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### Course: Class 12 Physics (India) > Unit 10

Lesson 4: Young's double slit experiment- Young's double slit introduction
- Young's double slit equation
- Young's double slit problem solving
- Fringe width in Young's double slit experiment
- Calculate path difference in YDSE
- Bright and dark fringes in YDSE
- Worked examples: Fringe width in double slit interference
- Double-slit experiment: fringe widths (one wavelength)
- Worked examples: Overlapping fringes (two wavelengths)
- Double-slit experiment: overlapping fringes (two wavelengths)

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# Young's double slit equation

Let's derive a formula that relates all the variables in Young's double slit experiment as we explore Young's Double Slit Equation. Uncover how light's path length difference relates to the angle of incidence, and how this leads to constructive and destructive interference. Apply this knowledge to the world of wave patterns, trigonometry, and the wonders of diffraction! Created by David SantoPietro.

## Want to join the conversation?

- Why these two angles in video are equal?(90 votes)
- David drew the first angle theta at a confusing point in this video. The angle theta should be redrawn away from the right angle (towards the left of the screen) from where it is now. If it is redrawn closer to the actual angle it represents in the triangle, one can see that two of the three right triangles are similar. Consequently, the angles are equal in these similar triangles.(12 votes)

- At2:05we have a right triangle with one of the purple lines being the hypothenuse. How is it equal to the other purple side? Hypothenuse is the largest side, wouldn't we rather have to construct an isosceles triangle?(69 votes)
- I agree. So this is essentially the consequence caused by the assumption we've made, that we are assuming that the distance from the slits to the wall >>> (way greater than)distance between slits. So the triangle is very long and thin.( the diagram is not drawn to scale. If the diagram were to be drawn to scale, d would be invisible, and so the base of the long and thin right triangle.) So basically, because we are using very long and thin triangle, it "approaches" isosceles triangle so to speak. The assumption lets us use basic trig, so it's useful, and like David said, the assumption is a fairly reasonable one. Exactly. I had the same question!(38 votes)

- At2:31

How on earth are the two angles equal? I've gone through all the answers on this page but am not satisfied! Help, please(12 votes)- First of all, if you measure these angles with a enough precision, they'll turn out to not be equal. However, they're very similar, and this can be shown with a real example. Suppose the 2 walls are 1,50 m apart, and the slits are 2,00 mm away from each other. If the point we're analyzing is 10,0 cm above the central axis (the central white line at the video), then we can easily find θ:

tan (θ) = 0.100/1.50 = 0.0667 , thus θ = 0.0666

Now let's find the angle x formed between the lower purple line and a line parallel to the central axis.

tan (x) = (0.100+0.001)/1.50 = 0.0673 , so x = 0.0672

To consider x = θ, we need to assume an error of 0.9% in our calculations, which is totally fine. (usually, errors can be even smaller by altering the initial conditions appropriately)

So, with x = θ, its easy to find that these two angles are equal, I encourage you to try it.(11 votes)

- At2:31How did he equate the two angles, I see no connection between the angles to make them equal..?(15 votes)
- The two thetas are equale since d <<<<L so the line in the middle and the two rays coming from the two openings are parallel now try to look at it in this point of view will find that its true that the two angles are equale to each others.(8 votes)

- what's going on at1:50? He's made a right triangle and said the long base and hypotenuse are the same length?? Shouldn't it be an isoceles?(13 votes)
- How would you measure the angles? I mean, we're talking about a light wave, what experimental apparatus would you expect to use to get an angle formed by light?(4 votes)
- You can see where the light has maxima (bright spots) and minima (dark spots) on the screen. The angle is measured from the slit to the screen.(4 votes)

- he tells here hat light is always diffracting,so how does this happen?is it like 1 phone splits into another two?or is it that the photons spread out?if it is that then is there a limit to how much a light beam can diffract?(4 votes)
- I think that the split needs to be wide enough for several photons to get through. Those which come through at the same time all have slightly different direction. And this difference in the direction of the momentum continues to be amplified over time/distance as it would have, had the electromagnetic wave not encountered the wll with slits. Hope that made sense :-)(2 votes)

- Could anyone please answer for me at1:58, if that's a right angle, why are the top path and the remainder bottom path equal?(3 votes)
- I struggled with this question as well. My understanding is that if the distance of the orange line is very small, then you can imagine the two lines almost being stack to each other. In other words, imagine if the orange line had a length of zero, then the lines must be the same. Anyway, that is how I made sense of it. There was a lot of handwaving in this demonstration, including how theta angles are congruent. Let me know what you think of my reasoning, not 100% if it is correct.(4 votes)

- At2:06, I dont understand how the two blue lines measure out to be equal. The two blue lines along with the orange line form a right angled triangle, so one of the blue lines becomes the hypotenuse and the other the base. And technically the hypotenuse should be the greatest side in a triangle.

So how is it possible?(2 votes)- If the screen is very far away, then they are so close to equal that we can pretend they are equal.(3 votes)

- I learned from school that to get a good interference pattern, one should do the YDSE using monochromatic light.

So, what happens if we do the experiment using white light?(1 vote)- Since white light is composed of many frequencies, each of those will generate their own interference pattern. In the end, you'll have kind of a rainbow pattern, I guess.(3 votes)

## Video transcript

- [Voiceover] Okay so that's all well and good, but we've got a problem. I told you these two slits
are so close together, maybe micrometers or nanometers apart, that how are we going to measure? How are we going to physically measure the difference in path length? If I go over to this
barrier, these two holes are gonna look like they're
at the exact same spot. That's how close they are, so
I need some way to determine the path length difference based on something I could measure. And that's where we're gonna
have to play a trick here. We're gonna have to figure
out a function for this path length difference
based on what angle I am at. So the basic idea is this, so
let me get rid of all this. And let me put it to you
this way, so let's say I draw a reference line that goes straight through the center. This centerline is my friend. This is gonna let me measure angles here. So I've got this line here
and let's say I wanted to measure to some point on
the wall what angle am I at. This is how I'm going to measure the angle from the centerline to some
point over here let's say. So my angle is going to be this, so this here would be my angle. And my question that I'm asking is based on this angle is there some way to determine the path length difference? That's the important thing here, how do I determine the path length difference. How is the path length
difference related to this angle? The way we can do it is this. If this screen is far away,
here's what I'm gonna do. I'm gonna draw a line from
the center of this bottom slit to that point and I'm gonna draw a line from the center of
this top slit to that point. And if this screen is far away,
significantly further away than these two holes are
spaced, which isn't too much of a problem because these holes are very close together, I
can draw a third line and this third line's
gonna look like this. Third line is gonna go
from here down, cut through this at a right angle and
if my screen's far away, what'll be true is that
if this is a right angle right here, then the remainder
of these paths will be equal. In other words, the path from here onward, from here forwards,
will be the same length as the path from here forwards. So what would the path length be? The path length difference would just be this piece down here. Whatever is left this would
be the path length difference. This is delta x in other words. So how do I find this? Well again, if I'm far
away here this angle here will equal this angle inside of here. So these two angles are the same. So now that I know that these two angles are the same it's just basic trigonometry. I've got a right triangle in here and I'm gonna redraw it over here. I'll just draw you a right triangle. So my right triangle looks like this. I've got this distance
between the holes, which is d. I'm gonna call that
distance d, the distance between the two holes,
center to center distance. And then I've got this other orange line. This represents that line I had to draw to make the right angle. And then I've got this path
length difference this way. So this is my triangle and this is supposed to be a right angle. This side is delta x, the
path length difference. The extra amount that
wave from the bottom hole had to travel compared to
the wave from the top hole. Well this is trigonometry,
here's my right angle. I can just say if I want a
relationship between these, I can say that sine of
theta, because this is theta and that theta is the same
as this theta over here. Sine of theta would be
opposite over hypotenuse. And the opposite to this theta is delta x, so I have delta x over the
hypotenuse in this case is d, this entire distance between the two holes because this side is the right angle. The hypotenuse never touches
the right angle side. The hypotenuse is this other side. So that's over d, so what's
the path length difference? The path length difference for a double slit is just d times sine of theta. So this is what I wanted. Now I know delta x is d sine theta. Now I can write the double slit formula. Let me get rid of this. The double slit formula looks like this. It says that M times
lambda equals d sine theta. And why, well remember delta
x for constructive points was integers times wavelengths, so zero, one wavelength, two wavelength and so on. And so in order to get
constructive points d sine theta, which is the path length
difference has to equal zero lambda, two lambda and this is the double slit formula,
it looks like this. What does it give you? This M is gonna be zero,
one, two and so on. The d is the distance between the two slits, that would be d. Theta is the angle from
the centerline up to the point on the wall where you
have a constructive point. And lambda is the wavelength, the distance between peaks of the wave. Now I mean theoretically
speaking you could plug in one halves for M and that
would give you the angles to the destructive points
because we know the delta x, the path length difference,
should just equal half lambdas to get to the destructive. So this can give you the
angles to constructive points and destructive
points if you plug in the correct M value, the
order, sometimes this is called the order of
the constructive point. This would be the zeroth order because the path length difference is zero. Sometimes this is called the first order because it is one wavelength difference. The next one might be
called the second order because it's two wavelength difference. You might object though,
you might still say "Wait this was no better because
d is really close together. "This d spacing right
here is extremely close. "We can't measure that well." But we can measure theta
and we can know that wavelength of a laser we send in. And we can count which order
we're at, so this is a quick way to figure out if you
had something with two holes in it you could
figure out how close they're separated even if you don't have a ruler that small, it's a quick way. Send some light in, you'll
get a diffraction pattern like this, an interference pattern. You measure the angle,
now I can figure out how close two holes are, two spacings. And you can do all kinds
of experiments to precisely determine how close two
holes are in some sort of crystal lattice or
a molecular structure. And it's determined by
Young's Double Slit Equation.