Deduce the relation between $n, u, v, R$ for refraction at a spherical surface, where the symbols have their usual meaning.

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Solution

Figure shows the formation of image $I$ of an object $O$ on the principal axis of a convex surface with the centre of curvature $C$ .
As the aperture is small the length of NM will be taken to be nearly equal to the length of the perpendicular from the point $N$ on the principal axis.
For small angles (assuming paraxial rays) we have,
$tan (∠ N O M) \approx ∠ N O M = \frac{M N}{O M}$
$tan (∠ N C M) \approx ∠ N C M = \frac{M N}{M C}$
$tan (∠ N I M) \approx ∠ N I M = \frac{M N}{M I}$
Here $i = ∠ N O M + ∠ N C M = \frac{M N}{O M} + \frac{M N}{M C}$
and $r = ∠ N C M - ∠ N I M = \frac{M N}{M C} - \frac{M N}{M I}$
For small angles, Snell's law, $n_{1} sin i = n_{2} sin r$ can be written as
$n_{1} i = n_{2} r$
$n_{1} (\frac{M N}{O M} + \frac{M N}{M C}) = n_{2} (\frac{M N}{M C} - \frac{M N}{M I})$
$n_{1} M N (\frac{1}{O M} + \frac{1}{M C}) = n_{2} M N (\frac{1}{M C} - \frac{1}{M I})$
$\frac{n_{1}}{O M} + \frac{n_{1}}{M C} = \frac{n_{2}}{M C} - \frac{n_{2}}{M I}$
$\frac{n_{1}}{O M} + \frac{n_{2}}{M I} = \frac{n_{2}}{M C} - \frac{n_{1}}{M C}$
$\frac{n_{1}}{O M} + \frac{n_{2}}{M I} = \frac{n_{2} - n_{1}}{M C}$
Applying the Cartesian sign convention, $O M = - u, M I = + v, M C = + R$
$\frac{n_{1}}{- u} + \frac{n_{2}}{v} = \frac{n_{2} - n_{1}}{R}$
$\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$

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