Long distance communication between two points on the earth is achieved through reflection of electromagnetic waves by ionosphere. Such waves are called sky waves. Sky wave propagation takes place up to frequency of about 40 MHz. A space wave travels in a straight line from transmitting antenna to the receiving antenna. It is used for line of sight (LOS) communication as well as satellite communication.
(a) The ionospheric layer acts as reflector for a certain range of frequencies (3 to 30 MHz). Electro-magnetic waves of frequencies higher than 30 MHz (upto 40 MHz) penetrate the ionosphere and escape.
(b) In television broadcast, microwave links and satellite communication, space wave mode of propagation is used.

Amplitude modulation index is the ratio of the modulating signal to the maximum amplitude of carrier wave.
Given, frequency of carrier wave,
v_c = 1.5 MHz = 1500 kHz
Frequency of sinusoidal (modulated) wave,
v_m  = 10 kHz
Amplitude of carrier wave, A_C = 50 V
Modulated index, \(\mu \) = 50%
= \(\frac { 50 }{ 100 } =\frac { 1 }{ 2 } \) 
Modulation index, \(\mu \) = \(\frac { { A }_{ m } }{ { A }_{ C } } \) 
\(\frac { 1 }{ 2 } =\frac { { A }_{ m } }{ 50 } \) 
A_m = 25 V
So, the amplitude of AM wave, A_m =25 V
As, we know, the sidebands are
USB = v_c + v_m  = 1500 + 10 = 1510kHz
LSB = v_c - v_m = 1500 -10 = 1490 kHz
There are the required frequencies of the sidebands produced. 

Here, \(\lambda =1.3\times { 10 }^{ -6 }m\)   \(v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 1.3\times { 10 }^{ -6 } } =2.3\times { 10 }^{ 14 }Hz\)
Max bit rate   = 2 v = 2 x 2.3 x 10¹⁴ = 4.6 x 10¹⁴ Hz
For optical fibre, bandwidth = 2 GHz = 2 x 10⁹ Hz
No of phone calls = \(\frac { 4.6\times { 10 }^{ 14 } }{ 2\times { 10 }^{ 9 } } =2.3\times { 10 }^{ 5 }\)

AS the range of T.V. transmission, \(d=\sqrt { 2hr } \)so by increasing height of tower \(h\) the distance \(d\) upto which T.V. coverage can be done will increase. That is why to have greater T.V. coverage transmission has to be made using tall antennas.

(a) Rita's father takes interest to understand the working of the mind of his daughter and make effort to satisfy her desire for knowledge.
(b) The diode used are LED(Light Emitting Diode).

(i)  Radio wave communication system.
(ii)  Use of scientific and technological advancement in service to mankind.
(iii) Patriotism and love for the country, presence of mind.
(iv) Television communication system.

Here, h = 125 m ;
R = 6400 km = 6.4 x 10⁶ m.
Area covered = \(\pi { d }^{ 2 }=\pi \times 2hR\ \left[ \because \ \sqrt { 2hR } \right] \)
= 3.14 x 2 x 125 x 6.4 x 10⁶
= 5024 x 10⁶ m² = 5024 km²

Range of frequency suitable for space wave propagation is 100 MHz to 220 MHz.

Modulation index is defined as the ratio of amplitude of modulating signal and amplitude of carrier wave
Alternatively,
 \(\mu =\frac { { A }_{ m } }{ { A }_{ c } } \)
The amplitude of modulating signal is kept less than amplitude of carrier wave to avoid/ minimize distortion/noise.

1. Analog signals are continuous variations of voltage or current.
2. Digital signals are those which can take only discrete (stepwise) values.

Here, h₁ = 20 m, h₂ = 25 m
(i) \(d=\sqrt { 2{ h }_{ 1 }R } =\sqrt { 2\times 20\times \left( 6.4\times { 10 }^{ 6 } \right) } =16\times { 10 }^{ 3 }m=16km\)
Area covered, \(A=\pi { d }^{ 2 }=\frac { 22 }{ 7 } \times { \left( 16 \right) }^{ 2 }\simeq 804.6{ km }^{ 2 }\)
(ii) Range, \({ d }_{ 1 }=\sqrt { 2{ h }_{ 1 }R } +\sqrt { 2{ h }_{ 2 }R } =\sqrt { 2\times 20\times \left( 6.4\times { 10 }^{ 6 } \right) } +\sqrt { 2\times 25\times 6.4\times { 10 }^{ 6 } } \)
\(=16km+17.9km=33.9km\)
Area covered, \({ A }_{ 1 }=\pi { d }_{ 1 }^{ 2 }=\frac { 22 }{ 7 } \times { \left( 33.9 \right) }^{ 2 }=3611.8{ km }^{ 2 }\)
% increase in area = \(\frac { { A }_{ 1 }-A }{ A } \times 100=\left( \frac { 3611.8-804.6 }{ 804.6 } \right) \times 100=\) 348.9%