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BartSMP [9]
4 years ago
10

A beam of red and blue light in air is incident at 30º on a glass surface. Which is true?

Physics
1 answer:
Alex17521 [72]4 years ago
4 0

Answer:

a. The blue light is bent toward the normal more than the red light.

Explanation:

When light travels from a less dense medium, in this case air into a denser medium (glass) it is bent towards the normal because of the reduction in speed due to the dense medium.In this scenario we have the same incident angle of 30º. therefore, we consider the wavelength of the light because at shorter wavelengths we have more refraction occurring as compared to longer wavelengths. We know that Blue light has a shorter wavelength as compared to red light, which means it will bend toward the normal more than the red light.

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A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel
marusya05 [52]

(a) 7.32\cdot 10^7 Hz

The frequency of an electromagnetic waves is given by:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of light

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Substituting into the equation, we find

f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz

(b) 4.60\cdot 10^8 rad/s

The angular frequency of a wave is given by

\omega = 2\pi f

where

f is the frequency

For this wave,

f=7.32\cdot 10^7 Hz

So the angular frequency is

\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s

(c) 1.53 m^{-1}

The angular wave number of a wave is given by

k=\frac{2\pi}{\lambda}

where

\lambda is the wavelength of the wave

For this wave, we have

\lambda=4.1 m

so the angular wave number is

k=\frac{2\pi}{4.1 m}=1.53 m^{-1}

(d) 1.03\cdot 10^{-6}T

For an electromagnetic wave,

E=cB

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

I=\frac{E^2}{2\mu_0 c}

where we have

E = 310 V/m is the amplitude of the electric field

\mu_0 is the vacuum permeability

c is the speed of light

Substituting into the formula,

I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2

(g) 1.53\cdot 10^{-8} kg m/s

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

\frac{dp}{dt}=\frac{A}{c}

where the <S> is the magnitude of the Poynting vector, given by

=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2

and where the surface is

A = 1.8 m^2

Substituting, we find

\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s

(h) 8.47\cdot 10^{-7} N/m^2

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p=\frac{}{c}

where we have

=254.2 W/m^2

c=3\cdot 10^8 m/s

Substituting, we find

p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2

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