Solution :

When the object O in the rarer medium lies close to the pole P of the convex refracting surface, the two refractive rays appear to diverge from a point I on the principal again So I is the virtual image of the point object O.

In $\Delta NOC, i=2+r$

In $\Delta NCI , r = \beta + r$

If all the rays are par axial then all the angle will be small , then

$\alpha = \tan \alpha = \large\frac{NM}{IM} = \frac{NM}{OP}$

$\beta = \tan \beta = \large\frac{NM}{IM} = \frac{NM}{IP}$

$r = \tan \gamma = \large\frac{NM}{MC} = \frac{NM}{PC}$

From Snell's law $ \large\frac{\sin i }{\sin r} = \frac{n_2}{n_1}$

But since i and r are small, $\large\frac{i}{r} =\frac{n_2}{n_1}$

or $n_1( \alpha + r)= n_2(\beta+r)$

$n_1 \bigg[ \large\frac{NM}{OP}+\frac{NM}{PC}\bigg]$$ = n_2 \bigg[ \large\frac{NM}{IP} +\frac{NM}{PC}\bigg]$

$n_1 \bigg[ \large\frac{1}{OP}+\frac{1}{PC}\bigg]$$ = n_2 \bigg[ \large\frac{1}{IP} +\frac{1}{PC}\bigg]$

=> $ \large\frac{n_1}{OP} -\frac{n_2}{IP} = \frac{n_2 - n_1}{PC} $

=> But $OP=-u, IP= -v$ an d$PC= +R$

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

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

(ii)

Since $f= \large\frac{1}{\mu -1}$

For $ \lambda_2 > \lambda_1 , \mu_2 < \mu_1$

Hence $f_2 > f_1$

Therefore When wave length of incident light is increased , focal length of the convex lens increases.

(iii)

According to lens maker's formula

$\large\frac{1}{f} $$ = (\mu-1) \bigg( \large\frac{1}{R_1}-\frac{1}{R_2}\bigg)$

(ie) $f \alpha \large\frac{1}{\mu-1}$

As $^{w} \mu g < ^a \mu g$, the focal length of the lens will increase when it is completely immersed in water .