Spherical Refracting Surfaces — Question 5

Question 5

Apparent depths through parallel layers add after index weighting

A point lies below two plane parallel transparent layers. The upper layer has thickness d1 and index n1; the lower has thickness d2 and index n2. The point is at the bottom of the lower layer. An observer is in index n0 above the top surface. Work paraxially.

1. Derive the apparent depth below the top surface by tracing a ray of small outgoing angle theta0.

2. Evaluate it for n0=1, d1=3 cm, n1=1.50, d2=8 cm, and n2=4/3.

Question 5 - Solution

Apparent depths through parallel layers add after index weighting

(a) Method and calculation

$$n_0\theta_0=n_1\theta_1=n_2\theta_2,\quad x=d_1\theta_1+d_2\theta_2.$$$$d_{\mathrm{app}}=x/\theta_0\Rightarrow\boxed{d_{\mathrm{app}}=n_0(d_1/n_1+d_2/n_2)}.$$

(b) Interpretation and further calculation

$$\boxed{d_{\mathrm{app}}=3/1.50+8/(4/3)=8\,\mathrm{cm}}.$$$$\boxed{d_{\mathrm{actual}}-d_{\mathrm{app}}=11-8=3\,\mathrm{cm}}.$$

Physical check

Each layer contributes its own reduced thickness; using one average index without appropriate weighting is generally incorrect.