i) Discuss Doppler’s effect in electromagnetic waves.
ii) A telescope has an objective of focal length 120 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope when it is used to view distant objects when (a) the telescope is in normal adjustment, and (b) the final image· is formed at the least distance of distinct vision (25 cm).

OR

A metallic rod of length land resistance R is rotated with a frequency v, with one end hinged at the center and the other end at the circumference of a circular metallic ring of radius /, about an axis passing through the center and perpendicular to the plane of the ring. A constant and uniform magnetic field ft parallel to the axis is present everywhere.
(a) Derive the expression for the induced emf and the current in the rod.
(b)Due to the presence of the. current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod.
(c)Hence, obtain the expression for the power required to rotate the rod.
 

I.Doppler effect is used to measure speed in RADAR sensors. When the fixed-frequency radio wave sent from the sender continuously strikes an object that is moving towards or away from the sender, the frequency of the reflected radio wave will be changed. This frequency shift is known as Doppler effect.

Magnifying power of the telescope:
$\begin{array}{l}\mathrm{m}=\frac{-{\mathrm{f}}_0}{{\mathrm{f}}_{\mathrm{e}}}\left(1+\frac{{\mathrm{f}}_{\mathrm{c}}}{\mathrm{D}}\right)\\ =\frac{-120}{2}\left(1+\frac{2}{5}\right)\\ =-60\end{array}$

OR

Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y direction.
Suppose the direction of the magnetic field is along +z direction.
Then, using Lorentz law, we get the following.

$\begin{array}{l}\overrightarrow{\mathrm{F}}=-\mathrm{e}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})\\ \mathrm{F}=-\mathrm{e}(\mathrm{v}\widehat{\mathrm{j}}\times \mathrm{B}\widehat{\mathrm{k}})\\ {\mathrm{F}}^{-}=-\mathrm{e}\left({\mathrm{v}}^{-}\times {\mathrm{B}}^{-}\right)\overrightarrow{\mathrm{F}}=-\mathrm{evB}\widehat{\mathrm{i}}\end{array}$
Thus, the direction of force on the electrons is along -x axis. 
So, the electrons will move towards the center i.e. the fixed end of the rod. This movement of electron will affect in current and thus it will generate an emf in the rod between the fixed end and the point touching the ring. 
Let θ be the angle between the rod inside the circle. 
$=\frac{1}{2}{\mathrm{\pi r}}^2\mathrm{\theta }=12{\mathrm{\pi r}}^2\mathrm{\theta }$
Induced emf.
$=\mathrm{B}\times \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{2}{\mathrm{\pi r}}^2\mathrm{\theta }\right)=\frac{1}{2}{\mathrm{\pi r}}^2\mathrm{B}\frac{\mathrm{d\theta }}{\mathrm{dt}}=\frac{1}{2}{\mathrm{\pi r}}^2\mathrm{B\omega }$