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(i) Derive the mathematical relation between refractive indices n₁ and n₂ of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n₁and a real image formed in the denser medium of refractive index n₂. Hence, derive lens maker's formula.
(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed?

Answers (1)

(i) consider a spherical surface with center of curvature at c and radius of curvature R.

i is very small and the curved portion considered is the part of a large circle
from $\Delta OMN$
$\tan < NOM= \frac{MN}{-u}---(1)$
From $\Delta MNC$
$\tan < NCM= \frac{MN}{R}---(2)$
from $\Delta MNI$
$\tan < NIM= \frac{MN}{v}---(3)$
since the angle is very small
$i= < NOM+< NCM= \frac{MN}{-u}+\frac{MN}{R}---(4)$
$< NCM= < NIM+r$
$r= < NCM-< NIM$
$= \frac{MN}{R}-\frac{MN}{v}---(5)$
from snells law
$n_{1}\sin i= n_{2}\sin r$
$n_{1}i= n_{2} r---(6)$
from (5), (6) & (4)
$n_{1}\left ( \frac{MN}{-u} +\frac{MN}{R}\right )= n_{2}\left ( \frac{MN}{R} -\frac{MN}{v}\right )$
$\frac{-n_{1}}{u}+\frac{n_{2}}{v}= \frac{n_{2}}{R}-\frac{n_{1}}{R}$
or
$\frac{n_{2}-n_{1}}{R}= \frac{n_{2}}{v}- \frac{n_{1}}{u}$
$\frac{n_{2}}{v}- \frac{n_{1}}{u}= \frac{n_{2}-n_{1}}{R}---(7)$
A spherical lens can be considered as a two spherical surfaces The image of left surface act as virtual object for the other surface

for surface ABC

Apply equation (7) for surface ABC
$\frac{n_{1}}{OB}+\frac{n_{2}}{BI_{1}}= \frac{n_{2}-n_{1}}{BC_{1}}---(8)$
for surface ADC

I₁ act as a virtual object for ADC
$\frac{-n_{2}}{DI_{1}}+\frac{n_{1}}{DI}= \frac{n_{2}-n_{1}}{DC_{2}}---(9)\left ( from(7) \right )$
for thin lens
$BI_{1}= DI_{1},DC_{2}= R_{2}$
adding (8) & (9)
$\frac{n_{1}}{OB}+\frac{n_{1}}{DI}= (n_{2}-n_{1})\left ( \frac{1}{BC_{1}}+\frac{1}{DC_{2}} \right )$
Suppose the object is at infinity
$OB\rightarrow \infty ,DI= f$
$\Rightarrow \frac{n_{1}}{f}= (n_{2}-n_{1})\left ( \frac{1}{BC_{1}}+\frac{1}{DC_{2}} \right )$
$BC_{1}= R_{1},DC_{2}= -R_{2}$
$\Rightarrow \frac{n_{1}}{f}=\left ( n_{2}-n_{1} \right )\left [ \frac{1}{R_{1}} -\frac{1}{R_{2}}\right ]$
is the lens makers formula
given u = -100 cm
ii) R = 20 cm
$\frac{n_{2}}{v}-\frac{n_{1}}{-100}= \frac{1\left ( n_{2} -n_{1}\right )}{20}$
$\frac{1\cdot 5}{v}+\frac{1}{100}= \frac{0\cdot 5}{20}$
$\frac{1\cdot 5}{v}= \frac{1}{40}-\frac{1}{100}$
$\frac{1\cdot 5}{v}= \frac{6}{400}$
$v= 100\, cm$