When radio waves try to pass through a city, they encounter thinvertical slits: the separations between the buildings. This causesthe radio waves to diffract. In this problem, you will see howdifferent wavelengths refract as they pass through a city andrelate this to reception for radios and cell phones. You will usethe angle from the center of the central intensity maximum to thefirst intensity minimum as a measure of the width of the centralmaximum (where nearly all of the diffracted energy isfound).
**Consider radio waves of wavelength lambda entering a city where thebuildings have an average separation of a.**
A.) Find the angle theta to the first minimum from thecenter of the central maximum (Express your answer in terms oflambda and a.):
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**Assume that the average spacing between buildings is a=20\;\rm m.**
B.) What is the angle theta_FM to the first minimum foran FM radio station with a frequency of 101 \rm MHz? (Express your answer numerically indegrees to three significant figures. Note: Do not write youranswer in terms of trignometric functions. Evaluate any suchfunctions in your working.)
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C.) What is the angle theta_cell for a cellular phonethat uses radiowaves with a frequency of 900 \rm MHz? (Express your answer indegrees to three significant figures.)
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D.) What problem do you encounter in tryingto find the angle theta_AM for an AM radio stationwith frequency 1000 \rm kHz?
i.) The angle becomes zero.
ii.) The angle can be given only in radians.
iii.) To find the angle it would be necessary to takethe arcsine of a negative number.
iv.) To find the angle it would be necessary totake the arcsine of a number greater than one.

Respuesta :

Answer:

b)  θ = 8.54°, c)  θ = 0.955°, d)The sine function cannot be greater than 1.  The correct answer is iv)

Explanation:

Diffraction phenomena are described by the equation, for constructive interference

     a sin θ = m λ                 m = ±1, ±2, ±3, ...

Where a is the width of the slit (20 m), λ is the wavelength and m integer

Let's look for the wavelength, for these we use the relationship

     c = λ f

     λ = c / f

     λ = 3  10⁸ / 101 10⁶

     λ = 2.97 m

a) Let's look for the angle for the first minimum found in m = 1

     θ = sin⁻¹ (m lam / a)

b) Let's calculate

     θ = sin⁻¹ (1 2.97 / 20)

     θ = sin⁻¹ (0.1485)

     θ = 8.54°

c)

    λ = 3 10⁸/900 10⁸

    λ = 0.333 m

    θ = sin⁻¹ (0.333 / 20)

    θ = sin⁻¹ (0.166666)

    θ = 0.955°  

d) repeat the calculations for f = 1000 kHz = 1 MHz

       λ = 3 10⁸/1 10⁶

       λ = 300 m

       θ = sin⁻¹ (300/20)

        θ = sin⁻¹ (15)

The sine function cannot be greater than 1.

The correct answer is iv)

Answer:

HE IS RIGHT !!!!!!!!

Explanation: