Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Long-range radio transmission is possible when the radio waves are reflected from the ionosphere. For this to happen the frequency of the radio waves must be in the range:
Long-range radio transmission via ionospheric reflection depends on the radio waves having a frequency that allows them to be reflected by the ionosphere rather than penetrating it or being absorbed. The ionosphere's ability to reflect waves is determined by the maximum electron density, $$ N_{\text{max}} $$, in electrons per cubic meter (m⁻³). The critical frequency $$ f_c $$, which is the maximum frequency that can be reflected, is given by the formula:
$$ f_c = 9 \sqrt{N_{\text{max}}} $$The electron density $$ N_{\text{max}} $$ varies with atmospheric conditions, time of day, and solar activity. Typically, $$ N_{\text{max}} $$ ranges from about $$ 10^{10} $$ to $$ 10^{12} $$ m⁻³. Let's calculate the critical frequencies for these extremes.
For $$ N_{\text{max}} = 10^{10} $$ m⁻³:
$$ f_c = 9 \times \sqrt{10^{10}} = 9 \times 10^{5} \text{ Hz} = 900 \text{ kHz} $$For $$ N_{\text{max}} = 10^{12} $$ m⁻³:
$$ f_c = 9 \times \sqrt{10^{12}} = 9 \times 10^{6} \text{ Hz} = 9 \text{ MHz} $$However, during periods of high solar activity, $$ N_{\text{max}} $$ can reach higher values, such as $$ 8 \times 10^{12} $$ m⁻³. For this density:
$$ f_c = 9 \times \sqrt{8 \times 10^{12}} = 9 \times \sqrt{8} \times 10^{6} = 9 \times 2\sqrt{2} \times 10^{6} $$Calculating $$ \sqrt{8} = 2\sqrt{2} \approx 2 \times 1.414 = 2.828 $$:
$$ f_c \approx 9 \times 2.828 \times 10^{6} = 25.452 \times 10^{6} \text{ Hz} \approx 25 \text{ MHz} $$Thus, the critical frequency can extend up to approximately 25 MHz under extreme conditions. For reflection to occur, the frequency must be less than or equal to $$ f_c $$. However, absorption in the lower ionospheric layers (like the D layer during the day) affects lower frequencies. Frequencies below about 5 MHz experience significant daytime absorption, limiting their use for long-range transmission. Therefore, the optimal range avoids severe absorption while staying below the maximum critical frequency.
Now, evaluating the options:
Hence, the correct answer is Option D.
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.