Maximum voltage, V_max = \(\frac { 100 }{ 2 } \)= 50 V
Minimum voltage, V_min = \(\frac { 20 }{ 2 } \)= 10 V
(i) Percentage modulation,
\(\mu =\frac { { V }_{ max }-{ V }_{ min } }{ { V }_{ max }{ +V }_{ min } }\times100\)
\(=\frac { 50-10 }{ 50+10 } \times100=\frac { 40 }{ 60 } \times100=66.67%\)
(ii) Peak carrier voltage,
\({ V }_{ c }=\frac { { V }_{ max }+{ V }_{ min } }{ 2 } =\frac { 50+10 }{ 2 } =30\ V\) 
(iii) Peak value of information voltage,
\({ V }_{ m }=\mu { V }_{ c }=\frac { 66.67 }{ 100 } \times 30=20 \ V\)

(i) The plot of amplitude versus can be shown in the figure below.

(ii) From figure, we note that frequency spectrum is not symmetric about \({ \omega }_{ c }\) .Crowding of spectrum is present for \(\omega <\omega _{ c }\)   
(iii) If more waves are to be modulated, then there will be more crowding in the modulating signal in the region.\(\omega <\omega _{ c }\)That will result more chances of mixing of signals
(iv) To accommodate more signals, we should increase bandwidth and frequency carries waves \({ \omega }_{ c }\). This shows that large carrier frequency enables to carry more information(i.e.more \(\omega _{ m }\))and the same will in turn increase bandwidth.

Give ,carrier wave frequency
\({ f }_{ c }=20MHz=20\times { 10 }^{ 6 }Hz\)
Bandwidth required for modulation is
\({ 2f }_{ m }=3kHz=3\times { 10 }^{ 3 }Hz\)
\(\Rightarrow \ \ { f }_{ m }=\frac { 3\times { 10 }^{ 3 } }{ 2 } =1.5\times { 10 }^{ 3 }Hz\) s
Demodulation by a diode is possible,if the condition 
Thus,\(\frac { 1 }{ { f }_{ c } } =\frac { 1 }{ 20\times { 10 }^{ 6 }Hz } =0.5\times { 10 }^{ -7 }s\) .....(i)
and \(\frac { 1 }{ { f }_{ m } } =\frac { 1 }{ 1.5\times { 10 }^{ 3 }Hz } =0.7\times { 10 }^{ -3 }s\)    ......(ii)
Now,gain through all the option of R and C one by one,we get
(i)  \(RC=1k\Omega \times 0.01\mu F\)
\(\\ ={ 10 }^{ 3 }\Omega \times ({ 0.01 }\times { 10 }^{ -6 }F)={ 10 }^{ -5 }s\)
Here, condition \(\frac { 1 }{ { f }_{ c } } <\) is satisfied  
Hence, it can be demodulated.
(ii) \((RC=1k\Omega \times 0.01\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -8 }F={ 10 }^{ -4 }s)\)
Here, condition\(\frac { 1 }{ { f }_{ c } }\) < is satisfied
 Hence, it  can be demodulated
(iii) \(RC=10k\Omega \times 1\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -6 }F={ 10 }^{ -2 }s\)
Here, condition\(\frac { 1 }{ { f }_{ c } } >RC,\) so this cannot be demodulated.

Given \(\lambda =800nm,c=3\times { 10 }^{ 8 }\quad m/s\)
Source frequency, \(v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 8\times { 10 }^{ -7 } } =3.75\times { 10 }^{ 14\quad }Hz\)
Now, total bandwidth of a channel
=1% of optical source frequency
=\(\frac { 1 }{ 100 } \times 3.75\times { 10 }^{ 14 }=3.75\times { 10 }^{ 12 }Hz\)
Number of channels
=\(\frac { Total \ bandwidth \ of \ channel }{ Bandwidth \ needed \ per \ channel }\)
(i) Number of channels for bandwidth of 8kHz
 \(=\frac { 3.75\times { 10 }^{ 12 } }{ 8\times { 10 }^{ 3 } } =4.7\times { 10 }^{ 8 }\)
(ii) Number of channels for video bandwidth of 4.5 MHz
 \(=\frac { 3.75\times { 10 }^{ 12 } }{ 4.5\times { 10 }^{ 6 } } =8.3\times { 10 }^{ 5 }\)

The distance travelled by the signal is 5km. Loss suffered in path of transmission = 2dB/km
So, total loss suffered in 5 km = \(-2\times 5=-10dB\)
Total amplifier gain = 10dB + 20dB = 30dB
Overall gain in signal = 30 - 10 = 30dB
According to the question, gain in dB = \(10{ log }_{ 10 }\frac { { P }_{ 0 } }{ { P }_{ 1 } } \)
\(\because \ 20=10{ log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } or\quad { log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } =2\)
Where\({ P }_{ i }=1.01mW\ and\ { P }_{ 0 }\) is the output power.  
\(\because \ \frac { { P }_{ 0 } }{ { P }_{ i } } ={ 10 }^{ 2 }=100\)
\(\\ \Rightarrow { P }_{ 0 }={ p }_{ i }\times 100=1.01\times 100\ or\ { p }_{ 0 }=101mW\)
Thus the output is 101 mW.