Question

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

Answer 1

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 .