Jenkins-White Optics: Relation between Prism/Deviation Angle and Rays

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Homework Help Overview

The discussion revolves around a problem from Jenkins-White's optics, specifically focusing on the relationship between the angles of a prism and the deviation of light rays. Participants are exploring trigonometric identities and the application of Snell's law in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various attempts to manipulate trigonometric identities to relate angles and the refractive index. There are questions about the definitions of variables like μ and n', and some participants express frustration over the complexity of the problem.

Discussion Status

Some participants have shared helpful trigonometric identities and suggested steps to approach the problem, while others are still grappling with the setup and relationships between the variables. There is no explicit consensus on a complete solution, and some interpretations are still being explored.

Contextual Notes

There are mentions of specific identities and relationships that need to be established, as well as a note that the problem is taken from a textbook, which may impose certain constraints on the approach. Participants are also navigating the challenge of missing information and the complexity of the trigonometric manipulations involved.

BlazenHammer
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Homework Statement
Show that, for any angle of incidence on a prism, following equation is true:
and and that the right-hand side reduces to μ' at minimum deviation.
Relevant Equations
sin[(A+d)/2]/ sin(A/2) = μcos[(r1-r2)/2]/ cos[(i-e)/2]

A = Angle of Prism
d = Total deviation by Prism
i= Angle Of First incident ray
e= Angle Of Final emergent Ray Corresponding to i
where
r1= Angle Of First refracted ray
r2= Angle Of Ray incident on second surface whose refracted ray makes angle e
μ= refractive index of prism relative to air
I've tried to attempt the first part of the problem(spent over an hour on this) as second part could be easily optained with some calculus ,I asked my friend but alas nobody could conjure the solution to this dangerous trigonometric spell.
It was just pages and pages of concoction of trigonometric manipulation with no end in sight so I think it is of no use to print my attempt here

My Thought Process: 1.) Begin with lhs, expand using sum trigonometric identity and substitute μ to arrive at something reducible to rhs
=>Failed
2.) Try to reverse engineer the rhs into lhs , do the same circus trick of substituting μ(in in trig expression) but nothing worked
=>Failed
3.) gOOGled but found nothing except out of print solution manuals available only in libraries :(

Regards,
BrazenHammer
 
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Please explain what μ and n' stand for. I don't want to guess.
 
Hello Kuruman,
I've corrected the mistake n' = μ (While i wrote it I did't knew about math symbols support :'))
μ= refractive index of prism relative to air
This problem is taken from Jenkins-White Fundamental of Optics Plane Surfaces Chapter(last problem)
 
I found the following trig identity to be helpful $$\sin u + \sin v = 2 \sin \left( \frac{u+v}{2} \right) \cos \left( \frac{u-v}{2} \right) $$
 
You need to
1. Find an expression relating A, r1 and r2.
2. Find an expression relating i, e, d and A.
3. Use Snell's law at each interface.
4. Look up the trig identity ##\sin\alpha + \sin\beta=\dots##
5. Put together all of the above equations. It pops right out.

On Edit: No need to do step 4; it has been provided by @TSny.
 
Thank you kuruman and TSny, I kind of made a monster out of a little moth.
Never knew this would be so simple ! :)
For others who are struggling with the same problem, here is complete solution:

## \sin{r_1} + \sin{r_2} =2\sin{\frac{r_1+r_2}{2}} \cos{\frac{r_1-r_2}{2}} - |##
## \sin{r_1} + \sin{r_2} =\frac{ \sin{i} + \sin{e}}{μ} ##
##\frac{ \sin{i} + \sin{e}}{μ} = 2\sin{\frac{i+e}{2}} \cos{\frac{i-e}{2}} - || ##

Equate (1) and (2) and the answer is right there :)
 
BlazenHammer said:
Thank you kuruman and TSny, I kind of made a monster out of a little moth.
Never knew this would be so simple ! :)
For others who are struggling with the same problem, here is complete solution:

## \sin{r_1} + \sin{r_2} =2\sin{\frac{r_1+r_2}{2}} \cos{\frac{r_1-r_2}{2}} - |##
## \sin{r_1} + \sin{r_2} =\frac{ \sin{i} + \sin{e}}{μ} ##
##\frac{ \sin{i} + \sin{e}}{μ} = 2\sin{\frac{i+e}{2}} \cos{\frac{i-e}{2}} - || ##

Equate (1) and (2) and the answer is right there :)
Wait a minute! This solution is only half completed. You are supposed to show that $$\frac{\sin\left(\frac{d+A}{2}\right)}{\sin\left(\frac{A}{2}\right)}=\mu\frac{\cos\left(\frac{r1-r2}{2}\right)}{\cos\left(\frac{i-e}{2}\right)}.$$What about the left hand side involving ##A##?