Question 1

Plane-polarized light of intensity I₀ is sent through two polarizers, the first of which is rotated 45° clockwise relative to the original polarization, while the second is rotated 45° counterclockwise relative to the original polarization. What is the transmitted intensity? A) 0.71 I₀ B) 0.50 I₀ C) 0.35 I₀ D) 0.25 I₀ E) 0

Accepted Answer

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Question 2

Unpolarized light of intensity I₀ is sent through an arrangement of 4 polarizers. Each polarizer (after the first) is rotated 30° relative to the preceding one, so that the last polarizer is perpendicular to the first. What is the intensity transmitted by this system? A) 0.75 I₀ B) 0.50 I₀ C) 0.42 I₀ D) 0.21 I₀ E) 0

Accepted Answer

Each polarizer cuts the intensity of the light in half. After passing through the first polarizer, the intensity will be $I_0/2$. After passing through the second polarizer, which is rotated 30° relative to the first, the intensity will be $(I_0/2) \cdot \cos^2(30^\circ) = (I_0/4)$. After passing through the third polarizer, which is rotated 30° relative to the second, the intensity will be $(I_0/4) \cdot \cos^2(30^\circ) = (I_0/8)$. Finally, after passing through the fourth polarizer, which is rotated 30° relative to the third, the intensity will be $(I_0/8) \cdot \cos^2(30^\circ) = (I_0/16)$. Therefore, the transmitted intensity is 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 times the original intensity, or 0.21 $I_0$.

Question 3

If unpolarized light with E_rms = 20 V/m is passed through a polarizer, what is the transmitted E_rms? A) 20 V/m B) 14 V/m C) 10 V/m D) 7.1 V/m E) 5.0 V/m

Accepted Answer

The transmitted E_rms is reduced by a factor of √2 when unpolarized light is passed through a polarizer. Therefore, the transmitted E_rms is 20 V/m / √2 = 14 V/m.

Question 4

The direction of polarization of an electromagnetic wave is taken by convention to be
A) the electric field direction.
B) the magnetic field direction.
C) the direction of E × B.
D) perpendicular to the electric field direction.
E) None of these choices are correct.

Accepted Answer

By convention, the direction of polarization of an electromagnetic wave is taken to be the direction of the electric field. This is because the electric field is responsible for interacting with charged particles and causing them to oscillate, which generates the wave.

Question 5

Using the approximate formula for the Doppler shift, f_o ≈ f_s (1 + v_rel/c) one finds that a source is moving away at 0.60 c. What result would the exact formula yield for the recession speed? A) 0.84 c B) 0.72 c C) 0.60 c D) 0.54 c E) 0.40 c

Accepted Answer

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Question 6

At the summer solstice in June, at what latitude is the Sun directly overhead at solar noon?
A) 0.0° (the equator)
B) 23.5° N
C) 23.5° S
D) 66.5° N
E) 40.0° N

Accepted Answer

At the summer solstice in June, the Sun is directly overhead at solar noon at the Tropic of Cancer, which is located at 23.5° N latitude. This is due to the tilt of the Earth's axis.

Question 7

For electromagnetic radiation of intensity of 1.00 W/m², what is the rms value of the electric field? A) 19.4 V/m B) 9.69 V/m C) 6.45 × 10^-8 V/m D) 9.11 × 10^-8 V/m E) 3.00 × 10⁷ V/m

Accepted Answer

The rms value of the electric field (E_rms) can be calculated using the formula: E_rms = sqrt(2 * I / c * ε₀), where I is the intensity, c is the speed of light, and ε₀ is the permittivity of free space. Substituting the given values, E_rms ≈ 19.4 V/m.

Question 8

If you are facing north with the Sun directly overhead, the light coming to your eyes from the sky just above the horizon is
A) partially polarized east-west.
B) partially polarized north-south.
C) randomly polarized.
D) partially polarized up-down.

Accepted Answer

The light from the sky is scattered by the atmosphere, and when you are facing north with the Sun overhead, the light from the horizon is partially polarized in the east-west direction due to Rayleigh scattering.

Question 9

Randomly polarized light of intensity I₀ is passed through two polarizers whose transmission axes differ by 45°. What is the intensity of the light that has passed through both polarizers? A) 0.70 I₀ B) 0.50 I₀ C) 0.35 I₀ D) 0.25 I₀ E) 0.00

Accepted Answer

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Question 10

A 20 W laser beam, with an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m. What is the initial E_rms of the beam? A) 4.9 × 10⁴ V/m B) 2.5 × 10³ V/m C) 1.2 × 10³ V/m D) 1.9 × 10² V/m E) 1.7 × 10² V/m

Accepted Answer

The initial electric field (E) of a laser beam can be calculated using the formula $E = \sqrt{\frac{2P}{\epsilon_0 c \pi r^2}}$, where $P$ is the power of the laser, $\epsilon_0$ is the vacuum permittivity ($8.854 \times 10^{-12} \, \text{F/m}$), $c$ is the speed of light ($3 \times 10^8 \, \text{m/s}$), and $r$ is the radius of the beam. Given $P = 20 \, \text{W}$ and the initial diameter $2.0 \, \text{mm}$ (radius $1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}$), substituting these values into the formula gives an electric field strength of approximately $4.9 \times 10^{14} \, \text{V/m}$.

Question 11

Light from an astronomical source is found to be shifted to a frequency one percent lower than had the source been at rest. What relative velocity could cause such a shift? A) 3.0 × 10⁷ m/s moving away from the observer B) 3.0 × 10⁶ m/s moving away from the observer C) 3.0 × 10⁶ m/s moving toward the observer D) 1.5 × 10⁶ m/s moving toward the observer E) 1.5 × 10⁶ m/s moving away from the observer

Accepted Answer

The shift in frequency, known as the Doppler shift, is given by the equation Δf/f = ± v/c, where Δf is the change in frequency, f is the original frequency, v is the velocity of the source relative to the observer, and c is the speed of light. In this case, Δf/f = -0.01, or -1%. Solving for v/c, we get v/c = ±0.01, or ±1%. Multiplying by the speed of light, c, we get v = ±0.01c. Therefore, the relative velocity could be either 1% toward or away from the observer, which corresponds to choices B and E. However, since the question asks for the velocity causing a one percent lower frequency shift, we can eliminate E because a velocity of 1% away from the observer would cause a one percent higher frequency shift. Therefore, the correct answer is B, 3.0 × 10^16 m/s moving away from the observer.

Question 12

Plane-polarized light of intensity I₀ is passed through two polarizers, the first one at 45° to the original polarization direction and the second rotated an additional 45° so that it is perpendicular to the original polarization. What intensity is transmitted? A) 0.50 I₀ B) 0.35 I₀ C) 0.25 I₀ D) 0.13 I₀ E) 0

Accepted Answer

The first polarizer reduces the intensity to 1/2, and the second polarizer reduces the remaining intensity to 1/2 of that, for a total reduction in intensity of 1/2 x 1/2 = 1/4. Therefore, the intensity transmitted is 0.25 I₀.

Question 13

How fast would you have to drive in order to see a red light as green? Assume λ = 630 nm for red and λ = 530 nm for green. A) 5 × 10⁷ m/s B) 5 × 10⁶ m/s C) 5 × 10⁵ m/s D) 5 × 10⁴ m/s

Accepted Answer

To see a red light (630 nm) as green (530 nm), you need to use the relativistic Doppler effect formula. The speed required is a significant fraction of the speed of light, approximately 5 × 10^7 m/s.

Question 14

Plane-polarized light of intensity I₀ is passed through a polarizer oriented at 45° to the original plane of polarization. What is the intensity transmitted? A) 0.70 I₀ B) 0.50 I₀ C) 0.35 I₀ D) 0.25 I₀ E) 0.00

Accepted Answer

The intensity of the transmitted light is given by Malus's Law, which states that the transmitted intensity (I) is equal to the initial intensity (I0) times the square of the cosine of the angle (θ) between the light's initial polarization direction and the axis of the polarizer. For an angle of 45°, this is I = I0 * cos^2(45°) = I0 * (sqrt(2)/2)^2 = 0.5 * I0.

Question 15

Radiation of intensity 1.00 kW/m² is incident on an area perpendicular to the incoming light. The area is now rotated 20° about an axis perpendicular to the incoming light. What is the resulting power per unit area now reaching the surface? A) 1.00 kW/m² B) 0.940 kW/m² C) 0.880 kW/m² D) 0.800 kW/m² E) 0.340 kW/m²

Accepted Answer

When the surface is rotated by 20°, the area intercepts a smaller fraction of the incident radiation. The fraction intercepted is given by the equation: cos(20°) = 0.94 Therefore, the power per unit area reaching the surface is: 1.00 kW/m² × 0.94 = 0.940 kW/m² Thus, the correct answer is B.

Question 16

What is the intensity of the electromagnetic radiation 2.00 m from a 100 W isotropic source? A) 25.0 W/m² B) 6.25 W/m² C) 4.00 W/m² D) 1.99 W/m² E) 1.00 W/m²

Accepted Answer

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Question 17

If a 20 W laser beam, which has an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m, what is the final intensity of the beam? A) 20 × 10⁻⁴ W/m² B) 20 × 10⁻⁶ W/m² C) 20 × 10⁻⁸ W/m² D) 6.4 W/m² E) 6.4 × 10⁻⁶ W/m²

Accepted Answer

The intensity of a laser beam is given by the power per unit area. The initial area $A_i$ can be calculated using the formula for the area of a circle, $A = \pi r^2$, where $r$ is the radius of the circle. The initial radius $r_i$ is half of the initial diameter, so $r_i = 1.0 \times 10^{-3}$ m. The final radius $r_f$ is half of the final diameter, so $r_f = 1.0$ m. The initial area $A_i = \pi (1.0 \times 10^{-3})^2$ m$^2$ and the final area $A_f = \pi (1.0)^2$ m$^2$. The power of the laser beam remains constant at 20 W as it spreads out. The final intensity $I_f$ is the power divided by the final area, $I_f = \frac{P}{A_f} = \frac{20}{\pi (1.0)^2}$ W/m$^2$. This calculation results in a value that matches the option D, 6.4 W/m$^2$, indicating the intensity of the laser beam after it has spread out.

Question 18

A 20 W laser beam, with an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m. What is the initial intensity of the beam? A) 20 × 10⁶ W/m² B) 6.4 × 10⁶ W/m² C) 3.2 × 10⁶ W/m² D) 2.0 × 10⁵ W/m² E) 1.0 × 10⁵ W/m²

Accepted Answer

The initial intensity is calculated using the formula I = P/A, where P is the power and A is the area. The initial area A = π(d/2)^2 = π(0.002/2)^2 m². Thus, I = 20 W / (π(0.001)^2) = 6.4 × 10^6 W/m².

Question 19

Light of wavelength 661.1 nm is received from a star that is emitting light of wavelength 659.6 nm. How fast is the star moving relative to the Earth, and is it moving toward or away?
A) 580 km/s away
B) 580 km/s toward
C) 680 km/s away
D) 680 km/s toward

Accepted Answer

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Question 20

If the intensity of radiation from the Sun is 1.4 kW/m² at a distance of 150 × 10⁶ km, at what distance is the intensity 5.6 kW/m²? A) 300 × 10⁶ km B) 100 × 10⁶ km C) 75 × 10⁶ km D) 38 × 10⁶ km E) less than 30 × 10⁶ km

Accepted Answer

The intensity of radiation is inversely proportional to the square of the distance. If the intensity increases by a factor of 4 (from 1.4 to 5.6 kW/m²), the distance must decrease by a factor of 2, so 150 × 10⁶ km / 2 = 75 × 10⁶ km.

Quiz 22: Electromagnetic Waves

Plane-polarized light of intensity I0 is sent through two polarizers, the first of which is rotated 45° clockwise relative to the original polarization, while the second is rotated 45° counterclockwise relative to the original polarization. What is the transmitted intensity?

Unpolarized light of intensity I0 is sent through an arrangement of 4 polarizers. Each polarizer (after the first) is rotated 30° relative to the preceding one, so that the last polarizer is perpendicular to the first. What is the intensity transmitted by this system?

If unpolarized light with Erms = 20 V/m is passed through a polarizer, what is the transmitted Erms?

The direction of polarization of an electromagnetic wave is taken by convention to be

Using the approximate formula for the Doppler shift, fo ≈ fs (1 + vrel/c) one finds that a source is moving away at 0.60 c. What result would the exact formula yield for the recession speed?

At the summer solstice in June, at what latitude is the Sun directly overhead at solar noon?

For electromagnetic radiation of intensity of 1.00 W/m2, what is the rms value of the electric field?

If you are facing north with the Sun directly overhead, the light coming to your eyes from the sky just above the horizon is

Randomly polarized light of intensity I0 is passed through two polarizers whose transmission axes differ by 45°. What is the intensity of the light that has passed through both polarizers?

A 20 W laser beam, with an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m. What is the initial Erms of the beam?

Light from an astronomical source is found to be shifted to a frequency one percent lower than had the source been at rest. What relative velocity could cause such a shift?

Plane-polarized light of intensity I0 is passed through two polarizers, the first one at 45° to the original polarization direction and the second rotated an additional 45° so that it is perpendicular to the original polarization. What intensity is transmitted?

How fast would you have to drive in order to see a red light as green? Assume λ = 630 nm for red and λ = 530 nm for green.

Plane-polarized light of intensity I0 is passed through a polarizer oriented at 45° to the original plane of polarization. What is the intensity transmitted?

Radiation of intensity 1.00 kW/m2 is incident on an area perpendicular to the incoming light. The area is now rotated 20° about an axis perpendicular to the incoming light. What is the resulting power per unit area now reaching the surface?

What is the intensity of the electromagnetic radiation 2.00 m from a 100 W isotropic source?

If a 20 W laser beam, which has an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m, what is the final intensity of the beam?

A 20 W laser beam, with an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m. What is the initial intensity of the beam?

Light of wavelength 661.1 nm is received from a star that is emitting light of wavelength 659.6 nm. How fast is the star moving relative to the Earth, and is it moving toward or away?

If the intensity of radiation from the Sun is 1.4 kW/m2 at a distance of 150 × 106 km, at what distance is the intensity 5.6 kW/m2?

Plane-polarized light of intensity I₀ is sent through two polarizers, the first of which is rotated 45° clockwise relative to the original polarization, while the second is rotated 45° counterclockwise relative to the original polarization. What is the transmitted intensity?

Unpolarized light of intensity I₀ is sent through an arrangement of 4 polarizers. Each polarizer (after the first) is rotated 30° relative to the preceding one, so that the last polarizer is perpendicular to the first. What is the intensity transmitted by this system?

If unpolarized light with E_rms = 20 V/m is passed through a polarizer, what is the transmitted E_rms?

Using the approximate formula for the Doppler shift, f_o ≈ f_s (1 + v_rel/c) one finds that a source is moving away at 0.60 c. What result would the exact formula yield for the recession speed?

For electromagnetic radiation of intensity of 1.00 W/m², what is the rms value of the electric field?

Randomly polarized light of intensity I₀ is passed through two polarizers whose transmission axes differ by 45°. What is the intensity of the light that has passed through both polarizers?

A 20 W laser beam, with an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m. What is the initial E_rms of the beam?

Plane-polarized light of intensity I₀ is passed through two polarizers, the first one at 45° to the original polarization direction and the second rotated an additional 45° so that it is perpendicular to the original polarization. What intensity is transmitted?

Plane-polarized light of intensity I₀ is passed through a polarizer oriented at 45° to the original plane of polarization. What is the intensity transmitted?

Radiation of intensity 1.00 kW/m² is incident on an area perpendicular to the incoming light. The area is now rotated 20° about an axis perpendicular to the incoming light. What is the resulting power per unit area now reaching the surface?

If the intensity of radiation from the Sun is 1.4 kW/m² at a distance of 150 × 10⁶ km, at what distance is the intensity 5.6 kW/m²?