- $tan \theta_i = \frac {MD}{CD} , tan 2\theta_i = \frac {MD}{QD} $
- For peraxial rays, point $M$ is close to $P$
- $\therefore C D \approx C P=R$, the radius of curvature llly $Q . D \approx Q . P$.
- Also, $\tan \theta_i \approx \theta_i, \tan 2 \theta_i$ $\approx 2 \theta_i$
- Gives $\theta_i=\frac{M D}{R}, 2 \theta_2=\frac{M P}{Q P} \rightarrow Q P=\frac{R}{2}$