Need help with this module 7 trial question (1 Viewer)

Redhotgrinspoon · May 23, 2022

Hi all,

Can anyone explain how they got the answer to this question.

jimmysmith560 · May 23, 2022

Would the following working help?

Given the frequency of radio wave is $f=75\:MHz\:=75\times 10^6\:Hz$ from the diagram, the path length of wave travel from P to Q to R is,
$L=PQ+QR=20+18=38m$

From the diagram, the path length of wave travel from P to R is $L'=PR=30m$

The difference between path length is $\Delta L=L-L'=38-30=8m$

Using $\lambda =\frac{\text{Speed of light (c)}}{\text{Frequency of radio wave}}$ , the wavelength of wave is $\lambda =\frac{3\times 10^8}{75\times 10^6}=4m$

The difference between path length is $\Delta L=2\times 4=2\lambda$

∴ option (B) is correct.

Need help with this module 7 trial question (1 Viewer)

Redhotgrinspoon

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