
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 $
Answer
522k+ views
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.
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.
Recently Updated Pages
The total number of structural isomers possible for class 12 chemistry CBSE

What is a parallel plate capacitor Deduce the expression class 12 physics CBSE

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Maltose on hydrolysis gives A Mannose +glucose B Galactose+glucose class 12 chemistry CBSE

