Question

Deduce the relation between n, u, V, and R for refraction at a spherical surface,
where the symbols have their usual meaning.
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Answer 1

Answer:

Hope these attachments help you...

Answer 2

To derive:

Lens formula for curved surfaces.

Derivation:

Object has been placed at O , image is formed at I , C is the Centre of Curvature of curved surfaces.

We shall make a small assumption that NM is approximately the length of the perpendicular from point N to principal axis.

$\tan(\angle NOM)\approx \angle NOM =\dfrac{MN}{OM}$

$\tan(\angle NCM)\approx \angle NCM =\dfrac{MN}{MC}$

$\tan(\angle NIM)\approx \angle NOM =\dfrac{MN}{IM}$

$\therefore \angle i = \angle NOM + \angle NCM$

$\boxed{=> \angle i = \dfrac{MN}{OM}+\dfrac{MN}{MC}}$

Again ,

$\therefore \angle r = \angle NCM-\angle NIM$

$\boxed{=> \angle r = \dfrac{MN}{MC}-\dfrac{MN}{IM}}$

Applying Snell's Law:

$\therefore\:n_{1}\angle i=n_{2}\angle r$

$=> n_{1}MN(\dfrac{1}{OM}+\dfrac{1}{MC})=n_{2}MN(\dfrac{1}{MC}-\dfrac{1}{IM}$

$=>\dfrac{n_{1}}{OM}+\dfrac{n_{1}}{MC}=\dfrac{n_{2}}{MC}-\dfrac{n_{2}}{IM}$

$=>\dfrac{n_{1}}{OM}+\dfrac{n_{2}}{IM}=\dfrac{n_{2}-n_{1}}{MC}$

Applying Sign Convention:

$\dfrac{n_{2}}{v}-\dfrac{n_{1}}{u}=\dfrac{n_{2}-n_{1}}{R}$.

So, final expression:

$\boxed{\bf{\dfrac{n_{2}}{v}-\dfrac{n_{1}}{u}=\dfrac{n_{2}-n_{1}}{R}}}$.