Question:
A radio antenna receiver works at \((10.5 \pm 4.5)\,MHz\) with an effective bandwidth of \(9\,MHz\). The corresponding wavelength bandwidth is:
A radio antenna receiver works at \((10.5 \pm 4.5)\,MHz\) with an effective bandwidth of \(9\,MHz\). The corresponding wavelength bandwidth is:
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Frequency and wavelength are inversely related: higher frequency means smaller wavelength.
Updated On: Jun 19, 2026
- 20 m
- 30 m
- 40 m
- 50 m
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The Correct Option is B
Solution and Explanation
Step 1: Find frequency range.
\[ f_{min} = 10.5 - 4.5 = 6\,MHz,\quad f_{max} = 10.5 + 4.5 = 15\,MHz \]
Step 2: Convert to wavelength range.
\[ \lambda = \frac{c}{f} \]
Step 3: Compute extreme wavelengths.
\[ \lambda_{max} = \frac{3 \times 10^8}{6 \times 10^6} = 50\,m \] \[ \lambda_{min} = \frac{3 \times 10^8}{15 \times 10^6} = 20\,m \]
Step 4: Bandwidth in wavelength.
\[ \Delta \lambda = 50 - 20 = 30\,m \]
Step 5: Final conclusion.
Thus, wavelength bandwidth is \(30\,m\).
Step 6: Final answer.
\[ \boxed{30\,m} \]
\[ f_{min} = 10.5 - 4.5 = 6\,MHz,\quad f_{max} = 10.5 + 4.5 = 15\,MHz \]
Step 2: Convert to wavelength range.
\[ \lambda = \frac{c}{f} \]
Step 3: Compute extreme wavelengths.
\[ \lambda_{max} = \frac{3 \times 10^8}{6 \times 10^6} = 50\,m \] \[ \lambda_{min} = \frac{3 \times 10^8}{15 \times 10^6} = 20\,m \]
Step 4: Bandwidth in wavelength.
\[ \Delta \lambda = 50 - 20 = 30\,m \]
Step 5: Final conclusion.
Thus, wavelength bandwidth is \(30\,m\).
Step 6: Final answer.
\[ \boxed{30\,m} \]
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