- #1

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## 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?