JEE Physics · Hard

Principles of Communication: Propagation MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

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PropagationHardQuestion 20294

Question

The Friis transmission equation gives received power P_r = P_t G_t G_r (λ/4πd)². For a satellite link at 6 GHz (λ = 5 cm), distance d = 36,000 km, with G_t = G_r = 10⁴ (40 dBi), and P_t = 100 W, the received power is:
  1. A
    ≈ 4.8 × 10⁻¹⁰ W
    Correct
  2. B
    ≈ 4.8 × 10⁻⁶ W
  3. C
    ≈ 4.8 × 10⁻¹² W
  4. D
    ≈ 48 W

Correct answer

≈ 4.8 × 10⁻¹⁰ W

Explanation

Pr = 100 × 10⁴ × 10⁴ × (0.05/(4π × 3.6×10⁷))² = 10¹⁰ × (0.05/4.52×10⁸)² = 10¹⁰ × (1.107×10⁻¹⁰)² = 10¹⁰ × 1.22×10⁻²⁰ ≈ 1.22×10⁻¹⁰ W ≈ 0.12 nW. Even with high-gain antennas and 100 W transmitter, received power is sub-nanowatt — explaining why satellite dishes need LNBs.