##
Problems on light as radiation

**Problem : **
Compute the energy density of a 2.0 mW laser beam with a diameter of 2mm.

The energy outputted by the laser per unit time (second) is 0.002 J/s. The area of the beam is

*Πr*^{2} = *Π*×(0.001)^{2} = 3.14×10^{-6} m

^{2}. We want the energy per volume, so just by dimensional
analysis we can say

*J*/*m*^{2} = *J*/*s*×1/*m*^{2}×*s*/*m*, so the energy density is given by

*u* = *power*×1/*area*×1/*c* = 0.002× = 2.12×10^{-6} J/m

^{3}.

**Problem : **
Write an expression (in SI units) for the magnetic field of a 10 MHz light wave traveling in the negative *z*
direction and an electric field amplitude of .

First let us consider the direction of the wave. We know that

must yield the
direction and

is given as in the positive

*x* direction. Thus

must point in the negative

*y* direction by the right hand rule. Thus we want an equation of the form

. The amplitude must be given by

*B*_{0} = *E*_{0}/*c* = 0.1/3.0×10^{8} = 3.33×10^{-10}.

*σ* = 2*Πν* = 2*Π*×10^{7} and

*k* = 2*Π*/*λ* = 2*Πν*/*c* = 2*Π*×0.033. So the
expression is:

**Problem : **
Consider the following electric and magnetic waves:

Show that these waves satisfy Maxwell's equations in free space. What conditions must be satisfied?

First we have

â??. which is clearly satisfied since
the magnetic field is traveling in the

*y*-direction. Similarly

â??.. Now

â??×, since the magnetic field only has a component in the

*x*-direction, and this is the only non-zero partial derivative. This should be equal to

*?ü*_{0}*ε*_{0} = . These components are equal if and only if

= *E*_{0}*?ü*_{0}*ε*_{0}. Similarly

â??×. This needs to be equal to

- = . This is true if

*E*_{0}/*v* = *B*_{0}. Now

*?ü*_{0}*ε*_{0} = 1/*c*^{2}, so the first
condition gives us that

*B*_{0}/*v* = *E*_{0}/*c*^{2}. Combining with the second condition we have

1/*v* = *E*_{0}/(*c*^{2}*B*_{0}) = *v*/*c*^{2}. Thus the condition is satisfied if

*v* = ±*c*. The two conditions required, then,
are

*v* = ±*c* and

*E*_{0} = *cB*_{0}.

**Problem : **
Consider a 60W light bulb. Assume that all radiation given off is in the form of light of
wavelength 650 nm. Calculate the number of photons given off per second.

Each second the total energy given off is 60 Joules. Each photon has an energy given by

*E* = *hν* where

*ν* = *c*/*λ* = 3.0×10^{8}/(650×10^{-9}) = 4.62×10^{14} Hz. So

*E* = 6.626×10^{-34}×4.62×10^{14} = 3.06×10^{-19} Joules per photon. Thus the total number of
photons must be

60/(3.06×10^{-19}) 1.96×10^{20} photons emitted every second!