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Problems on Interference

Problems on Interference

Problems on Interference

Problems on Interference

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.