Given, 

Object size = 2.5 cm

$$\begin{gathered} \mathrm{Object} \mathrm{distance}, \mathrm{u} = -27 \mathrm{cm} \\ \mathrm{Radius} \mathrm{of} \mathrm{curvature}, \mathrm{R} = 36 \mathrm{cm} \\ \mathrm{Focal} \mathrm{length}, \mathrm{f} =\frac{\mathrm{R}}{2} = \frac{-36}{2} = -18 \mathrm{cm} \\ \mathrm{Image} \mathrm{distance}, \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} = -\frac{1}{18}-\frac{1}{-27} \\ \Rightarrow \frac{1}{\mathrm{v}} = \frac{1}{27}-\frac{1}{18} =- \frac{1}{54} \\ \mathrm{i}.\mathrm{e}., \mathrm{v} = -54 \mathrm{cm} \end{gathered}$$

The negative sign indicates that the image is formed in front of the mirror. Thus, the screen should be placed at a distance of 54 cm in front of the mirror on the same side as that of the object. 

Now,
Magnification, $\boxed{\mathrm{m} = \frac{\mathrm{I}}{\mathrm{O}} = -\frac{\mathrm{v}}{\mathrm{u}}}$

Given,

$$\begin{gathered} \mathrm{Object} \mathrm{distance}, \mathrm{u} = -12 \mathrm{cm} \\ \mathrm{Focal} \mathrm{length}, \mathrm{f} = + 15 \mathrm{cm} \\ \mathrm{Size} \mathrm{of} \mathrm{the} \mathrm{object}, \mathrm{O} = 4.5 \mathrm{cm} \end{gathered}$$

Now using the mirror formula, 
                    $\boxed{\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} =\frac{1}{\mathrm{f}}}$ 
we have, 

             $$\begin{gathered} \frac{1}{\mathrm{v}} =\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ =\frac{1}{15}+\frac{1}{12} \\ = \frac{4+5}{60} \\ = \frac{9}{60} \\ \therefore \mathrm{v} = 60/9 = 6.7 \mathrm{cm} \end{gathered}$$

∴ Image is formed at 6.7 cm at the back of the mirror ( because v is positive)
Now,
Magnification, $\boxed{\mathrm{m} = \frac{\mathrm{I}}{\mathrm{O}} = -\frac{\mathrm{v}}{\mathrm{u}}}$ 

$\therefore$                       $$\begin{gathered} \frac{\mathrm{I}}{4.5} = -\frac{6.7}{-12} \\ \mathrm{I} = \frac{6.7 \times 4.5}{12} \\ = 2.5 \mathrm{cm}. \end{gathered}$$ 

Hence, the image formed is erect, and of course virtual. 

As, needle is moved farther away from the mirror then as a result, image moves away from the mirror (upto F) and keeps on decreasing in size.

Case I: When tank is filled with water.

Given, real depth = 12.5 cm;
          apparent depth = 9.4 cm 

Now, using the formula, 

                    $\boxed{\mathrm{\mu } = \frac{\mathrm{real} \mathrm{depth}}{\mathrm{apparent} \mathrm{depth}}}$

we have,
Refractive index,  $\mathrm{\mu } = \frac{12.5}{9.4} = 1.33$

Case II: When water in the tank is replaced by another liquid.

$$\begin{gathered} \mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{liquid}, \mathrm{\mu } = 1.63 \\ \mathrm{Real} \mathrm{depth} = 12.5 \mathrm{cm} \end{gathered}$$
Therefore,
$\mathrm{apparent} \mathrm{depth} = \frac{\mathrm{real} \mathrm{depth}}{\mathrm{\mu }}$ 

i.e., $\mathrm{apparent} \mathrm{depth} = \frac{12.5}{1.63} = 7.67 \mathrm{cm}$

Distance through which microscope has to be moved downward is = (9.4 – 7.67) cm = 1.73 cm.

(a)
Given, 

$$\begin{gathered} \mathrm{Angle} \mathrm{of} \mathrm{incidence}, \mathrm{i} = 60° \\ \mathrm{Angle} \mathrm{of} \mathrm{refraction}, \mathrm{r} = 35° \end{gathered}$$ 

Using Snell's law, we have 

Refractive index of glass with respect to air,     ${}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}} = \frac{\sin \mathrm{i}}{\sin \mathrm{r}} = \frac{\sin 60°}{\sin 35°}$

         $= \frac{0.8660}{0.5736} = 1.51$                                   

(b) 

$$\begin{gathered} \mathrm{Angle} \mathrm{of} \mathrm{incidence}, \mathrm{i} = 60° \\ \mathrm{Angle} \mathrm{of} \mathrm{refraction}, \mathrm{r} = 41° \end{gathered}$$ 

Using Snell's law, 

Refractive index of water wr.to air, ${}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{w}} = \frac{\sin \mathrm{i}}{\sin \mathrm{r}} = \frac{\sin 60°}{\sin 41°} = 1.32$ 

(c)
Angle of incidence in water, $\mathrm{i} =45°$

Refractive index of glass wr.to water,
 ${}^{\mathrm{\omega }}\mathrm{\mu }_{\mathrm{g}} = \frac{{}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}}}{{}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{w}}} =\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$
         $\frac{1.51}{1.32} = \frac{\sin 45°}{\sin \mathrm{r}} = \frac{0.7071}{\sin \mathrm{r}}$

$\therefore \sin \mathrm{r} = \frac{1.32 \times 0.7071}{1.51} = 0.6181$ 

$\Rightarrow$    $\mathrm{r} = 38.2°, which is the required angle of refraction of glass.$