Magnifying power of telescope,
$$\eqalign{ & MP = \frac{{\beta \left( {{\text{angle subtended by image at eye piece}}} \right)}}{{\alpha \left( {{\text{angle subtended by object on objective}}} \right)}} \cr & {\text{Also,}}\,\,MP = \frac{{{f_0}}}{{{f_e}}} = \frac{{150}}{5} = 30 \cr & \alpha = \frac{{50}}{{1000}} = \frac{1}{{20}}rad \cr & \therefore \beta = \theta = MP \times \alpha = 30 \times \frac{1}{{20}} = \frac{3}{2} = 1.5\,rad \cr} $$

Here $$u = - 2\,f, v = 2\,f$$
As $$\left| u \right|$$ increases, $$v$$ decreases for $$\left| u \right|$$ $$> f.$$  The graph between $$\left| v \right|$$ and $$\left| u \right|$$ is shown in the figure. A straight line passing through the origin and making an angle of 45° with the $$x$$ - axis meets the experimental curve at $$P(2\,f, 2\,f).$$

$$\frac{{\left| {{f_1}} \right|}}{{\left| {{f_2}} \right|}} = \frac{2}{3}$$
$${f_1}:$$  focal length of convex lens.
$$\eqalign{ & \frac{1}{f} = \frac{1}{{{f_1}}} - \frac{1}{{{f_2}}} \cr & \Rightarrow \frac{1}{{30}} = \frac{1}{{{f_1}}} - \frac{2}{{3{f_1}}} \cr & {f_1} = 10\,cm,{f_2} = - 15\,cm \cr} $$

If side of object square $$ = \ell $$
and side of image square $$ = \ell '$$
From question, $$\frac{{\ell '}}{\ell } = 9\,\,{\text{or}}\,\,\frac{{\ell '}}{\ell } = 3$$
i.e, magnification $$m = 3$$
$$u = - 40\,cm$$
$$v = 3 \times 40 = 120\,cm\,f = ?$$
From formula, $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{ & \frac{1}{{120}} - \frac{1}{{ - 40}} = \frac{1}{f}\,\,{\text{or,}}\,\,\frac{1}{f} = \frac{1}{{120}} + \frac{1}{{40}} = \frac{{1 + 3}}{{120}} \cr & \therefore f = 30\,cm \cr} $$

In an optical fibre, light is sent through the fibre without any loss by the phenomenon of total internal reflection as shown in the figure.

As light travels from denser to rarer medium, so refractive index $$\left( \mu \right) = \frac{{{\text{real depth}}}}{{{\text{apparent depth}}}}$$
Refractive index $$\left( \mu \right) = 1.5$$
Net apparent depth $$ = 2 + 5 = 7\,cm$$
And real depth $$ = {\text{apparent depth}} \times \mu $$
$$\therefore $$ Real depth $$ = 1.5 \times 7 = 10.5\,cm$$