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dedylja [7]
3 years ago
6

An astronomical source emits radio waves with a frequency of 450 MHz. a. If earth is 20 light years from the source, and we meas

ure an intensity of 8.5 x 10-10 W/m2 here on earth, what is the power of the source?
b. What is the amplitude of the electric field of the wave at our location?
Physics
1 answer:
hammer [34]3 years ago
6 0

Answer:

Explanation:

a )

Frequency n = 450 x 10⁶ .

20 light years = 20 x 9.461 x 10¹⁵m

Let power of source be P

Intensity at distance R  =  \frac{P}{4\pi R^2}

Substituting the given values

8.5 x 10⁻¹⁰ = \frac{P}{4\pi ((20\times9.461\times10^{15})^2}

P = 3822457 x 10²⁰ W.

b )

Half the power will be from electric and half will be from magnetic field.

Total power = 8.5 x 10⁻¹⁰ W

Half = 4.25 x 10⁻¹⁰ W .

power of electric field

= \frac{1}{2}\epsilon\times E_0^2\times c

ε is permittivity , E₀ is amplitude of electric field , c is velocity of light .

Putting the values

4.25 x 10⁻¹⁰ = .5 x 8.85 x 10⁻¹² x E² x 3 x 10⁸

E₀² = .32 x 10⁻⁶

E₀ = .565 x 10⁻³ W / s .

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T_2 = T_1\frac{P_2}{P_1}\frac{V_2}{V_1}

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A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi
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