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 a screen is 5.5 meters behind the double slit, you find the angular separation of each bright fringe to be 0.15 degrees. What is the wavelength of the laser? Okay. First, I want to discuss this big word right here, monochromatic. Okay? This means single-colored. Mono is Latin for 1. Chromo is Latin for color. Monochromatic, single-colored. This just means light at a single wavelength. Okay? And lasers are most typically monochromatic. Okay? But there are multichromatic lasers. Okay? 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 gonna do is we're gonna draw the situation because that's what we always do with these double slit problems. We're told that the screen is 5 and a half meters behind the double slit. And what we are told is that the 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 the third bright fringe etcetera. 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, the angular separation, is 0.15 degrees. Notice that first ray that I drew, this blue one right here, this has to be our theta1 angle because that's the angle for m equals 1. We're talking about the angle between the m equals 0 fringe and the m equals 1 fringe.
To find the angular location of bright fringes, our equation is:
sinθm=mλdWe know that the angle 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 sin( theta1 ), that angle that we want to find, is 1λ/d. λ is our unknown. We know what theta1 is. Right? It's 0.15 degrees. And for theta2, the angle at the location of the m equals 2, do you think it'd be 0.15 degrees? No. It's the entire sweep of this angle. So it's 0.15 + 0.15 which is 0.3. And you could also solve this problem by finding the theta2 angle. That would also tell you the same wavelength. Length. Alright? And you guys can double-check what I'm saying at the end of the problem if you want for the theta2 or the theta3 or the theta4. Okay? But for theta1 we know what d is. The separation is 0.2 millimeters. We know what theta1 is. It's 0.15 degrees. Wavelength is our unknown. So I'm gonna multiply the d up to the other side and this lambda is just d*sin(theta1) which is 0.2 millimeters, so times 10 to the negative 3 meters and theta is 0.15 degrees. So our wavelength is 5.24 times 10 to the negative 7 meters. Okay?
Now, you can leave it like this and be done. That's the answer. But I'm gonna rearrange this slightly. What I'm gonna do is I'm gonna increase the order of magnitude by 2. I move the decimal place over 2 points which is going to increase my order of 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 times 10 to the negative 9 is nanometer. So this is 524 nanometers. And you're gonna see most of your problems are gonna describe wavelength in nanometers because on the 100 of nanometer, about 450 nanometers to 750 nanometers, if I remember correctly. That is visible light. That's light that you can see. 450 nanometers is purple light, the lowest wavelength light, and 750 nanometers is red light, the highest wavelength light. Alright guys. That wraps up this problem. Thanks for watching.