Minimal angle of deviation through the prism

[diracdelta]

Homework Statement

Minimal angle of deviation δ_m of light after passing through prism is connected with index of refraction n and inner angle of prism Φ:

If index of refraction and inner angle of prism are directly measured, determine minimal angle of deviation,
n=(1.52 ± 0.04)
Φ=(30 ± 0.02)°

The Attempt at a Solution

When having dependent variables, we use formulae x=F(a,b,c,...)

x(medium) =F(a(medium), b(medium),...)
plus forumulae for uncertainty.

My question here is, how to express delta as a function of phi and n?

Thank you

 

[rude man]

Hint: use arc sin.

 

[diracdelta]

well, n * sin ( phi ) = sin 1/2 ( phi + delta )
arcsin (n* sin (phi) = 1/2 ( phi + delta )
2 * arcsin ( n* sin(phi))= ( phi + delta )
=> delta = 2*arcsin (n*sin (phi )) - phi

idem there other way?
i assume this one is correct tho

 

[rude man]

well, n * sin ( phi ) = sin 1/2 ( phi + delta )
arcsin (n* sin (phi) = 1/2 ( phi + delta )
2 * arcsin ( n* sin(phi))= ( phi + delta )
=> delta = 2*arcsin (n*sin (phi )) - phi

idem there other way?
i assume this one is correct tho

That's the way!

 

[diracdelta]

one more question.
i got result in a problem i.e.
x= 43.84
Mx = 0.01559

now, when i round it to one significant is this correct;

Mx= 0.02
x= 43

x = (43 +- 0.02)

 

[diracdelta]

or x = 40

 

[rude man]

one more question.
i got result in a problem i.e.
x= 43.84
Mx = 0.01559

now, when i round it to one significant is this correct;

Mx= 0.02
x= 43

x = (43 +- 0.02)

Not sure I can answer that question. I would have said x = 43.84 +/- 0.02 but I don't know the background.

OK, I didn't see the error bars on φ. and n. I thought you just wanted nominal δ_m.

Realize that φ and n are independent variables. δ depends on the material's n and on the prism angle φ.
So you need dδ as a function of dφ and dn. What would that be? (Hint: dδ is an "exact differential" of φ and n.)

 

[diracdelta]

I appreciate your help, but i solved it, the rest, by myself. thanks. :)