Problem :
What is the amplitude of the magnetic field of a light ray that has an irradiance of 0.00003 W/cm2.
We know that
I = 
. If we assume the magnetic field is harmonic then
< B2 > = < B02cos2(σt) >, but the time integral of the cosine squared is just
1/2 so this is equal to
B02/2. Thus substituting and rearranging the first equation,
B0 = 
= 
= 5.0×10-8 tesla.
Problem :
What is the position of the fourth maximum for a double-slit apparatus with slits 0.05 centimeters apart and
a screen 1.5 meters distant when performed with monochromatic red light of frequency 384×1012 Hz?
The wavelength of this light is
λ = c/ν = 7.81×10-7 meters. Just plugging to the
formula
ym = 
= 
= 9.38millimeters from the central bright maximum.
Problem :
In a Young's Double Slit experiment, what is the ratio of the irradiance at a distance 1 centimeter from
the center of the pattern, irradiance of each individual beam entering through the slits (assume the same
set up as before: light of frequency 384×1012Hz, 0.05 centimeters between the slits, and a
screen 1.5 meters away)?
The irradiance as a function of distance from the center of the pattern is given by
I = 4I0cos2
, where
I0 is the irradiance of each of the interfering
rays. Plugging into the formula:
I = 4I0cos2(
) = 1.77I0. Thus the ratio is just 1.77.
Problem :
A stream of electrons, each having an energy 0.5 eV is incident on two extremely thin slits 10-2 millimeters apart. What is the distance between adjacent minima on a screen 25 meters behind the slits (me = 9.11×10-31 kilograms, and 1eV = 1.6×10-19 Joules). Hint: use de Broglie's formula, p = h/λ to find the wavelength of the electrons.
We first need to calculate the wavelength of electrons with this energy. Assuming all this energy is kinetic we have
T = 
= 0.5×1.6×10-19 Joules. Thus
p = 
= 3.82×10-25 kgm/s. Then
λ = h/p = 6.626×10-32/3.82×10-25 = 1.74×10-9 meters. The distance between minima is the same as between any two maxima, so it will suffice to calculate the position of the first maximum. This is given by
y = = 
= 
= 4.34 millimeters.
Problem :
A Michelson interferometer can be used to calculate the wavelength of light by moving on of the mirrors and observing the number of fringes that move past a particular point. If a displacement of the mirror by λ/2 causes each fringe to move to the position of an adjacent fringe, calculate the wavelength of light being used if 92 fringe pairs pass a point when the mirror is shifted 2.53×10-5 meters.
Since for each
λ/2 moved one fringe moves to the position of an adjacent one, we can deduce that the total distance moved
D, divided by the number of displaced fringes
N must be equal to
λ/2. Thus:
D/N = λ/2. Clearly, then
λ = 2D/N = 
= 5.50×10-7 meters, or 550 nanometers.
