Question

please draw detailed pictures

Answer 1

Given that, frequency of broadcast = 90.3 MHz

wavelength will be given by:

wavelength = lambda = c/f

lambda = (3*10^8 m/sec)/(90.3*10^6 Hz) = 3.32 m

Now maximum signal will be produced during constructive interference, path difference is given by:

d*sin $\theta$ = m*lambda, where m = 1, 2, 3, ....

sin $\theta$ = m*lambda/d

$\theta$ = arcsin (m*lambda/d)

d = path difference = 9.0 m

when m = 1

$\theta$ = arcsin (m*lambda/d) = arcsin (1*3.32/9) = 21.65 deg

when m = 2

$\theta$ = arcsin (m*lambda/d) = arcsin (2*3.32/9) = 47.54 deg

when m = 3

$\theta$ = arcsin (m*lambda/d) = arcsin (3*3.32/9) = not possible, since sin function cannot have value greater than 1

So maximum signal will be detected at angles = 21.65 & 47.54 degree

Now minimum signal will be produced during destructive interference, path difference is given by:

d*sin $\theta$ = (m - 1/2)*lambda, where m = 1, 2, 3, ....

sin $\theta$ = (m - 1/2)*lambda/d

$\theta$ = arcsin ((m - 1/2)*lambda/d)

d = path difference = 9.0 m

when m = 1

$\theta$ = arcsin ((m - 1/2)*lambda/d) = arcsin (0.5*3.32/9) = 10.63 deg

when m = 2

$\theta$ = arcsin ((m - 1/2)*lambda/d) = arcsin (1.5*3.32/9) = 33.60 deg

when m = 3

$\theta$ = arcsin ((m - 1/2)*lambda/d) = arcsin (2.5*3.32/9) = 67.25 deg

when m = 4

$\theta$ = arcsin ((m - 1/2)*lambda/d) = arcsin (3.5*3.32/9) = not possible, since sin function cannot have value greater than 1

So minimum signal will be detected at angles = 10.63 & 33.60 & 67.25 degree

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