Image velocity (w.r.t mirror) $$ = - m \times {\text{object}}\,{\text{velocity}}\left( {{\text{w}}{\text{.r}}{\text{.t}}\,{\text{mirror}}} \right)$$
Here $$m = 1$$

The diameter of the sun, $$O = 1.39 \times {10^9}m$$    and its image is $$I.$$ As the distance of sun from lens $$u = 1.5 \times {10^{11}}m$$    and from paper is $$v = 0.1\,m.$$
Then by using relation
$$\eqalign{ & \frac{I}{O} = \frac{v}{u} \cr & \Rightarrow \frac{I}{{1.39 \times {{10}^9}}} = \frac{{0.1}}{{1.5 \times {{10}^{11}}}} \cr} $$
we get $$I = 9.2 \times {10^{ - 4}}m$$

The phenomenon of total internal reflection takes place during reflection at $$P.$$

$$\sin \theta = \frac{1}{{_g^w\mu }}\,......\left( {\text{i}} \right)$$
Now,
$$\eqalign{ & _g^w\mu = \frac{{_g^a\mu }}{{_w^a\mu }} = \frac{{1.5}}{{\frac{4}{3}}} = 1.125 \cr & \therefore \sin \theta = \frac{1}{{1.125}} = \frac{8}{9} \cr & \therefore \sin \theta \,{\text{should}}\,{\text{be}}\,{\text{greater}}\,{\text{than}}\,\frac{8}{9}. \cr} $$

Given, $$R = 20\,cm$$
$${h_0} = 2\,cm\,\,{\text{and}}\,\,u = - 30\,cm$$
In general we have assumed $$\mu = 1.5$$
$$\eqalign{ & {\text{As}}\,\,\frac{1}{f} = \left( {1.5 - 1} \right) \times \frac{2}{{20}} = \frac{1}{{20}} \cr & {\text{So,}}\,\,f = 20\,cm \cr & {\text{Now,}}\,{\text{as}}\,\,\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \cr & \frac{1}{{20}} = \frac{1}{v} + \frac{1}{{30}} \cr & \frac{1}{v} = \frac{1}{{20}} - \frac{1}{{30}} = \frac{{10}}{{600}} \cr & \therefore v = 60\,cm \cr & {\text{As}}\,\,m = \frac{v}{u} = \frac{{60}}{{ - 30}} = - 2 \cr & \Rightarrow m = \frac{{{h_i}}}{{{h_o}}} = - 2 \cr & \therefore {h_i} = - 2 \times 2 = - 4\,cm \cr} $$
Here, image is real, inverted, magnified and height of image is $$4\,cm.$$

To find the refractive index of glass using a travelling microscope, a vernier scale is provided on the microscope

Consider the situation shown is figure.Let radius of curvature of lens surfaces is $$R.$$ The combination is equivalent to three lenses in contact.
$$\eqalign{ & \therefore \frac{1}{{{f_{eq}}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} + \frac{1}{{{f_3}}} \cr & = \frac{2}{{{f_1}}} + \frac{1}{{{f_2}}}\,\,\left( {\because {f_1} = {f_3}} \right) \cr & {\text{Now,}}\,\,\frac{1}{{{f_1}}} = \frac{1}{{{f_3}}} = \left( {\mu - 1} \right)\left( {\frac{2}{R}} \right) = \frac{1}{f} \cr & \frac{1}{{{f_2}}} = \left( {{\mu _w} - 1} \right)\left( { - \frac{2}{R}} \right) = \left( {\frac{4}{3} - 1} \right)\left( {\frac{{ - 2}}{R}} \right) \cr & = \left( {\frac{1}{3}} \right)\left( { - \frac{2}{R}} \right) = \left( {\frac{{ - 2}}{R}} \right)\left( {\frac{1}{{2\left( {\mu - 1} \right)}}} \right) \cr & \Rightarrow \frac{1}{{{f_2}}} = \left( { - \frac{1}{3}} \right)\left( {\frac{1}{{\frac{3}{2} - 1}}} \right)\left( {\frac{1}{f}} \right) = - \frac{2}{3} \times \frac{1}{f} \cr & \therefore \frac{1}{{{f_{eq}}}} = \frac{2}{f} - \frac{2}{3} \times \frac{1}{f} = \frac{{6 - 2}}{{3f}} = \frac{4}{{3f}} \Rightarrow {f_{eq}} = \frac{{3f}}{4} \cr} $$

$$\eqalign{ & {\text{For}}\,{M_1}:V = - 60,{m_1} = - 2 \cr & {\text{For}}\,{M_1}:u = + 20,F = 10 \cr & \frac{1}{V} = \frac{1}{{20}} = \frac{1}{{10}} \Rightarrow V = 20 \cr & \therefore {M_2} = - \frac{{20}}{{20}} = - 1 \cr & \therefore M = {m_1} \times {m_2} = + 2 \cr} $$

$${\text{As}}\,{F_{{\text{average}}}} = \frac{{2IA}}{c}$$
where, $$I =$$  energy flux of light $$ = 25 \times {10^4}\,W{m^{ - 2}}$$
$$A =$$  Surface area of reflecting surface
and $$c =$$  speed of light $$ = 3 \times {10^8}m/s$$
$$\eqalign{ & \therefore {F_{{\text{average}}}} = \frac{{2 \times 25 \times {{10}^4} \times 15 \times {{10}^{ - 4}}}}{{3 \times {{10}^8}}} \cr & = 2.50 \times {10^{ - 6}}N \cr} $$

When $${r_2} = C,\,\angle \,{N_2}Rc = {90^ \circ }$$
Where $$C$$ = critical angle
As $$\sin C = \frac{1}{v} = \sin {r_2}$$

Applying snell’s law at $$'R'$$
$$\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,.....\left( {\text{i}} \right)$$
Applying snell's law at $$'Q'$$
$$\eqalign{ & 1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & {\text{But, }}{r_1} = A - {r_2} \cr & {\text{So, }}\sin \theta = \mu \sin \left( {A - {r_2}} \right) \cr & \sin \theta = \mu \sin A\cos {r_2} - \cos A\,\,\,\,.....\left( {{\text{iii}}} \right)\,\,\left[ {{\text{using}}\left( {\text{i}} \right)} \right] \cr} $$
From (i)
$$\cos {r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - \frac{1}{{{\mu ^2}}}} \,\,\,.....\left( {{\text{iv}}} \right)$$
By eq. (iii) and (iv)
$$\sin \theta = \mu \sin A\sqrt {1 - \frac{1}{{{\mu ^2}}}} - \cos A$$
on further solving we can show for ray not to transmitted through face $$AC$$
$$\theta = {\sin ^{ - 1}}\left[ {\mu \,{\text{sin}}\,\left( {A - {{\sin }^{ - 1}}\left( {\frac{1}{\mu }} \right)} \right.} \right]$$
So, for transmission through face $$AC$$
$$\theta > {\sin ^{ - 1}}\left[ {\mu \sin \left( {A - {{\sin }^{ - 1}}\left( {\frac{1}{\mu }} \right)} \right.} \right]$$

$$\eqalign{ & \frac{1}{f} = \frac{2}{{{f_e}}} + \frac{2}{{{f_m}}} \cr & {f_m} = \frac{R}{2} \cr & {f_m} = 0 \cr & \frac{1}{{{f_e}}} = \frac{{n - 1}}{R} \cr & {f_e} = \frac{{n - 1}}{R} \cr & {f_2} = \frac{R}{{2n}} \cr & {f_1} = \frac{R}{{2\left( {n - 1} \right)}} \cr & {\text{Ratio}}\,\,\frac{{{f_1}}}{{{f_2}}} = \frac{n}{{n - 1}} \cr} $$