**Concept:** Young's Double Slit Experiment

Hey guys in this video we're going to talk about Young's double slit experiment which is just an analysis of how light rays are going to interfere with one another after initially collimated light passes through a double slit let's get to it. A beam of light thats shown onto a double slit that just means it's aimed at a double slit was thought initially before diffraction is understood to produce a single spot of brightness, however when the experiment was performed that wasn't the case so in this image here on the left we have the expected outcome of the experiment if we shine this initially collimated light through a double slit we expect to get just uniform brightness at a single point on the screen the width of the slit. But what actually happens what was demonstrated is that there are multiple bright spots on the screen that this initially collimated light when it passes through the double slit does not produce a uniform brightness a constant brightness at one spot it actually produces a varying brightness that goes between that alternates between bright and dark along the screen and it produces a diffraction pattern that's what we call this. This diffraction pattern is due to the light interfering so this is due to interference, that as these waves pass through these slits they come out at different directions so maybe here you get constructed interference and you get a bright spot. Maybe down here you get destructive interference and you get a dark spot but that pattern of alternating brightness and darkness that diffraction pattern is produced entirely by those two light waves interfering with one another when they reach the screen and this is entirely due to diffraction because if without diffraction that collimated light encountering the slit goes through undisturbed it goes through collimated but with diffraction when that collimated light encounters the slit it spreads apart and comes out isotropic so you get all these different angles for the light coming through.Now each slit produces a ray in every direction alright isotropic the same in all directions this means that some rays are going to interact constructively you can see in the figure that here we have maximum and here on the top ray we have a minimum and when you have a maximum and a minimum or a peak and a trough you get destructive interference right down here you can see that both waves are at a maximum when both waves are a maximum or both waves are at a minimum you get this destructive interference and this depends upon where along the screen you want to look at the lights if the light happens to constructively interfere, that means that the amplitude is going to get larger you get bright what are called fringes up until this point I've been calling them spots but they are technically called fringes thats the physics term for this. Some of those light rays are going to also destructively interfere and they're going to produce a light with less amplitude or a dark fringe and we're going to get alternating spots or alternating dark and light fringes alternating spots of brightness and darkness now that's all conceptually but how do we actually apply math to figure out where these bright spots are with the bright fringes and where the dark fringes are. The bright fringes are located at angles given by this one formula that sine of the theta is M lambda over D now what's M? M is what we would call an indexing number and it's allowed to be 0,1,2,3 etc all positive integers and zero, the reason we have this so-called indexing number is because there is not a single angle where a brightness occurs of bright fringe but there are a multitude of angles where this occurs. I have on this figure here the diagram they are always going to draw when analyzing a double slit problem here we have the double slit where D is the separation between the two slits, and we're going to draw a dashed central line that lines up with the central brightness peak the M equals 0 peak and if we choose an arbitrary bright spot which is given by an arbitrary index theta M is the angle that the light has to travel at to reach that bright spot likewise we can also find the angles that dark spots occur at or dark fringes occur at this is indexed by a different letter N. It really doesn't matter that we use M and N maybe your book uses I and J maybe uses M twice that's also popular the indexes are different for dark fringes and bright fringes because they occur at different positions and the equation is slightly different. This is N plus one half lambda over D where D once again is the separation between the two slits N is the indexing number for the dark fringes and lambda is just the wavelength lambda has always been and you want to memorize this figure because you're always going to draw this figure when trying to solve double slit problems let's do a quick example. A 650 nanometer laser is shown through a double slit of ten millimeter separation what angle is the fourth brightest fringe located at if this double slit is 2.8 meters from the screen how far from the brightest fringe is the fourth brightest. So let's draw our little diagram of the double slit problem which is always what you want to start with. Now you're going to have a central bright fringe and then you're going to have a less bright one a less bright one and a less bright one and this is the first brightest, this is the second brightest, this is the third brightest and this is the fourth brightest. Now, remember that the equation for the angle for bright fringes is indexed by M we called it, so the question is what M does the fourth brightest fringe occur. Well M can be 1,2,3,4 etc zero is that first brightest fringe that central fringe the second brightest is M equals 1. The third brightest is M equals 2 And that fourth brightest is actually M equals 3 so this problem is written in a very specific way to teach you that just because this is the fourth brightest fringe doesn't mean you have to assume that M equals 4 in this case because M starts at 0, M actually equals three so we want to find that angle theta 3 that the N equals three fringe occurs at or the fourth brightest fringe occurs at so sine theta 3 is 3 lambda over D. Right which is going to be 3 Lambda the wavelength we're told its 650 nanometer laser, so 650 nanos times 10 to the -9 and D is that separation distance between the slits we're told that there's 10 millimeters of separation or 10 times 10 to the -3 and so plugging this into a calculator you get 0.000195 or you get that lambda 3 equals 0.011 degrees, something to note here is that this is an incredibly small angle the angles are always really going to be very very small alright and that's because actually to get this equation that we use we actually have to assume that the angles are always small this might be something that you read in your book this might be something that your professor points out but that's the reason why the angle should be small they're always going to be this this 0.011 degrees it's not even it's barely over 100 of a degree it's a very very small angle. Now the next part of the question is if the double slit is 2.8 meters. Away from the screen how far from the brightest fringe is the fourth brightest we want to know this distance how far is it and I'll call that distance Y. Look at this this is just a triangle so I'm just going to draw this triangle this angle is our theta 3 angle which is 0.011 degrees the base of the triangle is 2.8 meters that's how far the screen is from the double slit and a vertical distance sorry I'm actually going to call that Y sub three that's just that vertical distance of that third, that M equals 3 peak. The height of this triangle is going to be Y so clearly I can use trigonometry. I have the opposite edge and the adjacent edge so I can use the tangent and say tangent of our angle 0.011 degrees equals the opposite edge Y3 sorry over the adjacent edge which is 2.8 meters or we have that Y3, if I just multiply 2.8 meters up we have Y3 equals 2.8 meters times tangent 0.011 degrees which plugging into a calculator is just 0.00054 meters or 0.54 millimeters. So you see how small that angle is? Even though this horizontal distance is 2.8 meters because that angle so small because it's this tiny angle the height Y3 is only half a millimeter roughly. Alright guys that wraps up this talk about Young's double slit experiment. Thanks for watching.

**Problem:** A 450 nm laser shines light through a double slit of 0.2 mm separation. If a screen is placed 4 m behind the double slit, how wide are the bright fringes of the diffraction pattern?

**Example:** Unknown Wavelength of Laser through Double Slit

Hey guys, let's do an example about Young's double slit experiment a laser of unknown wavelength shines monochromatic light through a double slit of 0.2 millimeter separation if the screen is 5.5 meters behind the double slit you find that angular separation of each bright fringe to be 0.15 degrees what is the wavelength of the laser? First I want to approach I just want to discuss this big word right here monochromatic, this is fancy for single colored. Mono is latin for one chromo is latin for color monochromatic single colored this just means light at a single wavelength and lasers are most typically monochromatic but there are multichromatic lasers so it's specifically monochromatic the light is only emitted at a single wavelength and that wavelength is unknown that's what we want to find so the first thing we're going to do is we're going to draw the situation because that's what we always do with those double slit problems we're told that the screen is five and a half meters behind the double slit and what we're told is that that angular separation of each bright fringe is 0.15 degrees what does this mean well we have our central bright fringe right and then the second bright fringe and third bright fringe etc. What this is saying is the angle separating each bright fringe is 0.15 degrees if I were to draw a line through the next bright fringe that angle would also be 0.15 degrees if I were to draw a line through this next bright fringe right here this would also be 0.15 degrees the separation between every single bright fringe that angular separation is 0.15 degrees so notice that first ray that I drew this blue one right here this has to be our theta 0 angle because that's the angle sorry theta 1 my bad. For M equals 1 we're talking about the angle between M equals 0 fringe and the M equals 1 fringe so that is theta 1 and remember for the angular location of bright fringes our equation is sine of theta M equals M lambda D we know that the angle that we're looking for is of that second bright fringe the one just after the central bright fringe which occurs when M equals 1 so we have M equals 1 and so sine of theta 1 that Angle you want to find is one lambda over D lambda is our unknown.We know what theta 1 right it's 0.15 degrees just out of curiosity what do you guys thing theta 2 would be the angle at the location of the N equals 2 do you think it would be 0.15 degrees no it's the entire sweep of this angle so its 0.15 plus 0.15 which is 0.3 and you can also solve this problem by doing that by finding the theta 2 angle that would also tell you the same wavelength alright you guys can double check what I'm saying at the end of the problem if you want for the theta 2 or the theta 3 or the the theta 4 but for theta 1 we know what D is the separations 0.2 millimeters we know what theta one is its 0.15 degrees wavelength is our unknown so I'm going to multiply the D up to the other side and this lambda is just D sine Theta 1 which is 0.2 millimeters so times 10 to the -3 meters and theta is 0.15 degrees so our wavelength is 5.24 times 10 to the -7 meters now you can leave it like this and be done that's the answer but I'm going to rearrange this slightly what I'm going to do is I'm going to increase the order of magnitude by 2 I move the decimal place over two points which is going to increase my magnitude by 2 so I need to decrease my exponent by 2 alright if I'm gaining 2 in the decimal place I have to lose 2 in the power alright this time it's 10 to the -9 is nanometer so this is 524 nanometers and you're going to see most of your problems are going to describe wavelength in nanometers because on the hundreds of nanometers about 450 nanometers to 750 nanometers if I remember correctly that is visible light that's light you can see 450 nanometers is purple light the lowest wavelength light and 750 nanometers animators is red light the highest wavelength light. Alright guys sorry. That wraps up this problem thanks for watching.

A double slit is illuminated with monochromatic light of wavelength 6.00 x 10^{2} nm. The m = 0 and m = 1 bright fringes are separated by 3.0 cm on a screen which is located 10.0 m from the slits. What is the separation between the slits?

(a) 4.0 x 10^{-5} m

(b) 8.0 x 10^{-5} m

(c) 2.0 x 10^{-4} m

(d) 2.4 x 10^{-4} m

Watch Solution

A double slit is illuminated with monochromatic light of wavelength 6.00 x 10 ^{2} nm. The m = 0 and m = 1 bright fringes are separated by 3.0 cm on a screen which is located 10.0 m from the slits. What is the separation between the slits?

(a) 4.0 x 10^{-5} m

(b) 8.0 x 10^{-5} m

(c) 2.0 x 10^{-4} m

(d) 2.4 x 10^{-4} m

Watch Solution

You illuminate two very narrow slits in air by monochromatic coherent light and find that the first interference maximum on either side of the central bright spot is at an angle of ±12.0° from the center of the central bright spot. You then immerse the entire apparatus in a transparent liquid and find that the first maximum on either side of the central bright spot occurs instead at ±9.0°. What is the index of refraction of the liquid?

(a) 0.5

(b) 1.3

(c) 1.5

(d) 2.0

(e) 3.0

(f) 4.0

(g) none of the above

Watch Solution

When monochromatic light of wavelength (in air) of λ = 500 nm passes through two every narrow slits and the interference pattern is observed on a screen 2.0 m from the slits, the distance between adjacent maxima near the center of the screen is 3.0 mm. If the apparatus (slits and screen) is submerged in water, the distance between the adjacent maxima is

A) 3.0 mm

B) larger than 3.0 mm

C) smaller than 3.0 mm

Watch Solution

Two very narrow slits are illuminated by monochromatic coherent light in air that has wavelength 560 nm and the interference pattern is observed on a screen that is 4.00 m from the slits. The first interference minima are located at ∓58.0° on either side of the central bright band. You then immerse the slits and the space between the slits and the screen in a transparent liquid and illuminate the slits with the same light. Now the first minima on either side of the central bright band occur at ∓36.0°.

What is the refractive index of the liquid?

Watch Solution

Coherent light with wavelength *λ* = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed on a screen that is 4.00 m from the slits. Near the center of the screen the separation between adjacent maxima is 2.00 mm. The distance between the two slits is

(a) 2.40 mm

(b) 1.40 mm

(c) 1.20 mm

(d) 1.00 mm

(e) 0.400 mm

(f) none of the above answers

Watch Solution

A screen is illuminated by monochromatic light as shown in the figure below. The distance from the slits to the screen is *L*. Using the small angle approximation ( *θ* = sin *θ* = tan *θ*), what is the wave length if the distance from the central bright region to the second bright fringe is *y*.

Watch Solution

A screen is illuminated by 512 nm light as shown in the figure below. The distance from the slits to the screen is 6.7 m. How far apart y are the central bright region and the third bright fringe?

1. 1.52482 cm

2. 2.27754 cm

3. 2.74312 cm

4. 1.07854 cm

5. 1.05012 cm

6. 3.2389 cm

7. 3.78953 cm

8. 0.66491 cm

9. 2.37883 cm

10. 7.46025 cm

Watch Solution