AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, wavelength, λ=800‌nm
speed of light, c=3×108m∕s
Then, frequency, f=

c

λ

Substituting the values, we get
Source frequency, f=

3×104

800×10−9

‌Hz
=375×1014‌Hz
It is said in the question that only 1% of source frequency is available as signal bandwidth.
∴ Total available signal bandwidth
=1% of 375×1014
=

1

100

×3.75×1014‌Hz=3.75×1012‌Hz
Given that, the frequency bandwidth of one signal
=6‌MHz=6×106‌Hz
Hence, the number of channels
=

Total‌available‌signal‌bandwidth

Bandwidth‌of‌one‌signal

=

3.75×1012

6×106

=

1

16

×107