- #1
Pi Face
- 76
- 0
Homework Statement
(B) http://imgur.com/VL4aG
although I showed the rest of the problem for context if needed
Homework Equations
ΔL=αLΔT
The Attempt at a Solution
I'm a little confused by the wording of the problem. When we find the shift in ΔN, are we actually finding Δ(ΔN)?
Anyway, I started off by defining the original ΔN=2Lβ^2/λ as Ni, the initial ΔN value.
Then, I just replaced L with L+ΔL, which is the original arm length plus the change due to temperature. I called this new equation Nf
Nf=2(L+ΔL)β^2/λ=2(L+αLΔT)β^2/λ
Nf=2L(1+αLΔT)β^2/λ = 2Lβ^2/λ + 2Lβ^2ΔT/λ
Now to find the difference between the two, I subtract Ni from Nf
ΔN = Nf-Ni = 2Lβ^2/λ + 2Lβ^2ΔT/λ -2Lβ^2/λ = 2Lβ^2ΔT/λ
However, my answer for ΔN has an extra factor of β^2 in the numerator where the given expression in (B) does not. What did I miss? Am I not reading the question properly?