JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Semiconductors

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 :

(A) $375 \times 1 0^{7}$
(B) $75 \times 1 0^{7}$ ✓ Correct
(C) $375 \times 1 0^{8}$
(D) $75 \times 1 0^{9}$

Answer: (B)

Solution

Frequency of optical source at 1000 nm:

f = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{1000 \times 1 0 ^{- 9}} = 3 \times 1 0^{14}

Hz

Available channel bandwidth (2% of source frequency):

B = 0.02 \times 3 \times 1 0^{14} = 6 \times 1 0^{12}

Hz

Bandwidth per audio channel = 8 kHz =

8 \times 1 0^{3}

Hz

Number of channels:

N = \frac{6 \times 1 0 ^{12}}{8 \times 1 0 ^{3}} = \frac{6}{8} \times 1 0^{9} = 0.75 \times 1 0^{9} = 75 \times 1 0^{7}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Semiconductors

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