Three Proofs — Question 4

Question 4

Index-weighted transverse magnification determines longitudinal sensitivity

For a spherical refracting surface, a ray through the vertex obeys $n_1\theta_i=n_2\theta_t$. An object tip is at (-p,hO) and its image at (q,hI). Use paraxial slopes, then differentiate $n_1/p+n_2/q=(n_2-n_1)/R$ with fixed indices and radius.

1. Derive the transverse magnification.

2. Derive dq/dp and express it in terms of transverse magnification.

Question 4 - Solution

Index-weighted transverse magnification determines longitudinal sensitivity

(a) Method and calculation

$$\theta_i=-h_O/p,\quad\theta_t=h_I/q\Rightarrow-n_1h_O/p=n_2h_I/q.$$$$\boxed{m=h_I/h_O=-n_1q/(n_2p)}.$$

(b) Interpretation and further calculation

$$-n_1/p^2-(n_2/q^2)(dq/dp)=0\Rightarrow\boxed{dq/dp=-n_1q^2/(n_2p^2)}.$$$$\boxed{dq/dp=-(n_2/n_1)m^2}.$$

Physical check

Transverse magnification and axial sensitivity involve different index factors. The equal-index limit reduces to the familiar minus m squared.