**Problem : **
What is the amplitude of the magnetic field of a light ray that has an irradiance of 0.00003 W/cm^{2}.

We know that

*I* = . If we assume the magnetic field is harmonic then

< *B*^{2} > = < *B*_{0}^{2}cos^{2}(*σt*) >, but the time integral of the cosine squared is just

1/2 so this is equal to

*B*_{0}^{2}/2. Thus substituting and rearranging the first equation,

*B*_{0} = = = 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 384×10^{12} Hz?

The wavelength of this light is

*λ* = *c*/*ν* = 7.81×10^{-7} meters. Just plugging to the
formula

*y*_{m} = = = 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 384×10^{12}Hz, 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* = 4*I*_{0}cos^{2}, where

*I*_{0} is the irradiance of each of the interfering
rays. Plugging into the formula:

*I* = 4*I*_{0}cos^{2}() = 1.77*I*_{0}. 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 (*m*_{e} = 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* = 2*D*/*N* = = 5.50×10^{-7} meters, or 550 nanometers.