Spherical Refracting Surfaces — Question 8

Question 8

Index matching removes the imaging power of a curved boundary

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. A curved interface separates two media whose indices are exactly equal to n. A real object is at distance p from the vertex. The radius R is finite.

1. Find the image distance and transverse magnification.

2. Identify the laboratory image position and explain why the curved boundary has no optical effect in geometrical optics.

Question 8 - Solution

Index matching removes the imaging power of a curved boundary

(a) Method and calculation

$$n/p+n/q=0\Rightarrow\boxed{q=-p}.$$$$\boxed{m=-n(-p)/(np)=1}.$$

(b) Interpretation and further calculation

$$x_O=-p,\quad x_I=q=-p\Rightarrow\boxed{x_I=x_O}.$$$$n\sin\theta_i=n\sin\theta_t\Rightarrow\boxed{\theta_t=\theta_i}.$$

Physical check

The unchanged diverging rays have the original object as their backward origin; a mathematical interface with no index contrast does not redirect them.