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Public Radio station KXPR-FM in Sacramento broadcasts at 88.9 MHz. The radio waves pass between two tall skyscrapers that are 15.0 m apart along their closest walls.

(a) At what horizontal angles, relative to the original direction of the waves, will a distant antenna not receive any signal from this station?

(b) If the maximum intensity is 3.50 W>m2 at the antenna, what is the intensity at _5.00_ from the center of the central maximum at the distant antenna?

Short Answer

Expert verified

a) ±13.0,±27.7,±42.4,±64.1

b)2.08W/m22.08W/m2

Step by step solution

01

 Given.

f=88.9MHz=88.9×106Hz,a=15.0m,c=3.0×108m/s,I0=3.50W/m2,θ=±5.00

02

Ray diagram.

This is a view from above which means that the wave lines are horizontal, as the author mentioned.

03

Concept.

This problem is similar to the one-slit kind of problem whereas the distance between the two skyscrapers is the width of the whole a.

First of all, we need to find the wavelength which is given by

v=λf

In this case, v = c so

c=λf

Hence,

λ=cf (1)

We know that the angle of the minimum fringe in the one-slit experiments is given by

sinθm=mλα

Whereasm=±1,±2,±3,...

Substitute from (1)

sinθm=mλaf

Solving forθm

θm=sin-1mλaf

Now we can easily find the horizontal angles in which horizontal angles will a distant antenna will receive no signal from this station by substituting the values of m into the last equation. Thus, we will substitute the given and m's values.

04

Calculation.

When m=1,

θ1=sin-11×3.0×10815.0×88.9×106θ1=±13.0

When m=2,

θ2=sin-12×3.0×10815.0×88.9×106θ2=±27.7

When m=3,

θ3=sin-13×3.0×10815.0×88.9×106θ3=±42.4

When m=4,

θ4=sin-14×3.0×10815.0×88.9×106θ4=±64.1

When m=5,

θ5=sin-15×3.0×10815.0×88.9×106

Since the term of 2×3.0×10815.0×88.9×106is greater than one, so no angle available 2x3.0 x 105 here and this means that the last angle is for m=4.

05

 Step 5: Calculating intensity.

We know that the intensity is given by

I=I0sinβ2β22

We also know that

β=2πasinθλ

Substitute from (1)

β=2πafsinθc

Substitute into the intensity formula above, noting that 2 cancels

I=I0sinπafsinθc2πafsinθc2

Substitute the given

I=3.50×sinπ×15.0×88.9×106×sin5.003.0×108π×15.0×88.9×106×sin5.003.0×1082

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