Prove that $\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$
When refraction occurs of a convex spherical refracting surface and the ray travels from rearer to denser medium

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Solution

$tan α = \frac{M N}{O M} tan β = \frac{M N}{M I} tan γ = \frac{M N}{M C}$
For $△ N O C, i$ is the exterior angle
$i = ∠ N O M + ∠ N C M$
$i = \frac{M N}{O M} + \frac{M N}{M C}$
Similarly, $γ = \frac{M N}{M C} - \frac{M N}{M I}$
Applying snell's law
$n_{1} sin i = n_{2} sin r$
$\frac{n_{1}}{O M} + \frac{n_{2}}{M I} = \frac{n_{2} - n_{1}}{M C}$
as $O M = - u, M I = v, M C R$
$\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$ .

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Q. Define refraction of light waves.
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\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}

in object distance

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, image distance

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, refractive index of media

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and radius of curvature

(R)

Q. When a light rays enters a medium with refractive index

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from a medium with refractive index

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at curved interface with radius of curvature

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is given by

\frac{n_{2}}{V} - \frac{n_{1}}{U} = \frac{n_{2} - n_{1}}{R}

Now assume that the surface is plane and rewrite the formula with suitable changes.

In the adjoining figure, SS is a spherical surface separating two media of refractive indices n₁ and n₂ where n₁ > n₂. C is the centre of curvature of the spherical surface. An observer, keeping his eye beyond C in the medium of refractive index n₂ views the refracted image of an object AB placed as shown in the medium of refractive index n₁. The image will be: