1. Refraction at a Refracting Surface when Object lies in Rarer Medium :
Question : Prove that $\frac{-n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R}$
When image formed is real :
Consider a convex spherical refracting surface of refractive index $n_2$. Let it be placed in a rarer medium of refractive index $n_1$ ($n_2 > n_1$).
A point object O lies in rarer medium on the principal axis at a distance $u$ from the pole of the convex refracting surface. A ray of light from O incident on the convex surface at A. Let C be the centre of curvature, then AC is the normal to the convex surface. After refraction at A, the ray enters the denser medium and bends towards the normal. The refracted ray meets the principal axis at I which is the real image of the object O. The distance of the image I from the pole of the convex surface is $v$.
STEP 1. Determination of $i$ and $r$. Let $\alpha$, $\beta$ and $\gamma$ be the angles made by the incident ray, refracted ray and the normal respectively with the principal axis. Draw AN perpendicular on the principal axis.
$\triangle AOC$,
$i = \alpha + \gamma \quad \dots (i)$
$\triangle AIC$
$\gamma = r + \beta \quad \text{i.e.} \quad r = \gamma - \beta \quad \dots (ii)$
$\alpha$, $\beta$ and $\gamma$ are small angles, therefore, we can write $\alpha$, $\beta$ and $\gamma$ as $\tan \alpha$, $\tan \beta$ and $\tan \gamma$ respectively.
Then, eqns. (i) and (ii) can be written as
$i = \tan \alpha + \tan \gamma \quad \dots \text(iii)$
$r = \tan \gamma - \tan \beta \quad \dots \text(iv)$
Since,
$\tan \alpha = \frac{AN}{NO}$
$\tan \gamma = \frac{AN}{NC}$,
$\tan \beta = \frac{AN}{NI}...(v)$
Using equations (v), eqn. (iii) becomes
$i = \frac{AN}{NO} + \frac{AN}{NC}$
and eqn. (iv) becomes
$r = \frac{AN}{NC} - \frac{AN}{NI}$
Since aperture of the spherical surface is assumed to be small, so point N lies very close to point P.
$NO = PO, NC = PC, NI = PI$
Hence,
$i = \frac{AN}{PO} + \frac{AN}{PC} \quad \dots \text(vi)$
and
$r = \frac{AN}{PC} - \frac{AN}{PI} \quad \dots \text(vii)$
STEP 2. Using Snell's law,
$\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$
$n_1 \sin i = n_2 \sin r$
Since angles $i$ and $r$ are also small, so $\sin i = i$ and $\sin r = r$
$n_1 i = n_2 r \quad \dots \text(viii)$
Using eqns. (vi) and (vii) in eqn. (viii), we get
$n_1 \left( \frac{AN}{PO} + \frac{AN}{PC} \right) = n_2 \left( \frac{AN}{PC} - \frac{AN}{PI} \right)$
i.e.
$n_1 \left( \frac{1}{PO} + \frac{1}{PC} \right) = n_2 \left( \frac{1}{PC} - \frac{1}{PI} \right)$
$\frac{n_1}{PO} + \frac{n_1}{PC} =\frac{n_2}{PC} - \frac{n_2}{PI}$
i.e.
$\frac{n_1}{PO} + \frac{n_2}{PI} = \frac{n_2}{PC} - \frac{n_1}{PC} \quad \dots (ix)$
STEP 3. Applying new cartesian sign conventions, we find
$PO = -u, PC = R, PI = v$
Hence, eqn. (ix) can be written as:
$\frac{n_1}{-u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R} \quad \dots \text(x)$
It also holds for Convex and concave spherical surface for real and virtual object place in rarer medium.
2. Refraction at Spherical Refracting Surface when object lies in Denser Medium : Let an object O lie in the denser medium of refractive index $n_2$ on the principal axis. The formation of the real image I of the object O is
STEP 1. Determination of $i$ and $r$. Let $\alpha$, $\beta$ and $\gamma$ be the angles made by the incident ray, refracted ray and normal respectively with the principal axis. Draw AN perpendicular on the principal axis.
From $\triangle AOC$,
$\gamma = i + \alpha$
$i = \gamma - \alpha \quad \dots (i)$
From $\triangle AIC$,
$r = \beta + \gamma \quad \dots (ii)$
As per assumption, angles $\alpha$, $\beta$ and $\gamma$ are small, so $\alpha = \tan \alpha$, $\beta = \tan \beta$ and $\gamma = \tan \gamma$.
Hence, eqns. (i) and (ii) become
$i= \tan \gamma - \tan \alpha \quad \dots (iii)$
$r= \tan \beta + \tan \gamma \quad \dots (iv)$
Now,
$\tan \alpha = \frac{AN}{NO}$
$\tan \beta = \frac{AN}{NI}$
And
$\tan \gamma = \frac{AN}{NC} \quad \dots (v)$
Then eqn. (iii) becomes
$i = \frac{AN}{NC} - \frac{AN}{NO} \quad \dots (vi)$
and eqn. (iv) becomes
$r = \frac{AN}{NI} + \frac{AN}{NC} \quad \dots (vii)$
STEP 2. Using Snell's law,
$\frac{\sin i}{\sin r} = \frac{n_1}{n_2} \quad \text{or} \quad n_2 \sin i = n_1 \sin r$
Since angles $i$ and $r$ are small, so $\sin i = i$ and $\sin r = r$
$n_2 i = n_1 r \quad \dots (viii)$
Substituting the values of $i$ and $r$ from eqns. (vi) and (vii) in eqn. (viii), we get
$n_2 \left[ \frac{AN}{NC} - \frac{AN}{NO} \right] = n_1 \left[ \frac{AN}{NI} + \frac{AN}{NC} \right]$
$\frac{n_2}{NC} - \frac{n_2}{NO} = \frac{n_1}{NI} + \frac{n_1}{NC}$
$-\frac{n_2}{NO} - \frac{n_1}{NI} = \frac{n_2 - n_1}{NC} \quad \dots (ix)$
Since aperture of the spherical surface is assumed to be very small, so point N lies close to point P.
$NC = PC$, $NO = PO$ and $NI = PI$
Hence eqn. (ix) can be written as
$-\frac{n_2}{PO} - \frac{n_1}{PI} = \frac{n_2 - n_1}{PC} \quad \dots (x)$
STEP 3. Applying new cartesian sign conventions,}
$$PC = -R, \quad PO = -u, \quad PI = v$$
Hence eqn. (x) becomes
$-\frac{n_2}{-u} - \frac{n_1}{v} = \frac{n_2 - n_1}{-R}$
or
$\frac{n_2}{u} - \frac{n_1}{v} = \frac{n_1 - n_2}{-R}$
or
$-(-\frac{n_2}{u} + \frac{n_1}{v})=- \frac{n_1 - n_2}{R}$
$-\frac{n_2}{u} - \frac{n_1}{v}=\frac{n_1 - n_2}{R}$
Which is the required expression.