Question

At a certain location, electromagnetic wave 1 is measured to have an average electric field of Eav = 152 N/C, and electromagnetic wave 2 is measured to have an average magnetic field of Bav = 0.87 µT.

a. (5 points) Calculate the average intensity of each wave.

b. (5 points) If the electromagnetic waves are being measured 0.54 m from their light sources (assume that they are emitting light uniformly in all directions), what average power is each source emitting? By how many 60 W incandescent light bulbs would each light source need to be replaced to emit the same average power?

c. (5 points) Suppose an average human sun bathes in the light from wave 2 for 45 minutes. If we assume that this person exposes 0.45 m 2 of their skin to this light and that none of the light is reflected by the skin, how much energy is absorbed into their skin? If you could somehow utilize this energy to lift a 70. kg person on the surface of the Earth, how high could the person be lifted?

d. (5 points) Assume that both light waves are initially unpolarized. We decide to send the higher intensity light wave through a polarizer-analyzer combination so that its final intensity matches that of the lower (unpolarized) intensity light wave. If the transmission axis of the first polarizer is horizontal, how must the transmission axis of the analyzer be oriented so that the light intensities match? Express your angle as measured from the vertical axis

Answer 1

given

E_avg = 152 N/C,

and electromagnetic wave 2 is measured to have an average magnetic field of B_avg = 0.87 µT

= 0.87 x 10^-6 T

a )

I_avg = 1/2 _o C E_o²

= 1/2 C B_o² / _o

the average electric and magnetic field vectors varies as follows

E_avg = 2 E_o / , B_avg = 2 B_o /

E_o = E_avg / 2 , B_o = B_avg / 2

I_1,avg = 1/2 _o C ( E_avg / 2)²

= 1/2 x 8.85 x 10^-12 x 3 x 10⁸ x (3.14 x 152 / 2)²

I_1,avg = 75.6 W/m²

I_2,avg = 1/2 C B_o² / _o

= 1/2 C ( B_avg / 2)² / _o  

= 0.5 x 3 x 10⁸ x ( 3.14 x 0.87 x 10^-6 /2 )² / 4 x 3.14 x 10^-7  

I_2,avg = 222.92 W/m²

b )

P_avg = I_avg x ( 4r² )

P_1,avg = I_1,avg x ( 4r² )

= 75.6 x ( 4 x 3.14 x 0.54² )

P_1,avg = 277.32 W

P_avg = I_avg x ( 4r² )

P_2,avg = I_2,avg x ( 4r² )

= 222.92 x ( 4 x 3.14 x 0.54² )

P_2,avg = 816.86 W

P_1,avg/60 W = 4.62 nearly "5"

P_2,avg/60 W = 13.61 nearly "14"

so the first light source is need to replaced by 5 no of 60 W

the second light source is need to replaced by 14 no of 60 W

c )

E = P x t

E₂ = I_2,avg x area x time

= 222.92 x 0.45 x ( 45 x 60 )

E₂ = 270847.8 J

and also we have

E₂ = m g h

h = E₂ / m g

= 270847.8 / 70 x 9.8

h = 394.821 m

d )

here I = I_o cos²( 90 - )

= I_o sin²

I_o = I_2,avg / 2

I_1,avg = I_o sin²

= 1/2 I_2,avg sin²

sin² = 2 I_1,avg / I_2,avg

= sin^-1 ( 2 x 75.6 / 222.92 )^1/2

= sin^-1 ( 0.8235 )

= 55.436^o