Derive the following expression for the refraction at concave spherical surface: $\frac{μ}{v} - \frac{1}{u} = \frac{μ - 1}{R}$ .

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Solution

Let MPN be concabe spherical surface of a medium of refractive index $μ$ . P is the pole o is the centre of curvature PC is the principle axis. Let O be a point object or it's principle axis kept in a rarer medium first incident ray travelling through C is normal to the spherical surface therefor it will not deviate and goes along PX another mident ray OA will refract at Point A and it bends towards the normal. AB and PX are proceeded behind then at I, a virtual image will be form.
Let $α, β, γ$ are the ray angle normal with the principle axis respectively.
then by snell's law
$μ = \frac{s i n i}{s i n r}$ ........(1)
but here i and r are the small then we put.
$s i n i = i$ $s i n r = r$ in equation (1)
$μ = \frac{i}{r}$
$i = μ n$ ...........(2)
now by using exterior angle theorem in $Δ$ AOC
$γ = 1 + a$
$i = γ - α$ .........(3)
now in $Δ$ IAC by using exterior angle theorem
$γ = b + r$
$r h o = γ - β$ ..........(4)
putting value of i and r in equation (2)
$(γ - α) = μ (γ - β)$ ...........(5)
now angle $= \frac{a r c}{r a d i u s}$
$α = \frac{P A}{O P}$
$β = \frac{P A}{I P}$
now $γ = \frac{P A}{C P}$
using value of $α, β, γ$ in equation (5)
$\frac{P A}{P C} - \frac{P A}{P O} = μ (\frac{P A}{P C} - \frac{P A}{P I})$
$P A (\frac{1}{P C} - \frac{1}{P O}) = m . P A (\frac{1}{P C} - \frac{1}{P O})$
$\frac{1}{P C} - \frac{1}{P O} = μ (\frac{1}{P C} - \frac{1}{P I})$ .........(6)
now by using sign convention
$P C = - R$ $P I = V$
$P O = - u$
using these value in equation (6)
$(\frac{1}{- R}) - (\frac{1}{- u}) = μ (\frac{- 1}{R} + \frac{1}{V})$
$\frac{- 1}{R} + \frac{1}{u} = μ (\frac{- 1}{R} + \frac{1}{v})$
$\frac{- 1}{R} + \frac{1}{u} = \frac{μ}{R} + \frac{μ}{v}$
$\frac{- 1}{R} + \frac{μ}{R} = \frac{μ}{v} - \frac{1}{u}$
$\frac{μ - 1}{R} = \frac{μ}{v} - \frac{1}{u}$
This is the required expression for the refraction formula for the concave spherical surface.

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