For a sky wave propagation, the critical frequency is 10 MHz. What is the maximum usable frequency for a take-off angle of 30 degrees?

A. 11.54 MHz — The MUF is given by MUF = fc / cos(theta). For theta = 30 degrees, cos(30) = sqrt(3)/2 = 0.866. Thus, MUF = 10 / 0.866 = 11.54 MHz.

JEE Main · PhysicsHard

$For a sky wave propagation, the critical frequency is f_{c} = 10 MHz . What is the maximum usable frequency (MUF) for a take-off angle of θ = 3 0^{\circ} ?$

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Correct answer: A — $11.54 MHz$

MUF = f_{c} / cos (θ) . Given f_{c} = 10 MHz and θ = 3 0^{\circ}, MUF = 10/ cos (3 0^{\circ}) = 10/0.866 = 11.54 MHz .