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Astronomers observe a 60.0-MHz radio source both directly and by reflection from the sea as shown. If the receiving dish is 20.0 m above sea level, what is the angle of the radio source above the horizon at first maximum?

Direct
Radio
path
telescope
Reflected
path
Figure P36.45
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Transcribed Image Text:Direct Radio path telescope Reflected path Figure P36.45
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Step 1

Given:

The diagrammatic representation of the sea is shown as,

Advanced Physics homework question answer, step 1, image 1

Step 2

The wave's wave length is calculated as,

λ=cf

Substitute the given value of f=60 MHz is given as

λ=3×108 ms60×106 Hzλ= 5m

The distance is defined as,

RT=d, TA=AR=d2

Then,

TB=BR=d2cosθ

(or)

TB+BR=dcosθ(1)

Step 3

The path difference is shown as,

δ=TB+BR-TR

Substitute as,

δ=ddcosθ-d(or)δ=dsecθ-1(2)

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