Derive the relation between u, v and R for a convex spherical surface, when ray of light is going from rarer to denser medium. Draw appropriate diagram.

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Solution

one medium (rarer) and on the right side the second medium (denser). P is the pole of the spherical surface and C is its centre of curvature. According to ray diagram of fig, I is the image of the object O. The normal drawn from point M on principal axis is MP’.
According to Snell's law,
$_{1} n_{2} = \frac{s i n i}{s i n r}$
where $_{1} n_{2}$ = the refractive index of medium (2) with respect to medium (1).
$n = \frac{s i n i}{s i n r}$
(on taking n in place of $_{1} n_{2}$ temporarily ) If the angles i and r are very small, then
$s i n i \approx i$ and $s i n \approx r$
$∴ n = \frac{i}{r}$
or $i = n r$ ...........(1)
$∴$ In triangle, the exterior angle is equal to sum of the opposite two interior angles.
$∴$ From $Δ M O C, i = α + γ$ ..........(2)
And from $Δ M I C,$
$r = β + γ$ ...........(3)
Substituting the value of i and r in equation (1), then we get,
$(α + γ) = n (β + γ)$ ............(4)
If point M is not far from the principal axis, then
(i) Points P and P' are very close to each other and considered to be same point.
(ii) Angles $α, β$ and $γ$ will be small.
Therefore, $α \approx t a n α = \frac{M P}{O P} = \frac{M P}{- u}$
$β = t a n β = \frac{M P}{I P} = \frac{M P}{- v}$
and $γ = t a n γ = \frac{M P}{C P} = \frac{M P}{R}$
Now, substituting the values of $α, β$ and $γ$ in equation (4) then, we get
$(\frac{M P}{- u} + \frac{M P}{R}) = n (\frac{M P}{- v} + \frac{M P}{R})$
or $- \frac{1}{u} + \frac{1}{R} = - \frac{n}{v} + \frac{n}{R}$
or $\frac{n}{v} - \frac{1}{u} = \frac{n - 1}{R}$
Now again writing $_{1} n_{2}$ in place of n, we get
$\frac{_{1} n_{2}}{v} - \frac{1}{u} = \frac{_{1} n_{2} - 1}{R}$ ........(5)
This formula is known as 'Refraction Formula' of convex surface. If the absolute refractive indices of medium 1 and 2 be $n_{1}$ and $n_{2}$ respectively, then
$_{1} n_{2} = \frac{n_{2}}{n_{1}}$
$∴$ From equation $(5) \frac{n_{2}}{n_{1} v} - \frac{1}{u} = \frac{\frac{n_{1}}{n_{1}} - 1}{R} = \frac{n_{2} - n_{1}}{n_{1} R}$
or $\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$
This equation is true for all types of spherical surfaces.

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