Question

1. The U.S. Navy has long proposed the construction of extremely low-frequency (ELF waves) communications systems; such waves could penetrate the oceans to reach distant submarines.

Calculate the length of a quarter-wavelength antenna for a transmitter generating ELF waves of frequency 91 Hz.
km =

2. What are the wavelength ranges for the following?

(a) the AM radio band (540-1600 kHz)
m (maximum wavelength)
m (minimum wavelength)

(b) the the FM radio band (88-108 MHz)
m (maximum wavelength)
m (minimum wavelength)

3. The angle of incidence of a light beam in air onto a reflecting surface is continuously variable. The reflected ray is found to be completely polarized when the angle of incidence is 57.0

Answer 1

1)

?f = c = 2.998*10^8m/s (speed of light)
? = Wavelength in meters
f = frequency in Hz

? = c/f = 2.998*10^8/91 = 3.96e+6 meters
1/4 of that is 824175.82 meters or 824.176 km

2)

a)(540 kHZ) = 540 x 10 ^ 3 Hz, we can call that f1


the formula is, Speed = frequency * wavelength, v = f?
v = speed of light, radio wave. = 3x10^8 m/s
x= multiply
so 540 x 10 ^ 3 = 3x10^ 8 / ?

rearanging gives
lambda = 3x10^8 / 540x10^3 = 555.556 m

now for
f2 = 1600x10^3
same concept
lambda = 3x10^8 / 1600x10^3 = 187.5 m

min = 187.5m max is 555.556m

b). just do the same thing . the frequency is 88 Mhz, which is 88x10^6 Hz, so
v/f = ?
3x10^8 / 88x10^ 6 = 3.409 m

and the other f = 108x10^6 Hz
3x10^8 / 108x10^6 Hz = 2.778m

therefore
minimum is 2.778m and max is 3.409m

3)a)
From Brewster's law,

complete polarisation occurs when tan? = n2/n1
n1 = ref index of incident medium (air, n1 = 1.0),

? = angle in incident medium

Ref index or reflecting medium .. n2 = tan57 x 1.0

n2 = 1.5398 (1.54)

b)
By Snell's law,

n2 = sin i(air) / sin r = 1.5398

sin r = sin i / n2

sinr = sin57 / 1.5398 = 0.5446.

r = 33.0