# Astronomers observe a 60.0-MHz radio source both directly and by reflection from the sea as shown in Figure P36.45. 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? Figure P36.45

### Physics for Scientists and Enginee...

9th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305116399

Chapter
Section

### Physics for Scientists and Enginee...

9th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305116399
Chapter 37, Problem 37.69AP
Textbook Problem
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## Astronomers observe a 60.0-MHz radio source both directly and by reflection from the sea as shown in Figure P36.45. 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?Figure P36.45

To determine
The angle of the radio source above the horizon at first maximum.

### Explanation of Solution

Given info: The frequency observed by the astronomer is 60.0MHz and the distance of the receiving dish from the sea level is 20.0m .

The astronomer’s situation of the given Figure is shown below.

Figure (1)

One radio wave reaches the receiver R directly from the distance source at angle θ above the horizontal. The other wave undergoes phase reversal as it reflects from the water at P . The distance from P to R' , where R' is the mirror image of the telescope. So, the path difference is d .

From Figure (1), for the value of d in right triangle RQR' ,

sinθ=dRR'd=RR'sinθ=2(RS)sinθ

Here,

θ is the angle of the radio source above the horizon at first maximum.

d is the path difference.

RS is the distance of the receiving dish from the sea level.

Substitute 20.0m for RS to get the d .

d=2(20.0m)sinθ=(40.0m)sinθ

Formula to calculate the wavelength is,

λ=cf (1)

Here,

λ is the wavelength of the light.

c(3.8×108m/s) is the speed of the light in the vacuum

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