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Light waves: diffraction and interference
In Ch. 22 we learned light is a wave , and then in Ch. 23 we promptly ignored this fact! It took people a long time to notice it's a wave,
because λ is so small.
(Recall, violet light has λ=400 nm, red light is 700 nm)
Here is a "Ch. 23" picture of light coming in from the left, hitting a wall with a hole or slit in it, and forming a shadow on a distant wall.
This was basically Newton's idea (and is quite accurate!)
However, if the size of the slight is very small - somewhere on the same scale as the λ of light, then the picture is very different.
The lines I drew are NOT the same as in the previous picture, they are not "rays". Now we're trying to indicate the wave nature by drawing "wavefronts". You might think of this as connecting points where |E| is cresting. These lines are now perpendicular to the direction of travel of the waves. (Think of it as looking down at waves on the ocean, and what I'm drawing is the lines where the waves are highest) In the picture above, the waves move right.
(The image on the screen isn't quite right - we'll make a more accurate picture soon!)
Incident light
Big slit in wall
Distant screen
(Dark here - "shadow")
(Bright here)
(Dark here - "shadow")
Incident light waves
Small slit
Distant screen
Light spreads out all over (but not equal intensity everywhere)
Brief review of interference of waves (Phys 2010 Giancoli Ch. 11.11)
Imagine two traveling waves (same frequency, speed, and direction)
If they are "in phase" (meaning, the peaks are lined up) and if the waves are traveling at the same place and time, they will add up, they superpose , as waves always do. The resulting wave, the sum of the two, has the same f, c, and direction, but twice the amplitude. The waves have constructively interfered.
Now take those same two waves, but let them be exactly "out of phase" (meaning, the peak of one is at the same place as the trough of another!) Then the result when they superpose is cancellation, the "+" of a peak adds to the "-" of a trough giving zero, no wave at all!
The waves have destructively interfered.
You can have two waves heading towards a common point. At that point the waves will interfere. If they are "in phase", they will add up to double the amplitude. In this figure, both waves peak at the point where they meet - they add up: constructive interference.
If you were to shift one of the waves a little, so one peaks where the other "troughs" at the meeting point, they would instead be completely "out of phase", they would exactly cancel: destructive interference at that one point. [If they are just partly out of phase, they will add (interfere, superpose) but not to either extreme. Two waves of amplitude A can add up to a wave of anything between 0 and 2A.]
#1:
#2:
#1 + #2:
c
c
c
#1:
#2:
#1 + #2:
c
c
#1:
#2:
As an example, Huygen's principle fully predicts the phenomenon of refraction (!) The details are, as I said, tricky (again, Giancoli has somewhat better pictures), but the result is straightforward:
Waves always have λ f = speed, so
λ1 f1 = c/n λ2 f2 = c/n2.
Now, here's an important consequence of Huygen's ideas: f1 = f2. Why? As the incoming wave reaches the boundary and creates the outgoing wave, it is "jiggling" with frequency f1, and that's the same frequency that the new medium will jiggle with. (Frequency is just counting, it does not
depend on what medium you're in. But the resulting λ does depend
on the medium.)
The equations above tell us that if f1=f2, then
λ1 / λ2 = n2/n1 (can you see how I got that?)
That means in the picture above: λ2 is smaller because n2 is bigger.
Wavelength depends on speed. In a "fast medium" (smaller n), the
wavefronts are more spread out, λ is bigger.
The picture shows wavefronts, remember that the "rays" these represent are perpendicular to the wavefronts. (I have drawn two sample rays in the figure) The angle at the boundary is bent. Refraction!
Huygens principle predicts that if light hits small slits, you should be more likely to observe the "wave nature" of light. (Look back at the pictures on the previous page - the outgoing wave is more noticeably different from the incoming wave when the slit size is small) To really see this effect in the lab, it helps (a lot!) to have monochromatic light (one pure wavelength) and also coherent light (meaning nice incoming plane waves with all rays "in synch" with each other).
wave fronts (coming in)
λ
wave fronts (going out)
n 1
n2>n 1
λ
Young's Double Slit Experiment: The first observation of interference of visible light! Huygen's says each of the two slits will act as a point source for outgoing spherical waves.
On the right, light is emerging (heading off in all directions!) from each slit.
The incoming light is monochromatic (one λ),
and coherent, which means that if the E field is peaking at the top slit at some instant in time, it is also simultaneously peaking at the bottom slit.
What will we see on a screen far off to the right?
The old "particle" or "ray" model predicts this:
two bright spots, one directly being each slit.
This is experimentally dead wrong!! (At least, if the slits are very small
compared to λ.)
Huygens model of interfering waves predicts this - (We'll see why soon): And, this is what you really see. Never mind Giancoli Fig 24-9 and 10, for now! For two SMALL slits, each bright "bump" in the image is equally bright. You see a bunch of bright spots over there!! (We'll explain exactly why you get this pattern on the next page) They are not necessarily bright right behind the slits, either!
This is weird. "Shadows" are what we're used to, but the wave nature of light says it can effectively bend around corners. This phenomenon is called diffraction. It's not so intuitive (for light.)
wave fronts (plane wave)
c
λ
spherical waves
shadow
screen
shadow
shadow
Bright spot
Bright spot
screen
Bright
Bright
Bright
dark
dark
The question is: is the extra distance EXACTLY a whole number of full wavelengths? If it is, then the two waves are still in phase!
Think about the condition for waves to be in phase. If a wave travels through exactly ONE full wavelength, it's back in phase. Same if it travels through exactly TWO full wavelengths,...
So if the extra distance traveled by one ray compared to the other exactly equals λ (or 2λ, or 3λ, or mλ, where "m" is any integer) then
the two waves are still (back) in phase, and they constructively interfere, which means the spot we labeled "x" will be bright.
If, on the other hand, one wave travels through HALF a wavelength extra, it's exactly out of phase! That means the two waves destructively interfere, it would be totally dark at point "x". (Same
whether the path length difference is 0.5 λ, or 1.5λ, or 2.5λ, etc...)
We need to look at the picture again, this time a "zoom in" right down next to the slits...
The picture shows that the path length difference is d sin(θ).
The geometry is tricky! Think about it, compare to the previous pictures, see if you can really get it for yourself.
What we've just argued is that if that path length difference is an
INTEGER number of λ's, we get constructive interference, and thus
the spot x will be bright :
Constructive: d sin(θ) = m λ, (where m=any integer, 0, 1, 2, 3, ...)
Similarly, if the path length difference is a HALF integer number of
λ's, we get destructive interference, the spot x will be dark:
Destructive: d sin(θ) = (m+1/2) λ, (m=any integer, 0, 1, 2, 3,...)
d
This is d sin(θ) (Convince yourself!)
(screen is far away to the right...)
θ
θ
θ
Quick review of radians, and the "small angle" approximation:
R = radius s = "arc length" along circle
θ = s/R defines the angle in radians.
Full circumference is s = 2 π R. That means θ (all the way around) = s/R = 2 π R/R = 2 π rad. Or, 360 deg = 2 π rad.
Now look at a small wedge, and "square off" the curvy side: We know θ = s/R and we know tan θ = x/R and we know sin θ = x/(R+δ). (Those all come from "sohcahtoa") But, if θ is very small, we see s ≈ x (roughly, just look at the picture) and δ ≈ 0, too. Conclusion: if θ is very small, ≈ tan ≈ sin! (That would also mean x = R tan ≈ R ) Example: (try this on your calculator!) 1 deg = .017453 rad, sin(1 deg) = .017452 and tan(1 deg) = .017455. These are very close!
Finally, let's go back to our "2 slit interference" picture. I'm going to simplify the picture by only drawing ONE "average" ray, not both of them. (Remember, d is really tiny compared to L!) We already argued there are bright spots whenever d sin(θ) = m λ.
Note: = tan−^1 (x / L),
Since L is huge, we frequently have x/L small, and then θ≈ x/L So, for small x, and big L, d sin(θ) = m λ becomes d x/L (^) ≈ m λ, or BRIGHT SPOTS: x ≈ m λ L / d (m = 0, 1, 2, 3, ...) DARK SPOTS x ≈ (m +1/2) λ L / d (m = 0, 1, 2, 3, ...) [ If x/L is not small, just use the correct formula, d sin(θ) = mλ Use this to find θ, then solve for x = R tan(θ). ]
R
R
θ
s
R
R
θ s^
x
screen
d
x
L
θ
Bright dark
Example: Consider 2 slits, a distance d=0.01 mm apart. There is a screen L=2m away. Monochromatic red light (λ =700 nm) shines on the 2 slits. What will the brightness pattern look like over at the screen? Lingo: x2 is the "m=2" or "2nd order" line.
x 1 = 1st bright spot above the center. The "first order" line, or "m=1" line. x1 = 1* λ L / d Here, x 1 = (700E-9 m)*(2m)/(.01E-3 m) = .14 m
"zeroth" order, or m=0 line, in middle. (m=0, x=0)
x-1 = 1st bright spot below the center, "m=-1" line. x-1 = -1* λ L / d = -.14 m (Note the symmetry, |x-1| = x1.) Etc.
Note: θ1 = 1λ / d = .07 rad, θ2 = 2 λ / d = .14 rad, etc. (If x starts getting big, you should go back to the correct formula for the "angles of maxima", namely d sin(θ) = m λ. )
- If we'd used blue light (500 nm) instead of red, the pattern is a little different. x1 is slightly smaller, the bright spots are squeezed together more, towards the center (which is unchanged)
- If d was bigger, the pattern is also squeezed, in a similar way.
That's why it's so hard to notice this pattern in normal life. For visible light, λ is SO small, and d is so much bigger, that the bright spots all squeeze right next to each other - you can't even NOTICE that there are dark and bright spots. To make it worse, all the colors are normally present, with their bright spots in slightly different places... so it really just looks bright all over!
x 1
x-
x 2
x-
What if we used a grating with 1000 lines/cm instead of 2 slits? First, we have to think about the spacing between these lines... 1000 lines/cm means 1cm/1000 lines, or 1/1000 cm per line, or .01 mm per line - that's the same distance between lines as we had in the previous problem! ("lines" in a grating can be though of as analogous to "slits")
So the pattern is identical to that in the previous problem, except that the spots are brighter, and sharper. The x's (positions) of the bright spots are the same.
If you shine white light (rather than monochromatic) there are many λ's present. Each color has bright
spots at different positions x. So, the colors appear to get spread out on the screen. You get a little rainbow, a spectrum at each order. (One spectrum for each "peak" in the picture above)
Since x1 = λ L/d, measuring "x" for different colors can tell you
what the wavelength of each color is.
Some light sources have different colors, but not all colors. (E.g., Mercury arc-lamps used on highways look yellowish-white, but do not contain all wavelengths with anything near equal intensity)
A diffraction grating which spreads out colors allows you to quickly and easily see which colors are present in the source. You will see lines of different colors at each order. This is a unique "fingerprint" of the light source. This technique is called spectroscopy , and we use it to determine the chemical composition of materials. In this way, we can even deduce what elements are present in distant stars, by analyzing the light they emit!
x 1
x-
x 2
x-2 m=-
m=-
m=
m=
m=
