This is the second part of a 2-part lecture on the double-slit equation. In this part
we're just going to be using the equation that we proved in part 1 to solve problems.
This is the equation and what each variable means. You can pause the video if you need
to copy these down. I'm now going to show an animation that shows what each variable means
visually. So capital D is the distance from the diffraction grating to the wall. Lowercase
d is the distance between the double slits. So when light begins to pass through, constructive
interference occurs at very specific spots, and the integer value of n tells you how many points
of constructive interference you are away from the central point of constructive interference.
So that's what n represents in the equation. S is the distance between points of constructive
interference. So if we used 1 for the letter n then the S we would get would be the distance
from that central point to the first point of constructive interference. If we use 2 for n
in the equation we would get this full length to the 2nd point of constructive interference,
so s is always measured from that central point of constructive interference to the number of
points away equal to n. So if you plug n = 2 into that equation you would get this total length.
So looking at an actual double slit diffraction pattern, these are the n-values that go with each
point. Here's an example problem: red light of wavelength 600 nanometers passes through
double slit separated by 0.3 millimeters. If the wall is 4 meters away, how far will the second
point of constructive interference be from the center of the diffraction pattern? So I'm just
gonna write down the variables that I have. I'm trying to find s and specifically s when we're
using n = 2 so we'll just call that s-2, that's the distance from the center to the constructive
interference point. D, the distance from the wall to the diffraction grating, is 4 meters. and n is
2 because we're trying to find that second point of constructive interference. The wavelength is
6 times 10 to the negative seventh meters (nano stands for 10 to the negative ninth) so if you
convert that into scientific notation this is what you get, and we do want that always in meters to
do the calculation. And D is 3 times 10 to the negative fourth meters (again I'm just converting
these into meters and writing them in scientific notation) so now I have all my variables I can
plug them into my equation and when I do I get s = 0.16 meters, so that is going to be the
distance between the central point of constructive interference and the second point of constructive
interference, so that's how you use that equation. In example number 2, green light of wavelength 500
nanometers passes through double slits. If the wall is 8 meters away and the first point of
constructive interference is 0.2 millimeters away from the center, how far apart are the two slits?
So this time I'm given the distance from one point to the next 2 times ten to the negative fourth (or
0.2 millimeters) the distance from the diffraction grating to the wall is eight meters and is equal
to 1. This is just the first point away from the center. The wavelength of light here is 5 times
10 to the negative 7th meters and we're trying to find lowercase d, if we rearranged the equation
to solve for lowercase d instead of s this is what it looks like when I do that, and when I
plug in the variables for my problem this is what I get for a final answer of 0.02 meters. So that's
how far the slits are from each other in this problem. Example number 3 asks how many points
of constructive interference will occur between the center of the diffraction pattern and a point
4 centimeters away. If the lights wavelength is 400 nanometers, the distance to the wall is five
meters and the distance between slits is 0.2 millimeters, so because n measures the number
of points of constructive interference that's actually going to be what we're solving for here.
I'm just writing down the other information as its given. It's pretty clear by now how that works,
so I'm trying to solve for n and if I rearrange this equation this is what I get solving for n.
Plugging that in I get a final answer of n is equal to 4, so just visualizing that super-quick
if the top dot is the 0-point, the central point of constructive interference, and that s that we
have 4 times 10 to the negative 2 goes all the way to N equals 4. That actually means that five
points of constructive interference are going to occur along s so you can see that n is equal to
4 but s includes n = 0, so because of that there are actually 5 points of constructive interference
that occur there. In example number 4, if the wavelength of light doubles the distance to the
wall triples and the distance between slits is divided by 2. what will happen to the distance to
each maximum? So this is a proportional reasoning problem. So the original s is equal to this and
the new s is equal to these changed variables. If I isolate the original variables by taking out the
numbers, I get that the new distance is equal to 12 times the original distance, so that's
what's going to happen to the distance to each maximum. It's going to be multiplied by 12, that
would be how you would do a proportional reasoning problem with the double-slit equation. And that's
everything that you need to know about using the double-slit equation to find interference patterns