Polarization, Double Slit, Single Slit

In summary, the conversation discusses a test question that the speaker failed and is now seeking a detailed explanation of the answer. The question involves a 0.5mm diameter beam of light with a wavelength of 10nm that is passed through a polarizing double-slit. The speaker is asked to find the average intensity of the beam before and after it passes through the slits, as well as the distance between adjacent bright spots and dark fringes, the power of the beam incident on a screen, and the number of photons that hit the screen. They also need to explain what happens to the interference patterns when the light beam is replaced with a 100eV beam of electrons. The speaker mentions using the wrong equations and values in their
  • #1
durhamisnuts
2
0

Homework Statement



This is an old test question that I failed, but this material will resurface for the final so I just wanted a detailed description of what the full answer would've been. I'll include the rubric too so you'll know what I'm up against. Can you include explanations and pictures? I promise I already attempted it, and FAILED in class. I just don't have a scanner to scan you the exam.
Consider a 0.5mm diameter beam of light that has a wavelength of λ=10nm that is an equal mixture of horizontal and vertical polarizations. We pass this beam through a vertically oriented double-slit that also acts as a polarizer. The amplitude of the electromagnetic field that produces this beam is 100 N/C. Find the average intensity of this beam both before and after it passes through the two slits. Explain your reasoning very briefly. If we place a screen 10 cm from the slits, what is the distance between adjacent (double slit) bright spots? What is the distance between adjacent (single slit ) dark fringes? What is the power of the beam incident on the screen? How many photons per second hit the screen? Suppose we replaced the light beam with a 100eV beam of electrons. Explain what happens to the interference patterns when we do this.
a= .025mm d=.04mm
2pts for a correct equation leading to a correct answer for the average intensity of the before it reaches the slits
2pts for a correct value for the average intensity of the beam after it passes through the slits and briefly noting how you arrived at this
4pts for correct algebra and geometry leading to a correct algebraic equation for the distance between adjacent (double-slit) bright spots in terms of the distance to the screen, the slit separations, and the wavelength of the beam
1pt for the correct value for this separation
2pt for the correct equation for the distance between adjacent (single slit) dark fringes in terms of the distance to the screen, the slit width, and the wavelength of the beam
1pt for correct value of this separation
4pts for correctly noting the relationship between intensity and power and using this to derive an expression for the power in terms of the intensity and diameter of the beam
1pt for the correct value for the beam’s power incident on the screen
2pts for correctly noting the relationship between energy of the individual photons and their wavelength, and using this to derive an expression for the number of photons per second that impinge on the screen
1pt for the correct value for the number of photons that hit the screen
3pts for correctly determining the wavelength of the electron beam
2pts for correctly explaining what happens to the adjacent bright spots and dark fringes when we make this change



Homework Equations



k/c=30kg*m2/C2
tan theta=sintheta=theta for a small theta
rest energy/mass of an electron: mc2= 511000eV
theta=inverse sin of (nλ/a)
theta = inverse sin of (nλ/d)

The Attempt at a Solution



I found the intensity using I= cE2/8∏K, but I was wrong. I also said that becasue light is polarized the average intensity after will be the exact opposite of the intensity before, that is also wrong.
I used the wrong equation for the double slit portion. which then carried out for the majority of the problem.I also used the incorrect value for the area to find power, and I'm not sure why that was wrong-->.04mm
 
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  • #2
^2. I'm having trouble with the last part about the electrons. I know that for light the bright spots and dark fringes are a result of constructive and destructive interference, but I'm not sure how that applies to electrons.
 

Related to Polarization, Double Slit, Single Slit

1. What is polarization?

Polarization is a property of electromagnetic waves, such as light, where the oscillation of the electric field is restricted to only one direction. This can be achieved through various methods, such as passing the wave through a polarizing filter.

2. How does polarization affect light?

When light is polarized, it can only oscillate in one direction. This can have various effects, such as reducing glare and increasing contrast, depending on the application.

3. What is the double-slit experiment?

The double-slit experiment is a classic experiment in physics that demonstrates the wave-like nature of light. It involves passing light through two narrow slits and observing the interference pattern created by the overlapping waves.

4. How does the double-slit experiment relate to polarization?

The double-slit experiment can also be used to study polarization. By passing polarized light through the slits at different angles, the resulting interference pattern can reveal information about the polarization of the light.

5. What is the single-slit experiment?

The single-slit experiment is a variation of the double-slit experiment that involves passing light through only one narrow slit. This can also produce an interference pattern, but with a different shape and intensity compared to the double-slit pattern.

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