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

[BlazenHammer]

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

 

[kuruman]

Please explain what μ and n' stand for. I don't want to guess.

 

[BlazenHammer]

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)

 

[TSny]

I found the following trig identity to be helpful $$\sin u + \sin v = 2 \sin \left( \frac{u+v}{2} \right) \cos \left( \frac{u-v}{2} \right) $$

 

[kuruman]

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.

 

[BlazenHammer]

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 :)

 

[kuruman]

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 $A$?