In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx10^10 H_z` and amplitude `48V_m^-1`
(a) What is the wavelength o f the wave?
(b) What is the amplitude of the oscillating magnetic field.
(c) Show that the average energy density of the field `E` equals the average energy density of the field `B.[c=3xx10^8 ms^-1]`.

Explanation

Power rating of bulb, P=100W
It is give that about 5% of its power is converted into visible radiation,
`therefore` power of visible radiation,
`P'=(5)/(100)xx100=5W`
Hence, the power of visible radiation is 5W.
(a) Distance of a point from the bulb, d=1m
Hence , intensity of radiation at that point is given as:
`I=(P^(@))/(4pid^(2))`
`=(5)/(4pi(1)^(2))=0.398W//m^(2)`
(b) Distance of a point from the bulb, `d_(1)=10m`
Hence, intensity of radiation at that is given as:
`i=(P^(@))/(4pi(d_(1))^(2))`
`=(5)/(4pi(10)^(2))=0.00398W//m^(2)`