Solving Snell's Law: Find Incident & Refractive Angles

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Homework Help Overview

The discussion revolves around applying Snell's Law to determine the incident and refractive angles when only the deviation angle is provided. The context includes materials with known refractive indices, specifically air and glass, and a deviation angle of 30 degrees.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the incident angle, refractive angle, and deviation angle, with some suggesting trigonometric manipulations and the use of angle addition identities. Questions arise regarding the iterative methods for solving the equations and the handling of variables on both sides of the equations.

Discussion Status

Some participants have provided hints and guidance on manipulating equations, while others express frustration over the lack of direct solutions. There is an ongoing exploration of different approaches to solve for the refractive angle, with no explicit consensus reached.

Contextual Notes

Participants have noted the forum's rules regarding the provision of help, emphasizing the need for questioners to demonstrate their attempts at solving the problem before receiving assistance. There is also a mention of the original poster's feelings of irritation regarding the perceived lack of support.

zebedi
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Hi,
Can someone tell me how to find the incident and refractive angles when only the deviation angle is known?

For instance, I know that Snells law is n1 sin(i) = n2 sin(r)
I know that the materials are air (n1=1) and glass (n2=1.5)
And I know that the deviation angle d is 30deg.

I can summise that d = i - r
and r = arcsin( (n1/n2) x sin i )
therefore i = arcsin( (n1/n2) x sin i ) + d

How do I find i? Do I use an iterative method? How does that work?

Thanks in advance
Z
 

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You just need to use some trigonometric manipulations.
 
zebedi said:
Hi,
Can someone tell me how to find the incident and refractive angles when only the deviation angle is known?

For instance, I know that Snell's law is n1 sin(i) = n2 sin(r)
I know that the materials are air (n1=1) and glass (n2=1.5)
And I know that the deviation angle d is 30deg.

I can summise that d = i - r
and r = arcsin( (n1/n2) x sin i )
therefore i = arcsin( (n1/n2) x sin i ) + d

How do I find i? Do I use an iterative method? How does that work?

Thanks in advance
Z
Hello zebedi. Welcome to PF !

attachment.php?attachmentid=44030&d=1329489171.jpg


You have that d = i - r .

So, i = r + d .

Use that in Snell's Law to get: n1 sin( r + d) = n2 sin(r) .

Use the angle addition identity for sine . Solve for r .
 
You mean
sin(A+B) = sin(A)cos(B)+cos(A)sin(B) ?

Can you help out further because my maths still isn't good enough to solve it.
Thanks
 
That is the correct equation you need to use. And you are given d, so from there it is straightforward to solve for r.
 
BruceW wrote: "so from there it is straightforward to solve for r. "

That may be so for some, but for this ageing dunce it isn't! So if someone would be kind enough to help me out and show me how to solve it I'd be very appreciative.
 
SammyS said:
You have that d = i - r .

So, i = r + d .

Use that in Snell's Law to get: n1 sin( r + d) = n2 sin(r) .

Use the angle addition identity for sine .

Expand sin( r + d) and substitute into Snell's law. What do you get?


ehild
 
You get

n1(sinr cosd + cos r sind) = n2 sinr

but how do I solve it when there's an r on both sides?

And unless you're actually going to show me how to solve it then please don't bother posting!
 
I suggest you to read the rules of these Forums: We do not solve the problem instead of you. We give hint, we guide you. If you do not want that kind of help go to somewhere else.

ehild
 
  • #10
yes, you can do it yourself, zebedi. The next bit is to manipulate the equation, to get it to look like something you can solve for r.
 
  • #11
zebedi said:
...

And unless you're actually going to show me how to solve it then please don't bother posting!
You might like to read the rules for posting in the Homework section of this Forum. In particular:
...

NOTE: You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums. Once your question or problem has been responded to, do not go back and delete (or edit) your original post.
...

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

...​
 
  • #12
This whole process has been really irritating and a complete waste of my time.

I originally posted this question on the Classical Physics Forum, but someone moved it to here. If there was never any intention of answering the problem then why didn't someone say in the first place.

I'm not a kid struggling with my homework, I'm an employee struggling with a problem and I would have appreciated some real help - just like I dish out every day on other forums.
 
  • #13
zebedi
If you go back to your last equation and divide both side by Sinr it may become easier to sort out. Hope this is enough help to get you to the next stage.
 

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