REFRACTION AT A SPHERICAL SURFACE

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REFRACTION AT A SPHERICAL SURFACE

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Let us consider, LMN represents a ‘convex’ spherical surface of radius of curvature ‘R’ of refractive index ‘{n_2}’. Consider a point object ‘O’ on the principal axis situated in an optically rarer medium of {n_1}. \left( {{n_2}\, > \,{n_1}} \right).

A ray ‘OA’ incident at an angle ‘i’ is refracted along ‘AI’. Another ray ‘OM’ incident normally on the surface goes through undeviated and intersects at ‘I’ where final image is produced.

In ∆AOC, \angle i\, = \,\angle \alpha \, + \,\angle \gamma ………………….. (1)

In ∆ACI, \angle r\, = \,\angle \gamma \, - \,\angle \beta ………………….. (2)

From snell’s law, {n_1}\,\sin \,i\, = \,{n_2}\,\sin \,r

Since the rays are paraxial, i.e ‘A’ is close to ‘M’, so that angles i,\,r,\,\alpha ,\,\beta \,{\rm{and}}\,\gamma are small.

\sin \,i\,\, \simeq \,\,i\,\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,\,\sin \,\,r\, \simeq \,\,r

{n_1}\,i\,\, = \,\,{n_2}\,r ………………………… (3)

Put (1) and (2) in (3)

We have

\tan \,\alpha \, = \,\frac{h}{{OM}},\,\tan \,\beta \, = \,\frac{h}{{MI}}\,and\,\tan \,\gamma \, = \,\frac{h}{{MC}}

If ‘A’ is close to ‘M’,

Then OP ≃ OM, PI ≃ MI and PC ≃ MC

tan a ≃ a, tan b ≃ b and tan g ≃ g

\therefore \,\,\,\,\,\alpha \, = \,\frac{h}{{OP}},\,\,\,\beta \, = \,\frac{h}{{PI}}\,\,\,and\,\,\gamma \, = \,\frac{h}{{PC}}

Put in eq (4)

{n_1}\frac{h}{{OP}}\, + \,{n_2}\frac{h}{{PI}}\, = \,\left( {{n_2}\, - \,{n_1}} \right)\,\frac{h}{{PC}}

As per sign convention;

PI = + V, OP = –U and PC = +R

\therefore \,\,\,\,\,\,\,\,\frac{{{n_2}}}{{\rm{v}}}\, - \,\frac{{{n_1}}}{u}\, = \frac{{{n_2} - {n_1}}}{R}

NOTE:

The above hold good for concave refracting surface also, (proof will be same as done above).