Question 1

The rms Electric field strength at a distance r from a 25 W light source is found to be 4.33 N/C.What is the distance r?

Accepted Answer

The rms electric field strength, E, at a distance r from a point source of power, P, is given by:
E = sqrt(2Pr)/(4πε₀r²)
where ε₀ is the permittivity of free space.
Substituting E = 4.33 N/C and P = 25 W, we get:
4.33 = sqrt(2*25*r)/(4πε₀r²)
Squaring and rearranging, we get:
r = (2*25)/(4πε₀*4.33²) = 6.3 m (approx)

Question 2

Nexrad radar uses a frequency of 2.7 GHz to measure the speed of cloud systems.The radio waves from the transmitter are sent outward to the cloud system and reflected back,Doppler shifted.A Nexrad system observes clouds moving toward the transmitter at 21 m/s.What is the difference between the transmitted frequency and the frequency received after reflection off the cloud surface?

Accepted Answer

The Doppler shift formula for frequency is Δf = (2 * v * f) / c, where v is the velocity of the clouds (21 m/s), f is the transmitted frequency (2.7 GHz), and c is the speed of light (approximately 3 x 10^8 m/s). Plugging in these values gives a frequency shift of approximately 380 Hz.

Question 3

For an unpolarized EM wave,the electric field direction is

Accepted Answer

In an electromagnetic wave, the electric field is always perpendicular to the magnetic field, regardless of polarization.

Question 4

At the Earth's orbital distance,the intensity of the solar radiation is approximately 1400 W/m².What is the value for the intensity at Mercury's orbital distance,which is 38.7% the distance to the sun that the Earth is.

Accepted Answer

The intensity of solar radiation is inversely proportional to the square of the distance between the sun and the planet. Therefore,

Intensity at Mercury's distance = Intensity at Earth's distance * (Distance at Earth's orbit / Distance at Mercury's orbit)^2
= 1400 * (1 / 0.387)^2
= 1400 * 6.203
= 8682 W/m^2

Therefore, the closest choice is D, 540 W/m^2.

Question 5

You wish to read from a single lightbulb in a lamp at your desk.In order to get the same brightness on the page from a 25 W bulb as you can from 100 W bulb,how much closer do you need to bring the lamp to your book? Assume that the light bulbs are equally efficient.

Accepted Answer

The brightness is proportional to the power of the bulb and inversely proportional to the square of the distance. To achieve the same brightness with a 25 W bulb as a 100 W bulb, the distance should be reduced by a factor of √(100/25) = 2, which is half the distance.

Question 6

A magnetic dipole antenna is set up due west of a radio installation that utilizes an electric dipole antenna as a transmitter.What should the orientation of the antenna be used?

Accepted Answer

The answer of A magnetic dipole antenna is set up...

Question 7

At the Earth's orbital distance,the intensity of the solar radiation is approximately 1400 W/m².What is the value for the rms Electric field strength in solar radiation?

Accepted Answer

The intensity of solar radiation (I) is related to the rms electric field (E_rms) by the equation I = (cε₀/2)E_rms², where c is the speed of light and ε₀ is the permittivity of free space. Solving for E_rms with I = 1400 W/m² gives approximately 726 V/m.

Question 8

The sun has a radius of 6.955 $\times$ 10⁸ km and a rotational period of 25.38 days.What is the difference is there in the peak of the observed spectrum between light received from the approaching side of the Sun and from the receding side of the sun? Observations from the middle of the sun place the spectral peak at 480 nm.

Accepted Answer

The difference in the peak of the observed spectrum can be calculated using the formula:

Δλ/λ = v/c

where Δλ is the difference in wavelength, λ is the wavelength of the observed spectrum (480 nm), v is the velocity difference between the approaching and receding sides of the sun, and c is the speed of light.

The velocity difference can be calculated using the formula for rotational velocity:

v = 2πr/T

where r is the radius of the sun (6.955 x 10^8 m) and T is the rotational period (25.38 days = 2.192 x 10^6 seconds).

v = 2π(6.955 x 10^8)/(2.192 x 10^6) = 22,090 m/s

Substituting the values into the first formula:

Δλ/480 = 22,090/299,792,458

Δλ = 6.4 nm

Therefore, the best choice is B, with a difference of 6.4 nm between the peak of the observed spectrum received from the approaching and receding sides of the sun.

Question 9

At the Earth's surface,approximately 1000 W/m² is present due to solar radiation.At a latitude of 44.5 degrees North,the angle of the sun's rays,when the sun is highest in the sky on the winter solstice,are incident at an angle of 68 degrees from the vertical.What power is incident upon a 1.0 m $\times$ 1.5 m window in a vertical,south-facing wall at this time?

Accepted Answer

The correct answer is A.The power incident on the window is given by:P = I * A * cos(theta)where I is the solar radiation intensity, A is the area of the window, and theta is the angle of incidence.Plugging in the given values, we get:P = 1000 W/m^2 * 1.0 m * 1.5 m * cos(68 degrees) = 1390 W

Question 10

What power is produced by a light bulb that yields 12 mW on a 15 cm $\times$ 25 cm sheet of paper that is normal to the line directly toward the bulb,if the paper is 1.7 m from the bulb?

Accepted Answer

The answer of What power is produced by a light...

Question 11

The light from a distant star is that has a peak in its intensity spectrum at 620 nm and is moving at a rate of 0.024c towards Earth.What is the wavelength of the peak of its intensity spectrum,as observed on Earth?

Accepted Answer

The correct answer is A.The wavelength of the peak of the intensity spectrum of the star as observed on Earth is given by:$$\lambda_{obs} = \lambda_{rest} / (1 - v/c)$$where:$$\lambda_{obs}$$ is the observed wavelength$$\lambda_{rest}$$ is the rest wavelengthv is the velocity of the starc is the speed of lightSubstituting the given values into the equation, we get:$$\lambda_{obs} = 620 nm / (1 - 0.024) = 605 nm$$

Question 12

It is possible to detect Doppler shifts due to light on each side of a rapidly spinning star.Suppose it is seen that a nearby star,relatively stationary to us,is blueshifted in light on one side of the star and redshifted in the light from the other side.Suppose a star of radius 2.5 $\times$ 10⁸ km has light shifted from 525 nm to 527 nm on one side and to 523 nm on the other.What is the rotational period of the star?

Accepted Answer

The answer of It is possible to detect Doppler shifts...

Question 13

If the distance from a point source of EM radiation is doubled,the intensity is

Accepted Answer

According to the inverse square law, the intensity of EM radiation decreases as the square of the distance from the source increases. This means that if the distance is doubled, the intensity will decrease by a factor of 2^2 or 4. Therefore, the correct answer is C, reduced by a factor of four.

Quiz 22: Electromagnetic Waves