JEE Main 2022 · 29 Jun · Shift 1 · Physics Q50

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PhysicsSingle correct+4 / -1Question 50 of 90

Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission :

375\times10^7

75\times10^7

375\times10^8

75\times10^9

Answer and worked solution

Official answer: B

Bandwidth. $f=\frac{3\times10^8}{10^{-6}}=3\times10^{14}$ Hz; 2% is $6\times10^{12}$ Hz.

Channels. $\frac{6\times10^{12}}{8\times10^3}=7.5\times10^8=75\times10^7$ .

(B)

Topics:Elements of communication systems

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JEE Main 2022 · 29 Jun · Shift 1, Physics question 50