Question

When radio waves try to pass through a city, they encounter thin vertical slits: the separations between the buildings. This causes the radio waves to diffract. In this problem, you will see how different wavelengths diffract as they pass through a city and relate this to reception for radios and cell phones. You will use the angle from the center of the central intensity maximum to the first intensity minimum as a measure of the width of the central maximum (where nearly all of the diffracted energy is found).a. Find the angle θ to the first minimum from thecenter of the central maximum (Express your answer in terms λ and a.):b. What is the angle θFM to the first minimum foran FM radio station with a frequency of 101mMHz? (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.)c. What is the angle θcell for a cellular phonethat uses radiowaves with a frequency of 900MHz? (Express your answer indegrees to three significant figures.)d. What problem do you encounter in tryingto find the angle θAM for an AM radio stationwith frequency 1000kHz?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.

Answer 1

Answer:

a)    θ = λ / a , b)  θ = 0.294 rad , c) θ = 0.0333 rad , d) correct answer is IV

Explanation:

a) The equation that describes the diffraction process is

         a sin θ = m λ

Where a is the slit size, λ is the wavelength and m is an integer

In our case the firtz minimum occurs for m = 1

           a sin θ = λ

In general in the direction problems the angles are very small, we can approximate the sine function

           sin θ = θ

           a θ = λ

           θ = λ / a

To give a number we must have a specific separation of the buildings, suppose the separation given by the width of a street 10 m with value for a (a = 10m)

b) the speed of light is related to wavelength and frequency

        c = λ f

         λ = c / f

         λ = 3 10⁸/102 10⁶

        λ = 2.94 m

       θ = 2.94 / a

The angle is in radians, to observe the phenomenon depends the separation of the building, in general the maximum value of a = 2.94 m

The angle for a = 10m

       θ = 0.294 rad

c) let's look for the wavelength

       λ = 3 10⁸/900 10⁶

       λ = 0.333 m

       θ = 0.333 / a

For a = 10m

       θ = 0.0333 rad

d) f = 1000 kHz = 1 10⁶ MHz this case

       λ = 3 10⁸⁸/1 10⁶

       λ = 3 10² m

       θ = 300 / a

For a = 10 m

       θ = 300/10

       θ = 30

For this number, the originate sine function would give a value greater than 1, so it is not defined

The correct answer is IV