When white light from sun falls on raindrops, sometimes a band of different colours in form of a circular arc is seen in the sky. This is called the rainbow.
The reason of origin of rainbow is that the small drops of water behave like a prism for the white sunlight due to which refraction, dispersion and total internal reflection of white light occurs from the water drops.
The rainbow is not seen after every rain, but is seen only when the light rays of particular colour suffer minimum deviation after one or two total internal reflections inside the small water drops.

Velocity of image in mirror
$$\eqalign{ & \overrightarrow v = - 10\hat i + 10\hat j \cr & {\overrightarrow v _{{\text{rel}}}} = \overrightarrow u - \overrightarrow v = 20\hat i. \cr} $$

We know that in case of a convex lens when object is placed at $$C',$$ the image is obtained at $$C.$$ This situation is represented in the graph by the point corresponding to $$u = - 10\,cm, v = 10\,cm.$$
Therefore, $$R = 10\,cm$$
$$ \Rightarrow \,\,\frac{R}{2} = 5\,cm = f$$

Lens formula is
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \cr & \Rightarrow \,\,\frac{{\Delta f}}{{{f^2}}} = \frac{{\Delta v}}{{{v^2}}} + \frac{{\Delta u}}{{{u^2}}} \cr} $$
(for maximum error in $$f$$)
$$\eqalign{ & \Rightarrow \,\,\frac{{\Delta f}}{{25}} = \frac{{0.1}}{{{{\left( {10} \right)}^2}}} + \frac{{0.1}}{{{{\left( {10} \right)}^2}}}\left[ {\Delta u = \Delta v = 0.1\,{\text{from graph}}} \right) \cr & \Rightarrow \,\,\Delta f = 25 \times 0.1 \times 2 \times 0.01 = 0.05 \cr} $$
Therefore, the focal length $$ = \left( {5.00 \pm 0.05} \right)cm.$$

According to the given condition, the beam of light will retrace its path after reflection from $$BC.$$

So, $$\angle CPQ = {90^ \circ }$$
Thus, angle of refraction at surface $$AC$$
$$\angle PQN = \angle r = {90^ \circ } - {60^ \circ } = {30^ \circ }$$
By Snell's law
$$\eqalign{ & \mu = \frac{{\sin i}}{{\sin r}} \Rightarrow \sqrt 2 = \frac{{\sin i}}{{\sin {{30}^ \circ }}} \cr & \therefore \sqrt 2 \times \sin {30^ \circ } = \sin i \cr & \Rightarrow \sqrt 2 \times \frac{1}{2} = \sin i \cr & \Rightarrow \sin i = \frac{1}{{\sqrt 2 }} = \sin {45^ \circ } \cr & \therefore i = {45^ \circ } \cr} $$

Refractive index
$$\mu = \frac{{\sin i}}{{\sin r}},\mu = \frac{{\sin i}}{{\sin A}}$$
In case of small angle, $$\sin i \approx i$$   and $$\sin A \approx A$$

$$\eqalign{ & \mu = \frac{i}{A} \cr & i = \mu A \cr} $$

According to question,
Focal length of objective lens $$\left( {{F_0}} \right) = + 40\,cm$$
Focal length of eyepiece lens $$\left( {{F_e}} \right) = 4\,cm$$
Object distance for objective lens $$\left( {{u_0}} \right) = - 200\,cm$$
Applying lens formula for objective lens

$$\eqalign{ & \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{{ - 200}} = \frac{1}{{40}} \cr & \Rightarrow \frac{1}{v} = \frac{1}{{40}} - \frac{1}{{200}} = \frac{{5 - 1}}{{200}} = \frac{4}{{200}} \cr & \Rightarrow v = 50\,cm \cr} $$
Image will be form at first focus of eyepiece lens. So, for normal adjustment distance between objectives and eye piece lense (length of tube) will be
$$v + {F_e} \Rightarrow 50 + 4 \Rightarrow 54\,cm$$

$$\frac{{{{\left( {R.P} \right)}_1}}}{{{{\left( {R.P} \right)}_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{5}{4}$$

According to question, diagram is shown below.

$$\eqalign{ & {\text{So,}}\,\,\angle MON = {90^ \circ } - A \cr & {\text{and}}\,\,\angle r = {90^ \circ } - \left( {{{90}^ \circ } - A} \right) \cr & \Rightarrow \angle r = A \cr} $$
By Snell's law, $$\frac{{\sin i}}{{\sin r}} = \mu $$
$$\eqalign{ & \frac{{\sin \left( {2A} \right)}}{{\sin \left( A \right)}} = \mu \cr & \frac{{2\sin A\cos A}}{{\sin A}} = \mu \cr & \Rightarrow \mu = 2\cos A \cr} $$

Consider the activity $$A$$ to $$B,$$
Applying $${v^2} - {u^2} = 2as$$
$$\eqalign{ & {v^2} - {0^2} = 2 \times 10 \times 7.2 \cr & \Rightarrow v = 12\,m/s \cr} $$
The velocity of ball as perceived by fish is
$$v'{ = _w}\mu \times v = \frac{4}{3} \times 12 = 16\,m/s$$

The relation between velocity of light $$\left( c \right),$$  frequency $$\left( \nu \right)$$ and wavelength $$\left( \lambda \right)$$  is
$$c = v\lambda $$
Thus, wavelength $$\lambda = \frac{c}{\nu }$$
Given, $$c = 3 \times {10^8}\,m/s,\nu = 100\,Hz$$
$$\therefore \lambda = \frac{{3 \times {{10}^8}}}{{100}} = 3 \times {10^6}m$$