Spherical Refracting Surfaces — Question 3

Question 3

Reversing a refracting path reverses the radius sign

Use $n_1/p+n_2/q=(n_2-n_1)/R$ and $m=-n_1q/(n_2p)$ for paraxial refraction. Here p is positive on the incident side, q on the outgoing side, and R when the center lies on the outgoing side. In one direction, a surface takes light from index 1.00 to 1.50 with R=+10 cm, p=30 cm, and q=90 cm. Now place a source at that image and reverse light propagation.

1. Assign the reverse-direction indices, radius, and object distance, then calculate the new image distance.

2. Find the reverse magnification and compare its product with the original magnification -2.

Question 3 - Solution

Reversing a refracting path reverses the radius sign

(a) Method and calculation

$$\boxed{n_1^\prime=1.50,\ n_2^\prime=1.00,\ R^\prime=-10\,\mathrm{cm},\ p^\prime=90\,\mathrm{cm}}.$$$$\frac1{q^\prime}=\frac{1-1.50}{-10}-\frac{1.50}{90}=\frac1{30}\Rightarrow\boxed{q^\prime=30\,\mathrm{cm}}.$$

(b) Interpretation and further calculation

$$\boxed{m^\prime=-\frac{1.50(30)}{1(90)}=-1/2}.$$$$\boxed{m^\prime m=(-1/2)(-2)=1}.$$

Physical check

The same physical center lies on the opposite side relative to the reversed direction of propagation; keeping the old radius sign would break reversibility.