∙Find the difference in wavelength (λ1-λ2) for each of the following pairs of radio waves: (a) f

∙Find the difference in wavelength (λ1-λ2) for each of the...

$\bullet$ Find the difference in wavelength $\left(\lambda_{1}-\lambda_{2}\right)$ for each of the following pairs of radio waves: (a) $f_{1}=50 \mathrm{kHz}$ and $f_{2}=52 \mathrm{kHz}$
(b) $f_{1}=500 \mathrm{kHz}$ and $f_{2}=502 \mathrm{kHz}$

Key Concepts

Unit Consistency

Working with electromagnetic waves, especially in radio frequencies provided in kilohertz, requires careful unit conversion to maintain consistency (e.g., converting kilohertz to hertz when applying the wave equation). This ensures the accuracy of the calculations.

Wave Equation

This concept involves the relationship c = f * ?, where c is the speed of light, f is the frequency, and ? is the wavelength. It is the fundamental formula used to relate how changes in frequency affect the wavelength and is essential for calculating one when the other is given.

Inverse Proportionality of Frequency and Wavelength

Due to the constant speed of light, frequency and wavelength are inversely proportional. This means that as the frequency increases, the wavelength decreases at a rate that maintains the constant product, which is crucial for understanding how small changes in frequency lead to differences in wavelength.

Transcript

00:01 All right, here we've been told the frequencies of different wavelengths and of different electromagnetic waves, and we want to find their difference in wavelength.

00:12 So for all electromagnetic waves, the speed of the wave is going to be equal to the waves ' wavelength times its frequency.

00:22 So if we want to find the wavelength of a wave, you know that the wavelength is equal to the speed divided by the frequency.

00:31 Okay? if we want to know the difference in wavelengths, w1, wavelength 1 minus wavelength 2, we also can call this lambda 1 minus lambda 2, it's going to be equal to c over f1 minus c over f2.

00:50 The speed of electromagnetic radiation is always the same, but the frequencies and the wavelengths can be different.

00:58 Okay, this is the same thing as c times 1 over f1 minus 1 over f.

01:12 Okay, so this is 3, this is 3 times 10 to the 8th meters per second times for part a, frequency 1 is 50 kilohertz, so this is 1 over 5.

01:32 0 times 10 to the 3 hertz.

01:37 So this is in seconds, 1 over seconds is seconds, minus 1 over 52 times 10 to the 3 hertz.

01:51 Hertz is 1 over seconds, 1 over seconds, 1 over seconds.

01:55 I know units get kind of crazy sometimes.

01:58 So the answer to part a is 1 over 50 to the 3.

02:04 50 times 10 to the 3, so 1 over 5 times 10 to the 4th, minus 1 over 5 .2 times 10 to the 4th, times 3 times 10 to the 8th.

02:22 So this is going to be 230 .8 meters.

02:32 Okay, that's our answer to part a...

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