Focal length of combination of lenses placed in contact is
$$\frac{1}{F} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}}$$
For convex lens, $${f_1} = 25\,cm$$
For concave lens, $${f_2} = - 25\,cm$$
Hence, $$\frac{1}{F} = \frac{1}{{25}} + \frac{1}{{ - 25}} = \frac{1}{{25}} - \frac{1}{{25}} = 0$$
$$\therefore F = \frac{1}{0} = \infty $$
Hence, power of combination,
$$P = \frac{1}{F} = 0\,D$$

$$\eqalign{ & \sin {\theta _c} = \frac{1}{\mu } = \frac{3}{4} \cr & {\text{or, }}\tan {\theta _c} = \frac{3}{{\sqrt {16 - 9} }} \cr & = \frac{3}{{\sqrt 7 }} \cr & = \frac{R}{{12}} \cr} $$

$$ \Rightarrow \,\,R = \frac{{36}}{{\sqrt 7 }}cm$$

As parallel beam incident on diverging lens will form image at focus.
∴ $$v = - 25\,cm$$

The image formed by diverging lens is used as an object for converging lens,
So, for converging lens $$u = - 25 - 15 = - 40\,cm, f = 20\,cm$$
∴ Final image formed by converging lens
$$\frac{1}{V} - \frac{1}{{ - 40}} = \frac{1}{{20}}$$
or, $$V = 40\,cm$$   from converging lens real and inverted.

$$\eqalign{ & \frac{1}{F} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} \cr & \frac{1}{F} = \left( {{\mu _1} - 1} \right)\left( {\frac{1}{\infty } + \frac{1}{R}} \right) + \left( {{\mu _2} - 1} \right)\left( {\frac{1}{{ - R}} - \frac{1}{\infty }} \right) = \frac{{{\mu _1} - {\mu _2}}}{R} \cr & F = \frac{R}{{{\mu _1} - {\mu _2}}} \cr} $$

As wavelength of violet light $${\lambda _{{\text{violet}}}} < {\text{Wave}}$$
length of red light $${\lambda _{{\text{red}}}},$$
therefore, refractive index of violet light $${\mu _{{\text{violet}}}} > $$
Refractive index of red light $${\mu _{{\text{red}}}}.$$
Hence, $${f_{{\text{red}}}} > {f_{{\text{violet}}}}$$
So, focal length of lens for red light is maximum than other visible spectrum of light.

Focal length of the biconvex lens is $$15\,cm.$$  A small object is placed at a distance of $$30\,cm$$  from the lens i.e. at a distance of $$2\,f.$$  Therefore the image should form at $$30\,cm$$  from the lens at $${I_1}.$$

But since the ray strike the plane mirror before reaching $${I_1},$$ the image $${I_1}$$ acts as the virtual object for reflection on plane mirror kept at a distance of $$20\,cm$$  from it. It should produce an image $${I_2}$$ but as the ray encounters the lens, it gets refracted and the final image is formed at $${I_3}.$$ For the last refraction from the biconvex lens, $$u = 10\,cm.$$
Applying lens formula $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{ & \Rightarrow \,\frac{1}{\nu } - \frac{1}{{10}} = \frac{1}{{15}} \cr & \Rightarrow \,\frac{1}{\nu } = \frac{1}{{15}} + \frac{1}{{10}} = \frac{{25}}{{150}} \cr & \Rightarrow \,\nu = 6\,cm \cr} $$
Therefore a real image is formed at a distance of $$16\,cm$$  from the plane mirror.

$$\eqalign{ & \frac{1}{{{f_a}}} = \left( {\frac{{1.5}}{1} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,\,\,.....\left( {\text{i}} \right) \cr & \frac{1}{{{f_m}}} = \left( {\frac{{{\mu _g}}}{{{\mu _m}}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right) \cr & \frac{1}{{{f_m}}} = \left( {\frac{{1.5}}{{1.6}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii),
$$\eqalign{ & \frac{{{f_m}}}{{{f_a}}} = \left( {\frac{{1.5 - 1}}{{\frac{{1.5}}{{1.6}} - 1}}} \right) = - 8 \cr & {P_a} = - 5 \cr & = \frac{1}{{{f_a}}} \cr & \Rightarrow \,\,{f_a} = \frac{1}{5} \cr & \Rightarrow \,\,{f_m} = - 8 \times {f_a} \cr & = - 8 \times - \frac{1}{5} \cr & = \frac{8}{5} \cr & {P_m} = \frac{\mu }{{{f_m}}} \cr & = \frac{{1.6}}{8} \times 5 \cr & = 1\,D \cr} $$

Resolving limit of telescope is given by $$\theta \propto \frac{x}{D} = \frac{\lambda }{d}$$
$$\lambda =$$  wavelength
$$d =$$  diagram of objective lens of telescope
$$D =$$  Distance between object and telescope
$$\eqalign{ & \Rightarrow x = \frac{{\lambda D}}{d} \cr & {\text{Given,}}\,\,\lambda = 5000\,\mathop {\text{A}}\limits^ \circ = 5000 \times {10^{ - 10}}m, \cr & D = 1\,km = 1000\,m, \cr & d = 10\,cm = 0.1\,m \cr & {\text{Hence,}}\,\,x = \frac{{5000 \times {{10}^{ - 10}} \times 1000}}{{0.1}} \cr & = 5 \times {10^{ - 3}}m \cr & = 5\,mm \cr} $$

$$\eqalign{ & \frac{1}{f} = \left( {_g^m\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right) \cr & {\text{Now, }}_g^m\mu = \frac{{_g\mu }}{{_m\mu }} = \frac{{1.5}}{{1.75}} \cr} $$
For concave lens as shown in figure in this case
$${R_1} = - R\,\,{\text{and }}{R_2} = + R$$

$$\eqalign{ & \therefore \,\,\frac{1}{f} = \left( {\frac{{1.5}}{{1.75}} - 1} \right)\left( { - \frac{1}{R} - \frac{1}{R}} \right) \cr & = + \frac{{0.25 \times 2}}{{1.75\,R}} \cr & \Rightarrow \,\,f = + 3.5\,R \cr} $$
NOTE : The positive sign shows that the lens behaves as convergent lens.

For the prism as the angle of incidence $$(i)$$  increases, the angle of deviation $$\left( \delta \right)$$  first decreases goes to minimum value and then increases.