Answer:
a) # _photon = 2.5 10¹⁸ photons / s, b) E = 10⁻² N / C, c) B = 3 10⁻¹¹ T
d) r= 2 10⁹ m
Explanation:
a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation
E₀ = h f
c = λ f
E₀ = h c /λ
E₀ = 6.63 10⁻³³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻⁻¹⁹ J
Let's use a direct ratio rule to find the number of photons
#_foton = E / Eo
#_fototn = 1 / 3.978 10⁻¹⁹
# _photon = 2.5 10¹⁸ photons / s
b) The intensity received by the detector is related to the electric field
I = E²
Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space
I = P / A
P = I A
Let's use index 1 for the point on the bulb and index 2 for the point on the detector.
The area of a sphere is
A = 4π r²
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
I₂ = I₁ r₁²/r₂²
I₂ = I₁ 1 / 100²
I₂ = I₁ 10⁻⁴
we must know the intensity at the output of the bulb suppose that I₁ = 1 J
I₂ = 10⁻⁴ J
let's look for the electric field
E =√I
E = √10⁻⁴
E = 10⁻² N / C
c) for the calculation of the magnetic field we use that the field is in phase
E / B = c
B = E / c
B = 10⁻² / 3 10⁸
B = 3 10⁻¹¹ T
d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area A = 4π R² where R = 100 m how many photons are there in the area of the detector r = 1 cm, A’= 10⁻⁴ m²
#_photons = 2.5 10¹⁸ A_detector / A_sphere
#_photons = 2.5 1018 10-4 / 4π 10⁴
#_photons = 2 10⁹ photons in the detector area
for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m
