Deriving the Deflection Equation for a Prism

In summary, the person is stuck on how to move a form of the deflection equation for a prism into the requested equation for a prism with a specific apex angle. They have simplified the equation for the displacement angle into terms of θ, n, and α. They have tried every trigonometric identity they can think of but none of them provide the required equation.
  • #1
Axis001
12
0

Homework Statement



The problem I am having troubles with is proving the deflection of a ray by a prism can be represented by a specific equation : θd = θ - α + arcsin(sin(α)√(n2 - sin2(α)) - sin(α)cos(α)). I have derived another form of the deflection equation for a prism and I am attempting to move it into the form requested but I am getting stuck on some trigonometry.

Homework Equations



α is the apex angle of the prism and θ is the angle of incidence on the prism.

The Attempt at a Solution



See attachments.
 

Attachments

  • 2012-08-25_21-16-59_758.jpg
    2012-08-25_21-16-59_758.jpg
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Last edited:
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  • #2
Hello, Axis001. Could you please define the symbols, especially the meaning of θ and α?
 
  • #3
My apologies not sure how I overlooked defining my variables. α is the apex angle of the prism and θ is the angle of incidence on the prism.
 
  • #4
Is your attachment showing your own work? Are you trying to find a mistake in the attachment?

I'm guessing that θ1 and θ'1 are the angles of incidence and refraction, respectively, at the first refraction; whereas, θ'2 and θ2 are the angles of incidence and refraction, respectively, at the second refraction. (That is, the primes denote angles inside the prism.)
 
  • #5
The work in the attachment is mine and your assumption of origin of angles of refraction is correct. I have simplified the equation for the displacement angle I originally found θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ))) into terms of simply θ, n, and α. The last step shown in the attachment is where I am currently stuck. I currently have the displacement of the prism at θd = θ - α + arcsin(nsin[sin(α)√(n2 - sin2(α)) - sin(θ)cos(α))] and I simple cannot figure out how to reduce it to the required form of θd = θ - α + arcsin(sin(α)√(n2 - sin2(α)) - sin(θ)cos(α)). I have tried every single trigonometric identity I can think of and none of them give the reduced equation I am looking for.
 
Last edited:
  • #6
In the attachment, the third to last equation looks ok. However, in going to the final expression there appears to be two mistakes:
(1) the cosine factor is not quite simplified correctly (going from the next to last line to the last line)
(2) there’s a place where you took no/n1 to be n instead of 1/n (going from the third to last line to the next to last line)
 
  • #7
Firstly thank you for your reply I greatly appreciate it.

(2) there’s a place where you took no/n1 to be n instead of 1/n (going from the third to last line to the next to last line)

Ah yes I'm not sure how I missed.

(1) the cosine factor is not quite simplified correctly (going from the next to last line to the last line)

With the correction to the incorrectly stated index of refraction the correct reduction from the second to last line to the last line should be:

θd = θ - α + arcsin(nsin[sin(α)√(n2 - sin2(α)) - (1/n)sin(θ)cos(α)])

Which leaves me with the problem of the nsin(...) term in the equation. The required equation θd = θ - α + arcsin(sin(α)√(n2 - sin2(α)) - sin(θ)cos(α)) which is just the equation I currently have with the nsin(...) term eliminated from it. I have tried using sum property for sines again to see if it generates the required equation but it just provides a useless and incredibly complicated new equation.
 
  • #8
Note that cos[sin-1(β)] = √(1- sin2[sin-1(β)]) = √(1-β2)
 
  • #9
Ok from "start" to current step with the corrections you helped me find this is what I have.

θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ)))
θd = θ - α + arcsin(nsin[sin(α)cos[arcsin((1/n)sin(θ))] - cos(α)sin[arcsin((1/n)sin(θ))])
θd = θ - α + arcsin(nsin[sin(α)sqrt[1 - (sin2(θ))/n2] - (1/n)cos(α)sin(θ))
θd = θ - α + arcsin(nsin[(1/n)sin(α)sqrt[n2 - sin2(θ)] - (1/n)cos(α)sin(θ))
θd = θ - α + arcsin[nsin[(1/n)[sin(α)sqrt[n2 - sin2(θ)] - cos(α)sin(θ)]]]
 
  • #10
Axis001 said:
Ok from "start" to current step with the corrections you helped me find this is what I have.

θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ)))
θd = θ - α + arcsin(nsin[sin(α)cos[arcsin((1/n)sin(θ))] - cos(α)sin[arcsin((1/n)sin(θ))])
θd = θ - α + arcsin(nsin[sin(α)sqrt[1 - (sin2(θ))/n2] - (1/n)cos(α)sin(θ))
θd = θ - α + arcsin(nsin[(1/n)sin(α)sqrt[n2 - sin2(θ)] - (1/n)cos(α)sin(θ))
θd = θ - α + arcsin[nsin[(1/n)[sin(α)sqrt[n2 - sin2(θ)] - cos(α)sin(θ)]]]

Oh, I overlooked another mistake which I highlighted in red above. Should those sin's be there?
 
  • #11
TSny said:
Oh, I overlooked another mistake which I highlighted in red above. Should those sin's be there?

Yes the nsin(...) terms should be there and it is what is giving me such a head ache since I just simply cannot get it out of the equation. A form of the deflection equation that I know is θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ))). It is easily derivable from applying Snell's Law at both "sides" of the prism and the expression of θd = θ1 + θ2 - α. Solving for θ2 into a expression of θ1 and α will result in the deflection equation I started this with but it results with a nsin(...) term I just can't seem to get out of my equations.
 
  • #12
Axis001 said:
Yes the nsin(...) terms should be there

Hmm. Which is correct:

sin(α - β) = sinαcosβ-cosαsinβ

or

sin(α - β) = sin[sinαcosβ-cosαsinβ]
 
  • #13
:eek: I need to go to sleep. Thank you so much I really appreciate it.
 
  • #14
Axis001 said:
:eek: I need to go to sleep.

Me too. :zzz:
 

Related to Deriving the Deflection Equation for a Prism

1. What is the purpose of a prism in ray deflection?

The purpose of a prism in ray deflection is to split white light into its component colors by refracting and reflecting light at different angles, allowing for the study and analysis of the different wavelengths of light.

2. How does a prism deflect light?

A prism deflects light by changing the direction of the light's path through refraction. When light enters the prism, it slows down and bends towards the base of the prism. Then, as it exits the prism, it speeds up and bends away from the base, resulting in a change in the direction of the light.

3. What factors affect the amount of deflection in a prism?

The amount of deflection in a prism is affected by the angle of incidence (the angle at which the light enters the prism), the angle of refraction (the angle at which the light exits the prism), and the refractive index of the material the prism is made of.

4. Can a prism deflect light in multiple directions?

Yes, a prism can deflect light in multiple directions. This is because the amount of deflection depends on the angle of incidence, and the angle of incidence can be changed by rotating the prism.

5. What is the difference between dispersion and deviation in ray deflection?

Dispersion refers to the separation of white light into its component colors, while deviation refers to the change in direction of light as it passes through the prism. In other words, dispersion is the result of deviation in a prism.

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