JEE Main · 2019 · 09 Jan 2019 · Shift 2 · EM Waves

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In a communication system operating at wavelength 800 nm , only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light

c = 3 \times 1 0^{8} m / s

h = 6.6 \times 1 0^{- 34} J - s

)

(A) $3.75 \times 1 0^{6}$
(B) $3.86 \times 1 0^{6}$
(C) $6.25 \times 1 0^{5}$ ✓ Correct
(D) $4.87 \times 1 0^{5}$

Answer: (C)

Solution

First find how much bandwidth the optical carrier offers. Its frequency is

f = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{800 \times 1 0 ^{- 9}} = \frac{3}{8} \times 1 0^{15} Hz

. Only one percent is usable, so

Δ f = 0.01 \times \frac{3}{8} \times 1 0^{15} = \frac{3}{800} \times 1 0^{15} Hz

. Each TV channel needs

6 MHz = 6 \times 1 0^{6} Hz

, so the number of channels is

n = \frac{Δ f}{6 \times 1 0 ^{6}} = \frac{3 \times 1 0 ^{15}}{800 \times 6 \times 1 0 ^{6}} = 6.25 \times 1 0^{5}

. The Planck constant quoted in the data is a red herring here.

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · EM Waves

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