Modulation index
\({ m }_{ a }=\frac { { E }_{ max }-{ E }_{ min } }{ { E }_{ max }+{ E }_{ min } }\)
\({ E }_{ min }=\frac { 4 }{ 2 } =2mV\)
\(Here,\quad { E }_{ max }=\frac { 16 }{ 2 } =8\quad mV\)
\(\therefore \ { m }_{ a }=\frac { 8-2 }{ 8+2 } =0.6\)

Optical source frequency, \(v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 800\times { 10 }^{ -9 } } =3.75\times { 10 }^{ 14 }Hz\)
Bandwidth of channel = 1% of source frequency = \(\frac { 1 }{ 100 } \times 3.75\times { 10 }^{ 14 }=3.75\times { 10 }^{ 12 }Hz\)
Number of channels for video T.V. signal = \(\frac { 3.75\times { 10 }^{ 12 } }{ 4.5\times { 10 }^{ 6 } } =8.3\times { 10 }^{ 5 }\)

Here, h = 289 m, R = 6.4 x 10⁶ m, N_max = 10¹² m^-3
For space wave propagation, maximum distance covered is given by
\(d=\sqrt { 2Rh } =\sqrt { 2\times \left( 6.4\times { 10 }^{ 6 } \right) \times 289 } =60.83\times { 10 }^{ 3 }m\simeq 61km\)
Since the distance between transmitter and receiver is 100 km, hence for the given frequency signal of 8 MHz, the propagation is not possible via space wave propagation.
For sky wave propagation, the critical frequency
\({ v }_{ c }=9\left( { N }_{ max } \right) ^{ 1/2 }=9\left( { 10 }^{ 12 } \right) ^{ 1/2 }=9\times { 10 }^{ 6 }Hz=9MHz\)
Since the frequency of signal 8 MHz is less than v_c  hence its propagation via sky wave is possible.

For D-region, N = 10⁹ m^-3 Here, v = 60 x 10⁶ Hz
Refractive index of the atmospheric layer is
\(\mu =\left[ 1-\frac { 81.45N }{ { v }^{ 2 } } \right] ^{ 1/2 }=\sqrt { 1-\frac { 81.45\times { 10 }^{ 9 } }{ \left( 60\times { 10 }^{ 6 } \right) ^{ 2 } } } \approx 1\)
Now, \(\mu =\frac { \sin { i } }{ \sin { r } } =1\) or  \(\sin { i } =\sin { r } \) or r = i = 30°

Selectivity of radio receiver is the ability of a receiver to accept the wanted signal and to reject the unwanted signals. Sensitivity of a radio receiver is its ability to amplify the desired weak signal.