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Maximum usable frequency (MUF) IN F-region layer is $x$, when the critical frequency is \[60\] MHz and the angle of incidence is \[{70^ \circ }\], then $x$ is
A. $150\,MHz$
B. $170\,MHz$
C. $175\,MHz $
D. $190\,MHz $

Last updated date: 24th Jul 2024
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Hint:Critical angle is defined as an angle of incidence from denser to rarer medium at which the refracted ray moves along the interference of two mediums. Critical frequency, more than this frequency will lead the signals to penetrate from the ionosphere and less than this reflects in the same medium.

Complete step by step answer:
Snell’s law defines the relation between the angle of incidence and angle of refraction whenever the wave may be light or signal passes through an interference boundary of two mediums.
\[{\mu _1}\sin i = {\mu _2}\sin r\]
Given in the question: Maximum usable frequency in F-region is $x$, Critical frequency is \[60\] MHz and angle of incidence is \[{70^ \circ }\].
\[{\mu _2} = \sqrt {1 - \dfrac{{81.45 \times N}}{{{\nu ^2}}}} \]
\[\nu \] is the maximum usable frequency
\[{\text{N}}\] is the electron density

For maximum usable frequency angle of refraction becomes \[r = {90^ \circ }\] and angle of incidence is 1. Substituting in Snell’s Law:
\[1\sin i = {\mu _2}\sin {90^ \circ }\]…….(1)
The incidence angle is critical angle
\[{i_c} = {\sin ^{ - 1}}{\mu _2}\]
Squaring equation (1)
\[\sin {i_c}^2 = {\mu _2}^2\]
$\Rightarrow \sin {i_c}^2 = 1 - \dfrac{{81.45 \times N}}{{{\nu ^2}}} \\
\Rightarrow 1 - \sin {i_c}^2 = \dfrac{{81.45 \times N}}{{{\nu ^2}}} \\
\Rightarrow \nu = \sqrt {81.45 \times N} \times \sec {i_c} \\$
We know that critical frequency is \[{\nu _c} = \sqrt {81.45 \times N} \]
Maximum usable frequency can be written in the form:
\[\nu = {\nu _c} \times \sec {i_c}\]
Substituting the values in the equation:
\[x = 60 \times {10^6} \times \sec {70^ \circ }\]
\[\Rightarrow \sec {70^ \circ } \approx 2.92\]
\[\therefore x = 175.43\,MHz \approx 175\,MHz\]
Thus, the maximum usable frequency in the F-region is \[175\,MHz\].

Hence, the correct answer is option C.

Note: Maximum usable frequency, when the frequency of signal increases it has more changes to penetrate from the ionosphere and travel into outer space. It is the frequency when a radio communication starts to lose the signal.