Interference from Thin Films — Question 6

Question 6

An air wedge turns equal thickness increments into equally spaced fringes

Two glass plates form a thin air wedge with $t(x)=\alpha x$, where $\alpha\ll1$. In near-normal reflection there is one relative phase reversal, so dark fringes obey $2t=m\lambda_0$. At $\lambda_0=600\,\mathrm{nm}$, neighboring dark fringes are $1.50\,\mathrm{mm}$ apart.

1. Find the wedge angle and the gap at $x=30.0\,\mathrm{mm}$.

2. If a liquid of index 1.50 fills the gap without changing its shape and both glass indices exceed 1.50, find the new dark-fringe spacing and the dark order at that same x.

Question 6 - Solution

An air wedge turns equal thickness increments into equally spaced fringes

(a) Method and calculation

$$2\alpha x_m=m\lambda_0\Longrightarrow\Delta x=\lambda_0/(2\alpha).$$$$\boxed{\alpha=2.00\times10^{-4}\,\mathrm{rad},\quad t(30.0\,\mathrm{mm})=6.00\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$2n\alpha x_m=m\lambda_0\Longrightarrow\boxed{\Delta x^\prime=\lambda_0/(2n\alpha)=1.00\,\mathrm{mm}}.$$$$\boxed{m=2nt/\lambda_0=30}.$$

Physical check

The lower gap interface still reflects from lower to higher index after filling, so the single relative phase reversal remains.