We're beginning double-slit problems right now. This is not a difficult topic, especially for AP physics B students who have spent all year learning how to learn physics. They generally pick up quickly what the variables in dsinθ=mλ mean. They have a bit of trouble with a deep understanding of m. At first, I just get them to see that m = 0 at the central maximum, m = 0.5 at the first dark spot, m = 1 at the first bright spot, etc. Once they have facility with getting physically reasonable answers out of the relevant equation, we talk about m as the number of wavelengths in the path difference between two waves. The meaning of m is the only true conceptual challenge in double slits.
Getting correct answers out of dsinθ=mλ can sometimes be a chore, though. It's easy to think that such a straightforward calculation doesn't require careful thought; and students hopefully have been trained by now that plugging into the calculator is of minor importance compared to a conceptual understanding of the topic at hand. Nevertheless, I usually need to remind the class that all distance quantities in this equation, or in its small-θ companion x = mλL/d, must be in METERS.
Here's today's quiz. Note that I ask for no problem solving or calculation; I just make sure everyone knows what the variables mean, and how to convert to meters. The most common mistake on this quiz: 620 nm can be written as 620 x 10-9 m, or as 6.20 x 10-7 m, but NOT as 6.20 x 10-9 m.
GCJ
A red laser with wavelength 620 nm in air shines through two slits which are separated by 0.50 mm. On a screen 2.0 m away from the slits, the laser makes an interference pattern. The brightest spot is located directly in front of the two slits. You are asked to find the location of the nearest bright spot to the central maximum.
1. What is the relevant equation?
2. Assign a value to the following variables. USE UNITS OF METERS FOR ALL DISTANCE QUANTITIES!!! Indicate the variable that is not given in the problem statement; don’t bother to solve for it.
d =
θ =
m =
λ =
I've had more success treating m as an integer that essentially counts the wavelength shifts, and therefore counts the bright and dark fringes. In this case, for the dark fringes, the equation becomes dsinθ=(m+1/2)λ. When we get to single slit interference, with the exception of the central maximum, the equations just reverse.
ReplyDeleteYes. This is the way most textbooks approach m. I prefer to treat m as the number of wavelengths in the path difference; this way, m = 0 is at the central maximum, m = 0.5 is at the first dark spot, m=1 is at the first bright spot, and so on. Either way works. Yes, my approach could cause confusion with the single slit, but I've rarely seen a quantitative question on the AP exam with the single slit that didn't work out fine with my approach.
ReplyDeleteThis is as good a reminder as any -- I'm just telling you what works for me, not what is right. :-)