Spherical Refracting Surfaces — Question 1

Question 1

A single curved interface can form an enlarged real image

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. Light passes from air n1=1.00 into glass n2=1.50 across a surface with R=+10 cm. The real object is at p=30 cm and has height 3 mm.

1. Find the image distance and classify the image.

2. Find magnification and signed image height, retaining the index ratio.

Question 1 - Solution

A single curved interface can form an enlarged real image

(a) Method and calculation

$$\frac{1.50}{q}=\frac{0.50}{10}-\frac1{30}=\frac1{60}\Rightarrow\boxed{q=90\,\mathrm{cm}}.$$$$q>0\Rightarrow\boxed{\text{real image in glass}}.$$

(b) Interpretation and further calculation

$$\boxed{m=-\frac{1(90)}{1.50(30)}=-2}.$$$$\boxed{h_I=-2(3)=-6\,\mathrm{mm}}.$$

Physical check

The transverse magnification for one refracting surface is not simply minus q/p when the two media differ.