The wavelength of these signals are :
AM Radio Signals → 188 m < λ < 545 m
FM Radio Signals → 2.78 m < λ < 3.41 m
[tex]\texttt{ }[/tex]
Let's recall the speed of wave formula as follows:
[tex]\boxed {v = \lambda \times f}[/tex]
where:
v = speed of wave ( m/s )
λ = wavelength ( m )
f = frequency of wave ( Hz )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
maximum frequency of AM radio signals = f_max = 1600 kHz
minimum frequency of AM radio signals = f_min = 550 kHz
maximum frequency of FM radio signals = f_max = 108 MHz
minimum frequency of FM radio signals = f_min = 88.0 MHz
speed of wave = v = 3.00 × 10⁸ m/s
Asked:
wavelength = λ = ?
Solution:
AM Radio Signals
[tex]550 \texttt{ kHz} < f < 1600 \texttt{ kHz}[/tex]
[tex]550 \texttt{ kHz} < \frac{v}{\lambda} < 1600 \texttt{ kHz}[/tex]
[tex]550 \texttt{ kHz} < \frac{3.00 \times 10^8}{\lambda} < 1600 \texttt{ kHz}[/tex]
[tex]\frac{3.00 \times 10^8}{1600 \times 10^3} \texttt{ m} < \lambda < \frac{3.00 \times 10^8}{550 \times 10^3} \texttt{ m}[/tex]
[tex]\boxed {188 \texttt{ m} < \lambda < 545 \texttt{ m}}[/tex]
[tex]\texttt{ }[/tex]
FM Radio Signals
[tex]88.0 \texttt{ MHz} < f < 108 \texttt{ MHz}[/tex]
[tex]88.0 \texttt{ MHz} < \frac{v}{\lambda} < 108 \texttt{ MHz}[/tex]
[tex]88.0 \texttt{ MHz} < \frac{3.00 \times 10^8}{\lambda} < 108 \texttt{ MHz}[/tex]
[tex]\frac{3.00 \times 10^8}{108 \times 10^6} \texttt{ m} < \lambda < \frac{3.00 \times 10^8}{88.0 \times 10^6} \texttt{ m}[/tex]
[tex]\boxed {2.78 \texttt{ m} < \lambda < 3.41 \texttt{ m}}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Light
