle between two rays can be used in various fields, such as engineering, physics, and navigation. For example, in engineering, it can be used to calculate the force required to move an object in a specific direction. In navigation, it can be used to determine the heading of a ship or plane.

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