Two narrow slits 0.070 mm apart are illuminated by a very brig...

Question

Two narrow slits 0.070 mm apart are illuminated by a very bright 488-nm light source forming an interference pattern on a screen 4.0 m away. Calculate (a) the distance between the m = 0 and m = 1 lines in the pattern and (b) the distance between the m = 100 and m = 101 lines.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first stop, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A double slit experiment is conducted with a light source of wavelength 550 nanometers. The slits are spaced 0.50 millimeters apart and the screen is located 3.0 m from the slits determine I the separation between the central maximum and the first order maximum. And I I, the separation between the 150th and the 151st order maximum. So that's our angle. It appears our angle. Our final answer or I should say answers that we're ultimately trying to solve for is we're trying to figure out an answer for part I and II I is asking us to figure out the separation between the central maximum and the first order maximum. And our second answer I I is asking us to figure out the separation between the 150th and the 151st order maximum. Awesome. Now that we know what we're solving for. Let's read off our multiple choice answers to see what our final answer pair might be. Noting that our answers for part I and I, I are all in units of millimeters starting with I. And then reading I, I second A is 3.3 and 3.3. B is 3.3 and 3.4. C is 6.6 and 3.3 and D is 6.6 and 3.4. OK. First off, in order to help us solve for this problem, let's write down all of our known variables, the variables that are given to us by the prog itself. We're told that the wavelength which we'll call lambda and the wavelength for this particular prong or rather the wavelength of light to be specific is equal to 550 nanometers. And this is also equal to 550 multiplied by 10 to the power of negative 9 m. So I've gone ahead and done the conversion for you and I'm gonna do all the conversions for you to save some time while solving this particular problem. But if you wanna do any of these conversions on your own, you can recall and use dimensional analysis to do those conversions on your own or you could just quickly look them up. Awesome. We also know the slit separation which we call D and the slit separation is equal to 0.50 millimeters which is equal to 0.50 multiplied by 10 to the power of negative 3 m. And we also know the distance to the screen from the slits which we'll call capital L. And I'll just say L from now on and L is equal to 3.0 m. Fantastic. So now our next step is we need to calculate the position of the M equals one maximum. So the first order maximum. So as we should recall and note that for a double slit interference pattern, the secondary maxima are given by the following equation which we call this equation one which states that D multiplied by sine of theta is equal to M multiplied by Lambda. So now let's make a little note here off to the side in blue that for small angles using small angle approximations of theta is approximately equal to tangent of theta which is also approximately equal to theta, which is equal to Y divided by L fantastic. Where why in this case, why is the distance from the central maximum? Thus, we can go ahead and use this information to safely write that Dy one, Dy one is divided by L which is all equal to M multiplied by lambda. Awesome. So when we rearrange this equation to isolate and sulfur, Y one, Y one will be equal to M multiplied by Lambda multiplied by L all divided by D fantastic. So now all we need to do is plug in all of our known variables and solve for the numerical value of Y one. So let's do that. So Y one is equal to and we know that M in this case is one means we're dealing with a first order maximum since M is some integer value. So M is one and then we need to multiply it by our wavelength which is 550 multiplied by 10 to the power of negative 9 m. And then we need to multiply it by L which is 3.0 m. Fantastic. Then we need to divide everything by D which is 0.50 multiplied by 10 to the power of negative 3 m. Fantastic. So when we plug that into our calculator, we will find that Y one is equal to when we round to one decimal place, 3.3 multiplied by 10 to the power of negative 3 m. But we have a bit of a problem here. We want all of our multiple choice answers to be in units of millimeters. So we need to convert our final answer from meters to millimeters. So when we do that, we will find that it's approximately equal to 3.3 millimeters. Awesome. And that is our final answer. For part I hooray, we did it. And once again, you can use dimensional analysis to do this conversion on your own. Awesome. So now we could start solving for part I I our second answer. Fantastic. So once again, quickly recap part I so the separation between the central maximum and the first order maximum is 3.3 millimeters. So now switching gears. So our next step in order to help us solve. For part I I, we need to evaluate the position of the mth maximum. So now we need to note that since the order is great as we should recall, we do not consider the small angle approximation. So once again, as we should recall, since the order is a great or a large number or a large value, I should say we do not consider small angle approximation. Therefore, we need to recall and use the following equation that states that D multiplied by s of theta is equal to M multiplied by lambda fantastic. And rearranging this equation to isolate and sulfur theta, all by itself will give us that theta is equal to inverse sign of M multiplied by lambda all divided by D fantastic. So, and let's call this equation three fantastic. And let, so we don't get confused. Yeah, we'll just call that one equation three. And then Y one, let's call that one equation two. So in case we need to come back to that Awesome. Or more specifically, let's put it on our empty skeleton one. So Y one is our, our Y one equation is our equation two and equation three is our equation for theta fantastic. So moving right along here. So now the mth order maximum is at the position from the central maximum which we'll call, ym. So as we should recall, we should be able to write that YM is equal to L multiplied by tangent of theta. So we can plug in our value for theta. And when we do, we can rewrite Ym as YM equals L multiplied by tangent of. And let's use a bracket tangent of sine inverse or inverse sine of M multiplied by lambda divided by D Fantastic. And let's call this equation four. So Ym is our fourth equation fantastic moving right along here. So now we need to figure out the distance between M equals 150 M equals 151 maxima. So therefore, as we should recall and note the separation between Y 151 and Y 150 is delta Y. The change in Y is equal to Y 151 minus Y 150. Fantastic. So now we need to plug in our known variables into this equation and solve for the numerical value of delta Y because delta Y is the final answer we're trying to solve for, for part I I our second answer. So when we do that, we will find that delta Y is equal to L multiplied by tangent of drum roll sign or inverse sign of 151. So 151 multiplied by Lambda, divided by deep. Fantastic. And don't forget your bracket minus L multiplied by tangent of inverse sign of 150. So 150 because that's our order that we're focusing on 150. So 150 then we need to multiply that by Lambda and then we need to divide it by D and don't forget your parenthesis and your brackets. Fantastic. So now is the easy part. Now we just plug in all of our known variables into our equation to solve for the numerical value of delta Y. So delta Y is equal to our value for L which is 3.0 m multiplied by tangent of the inverse sign. So tangent of inverse sign, so tangent of inverse sign of 151. So 151 multiplied by our wavelength which is 550. So 550 multiplied by 10 to the power of negative 9 m divided by our value for D which is zero point 50. So 0.50 multiplied by 10 to the power of negative three. And don't forget that its units of measurement are in meters. Fantastic. So now we need to subtract tangent up. And so instead of writing the, you know the, the L value for each time, we're just multiplying the whole expression by that to make it a little bit simpler just in case you're like wondering why aren't we doing it a second time for L? That's because we moved the L out to the outside of the brackets. OK. So moving right along here, so tangent of inverse sign of, and then we need to take our different order. Now, since we're focusing on 150 multiplied by our wavelength, which is 550 multiplied by 10 to the power of negative nine. So negative nine and don't forget that our units of measurement are in meters, divide everything by our D value which don't forget your parentheses, which is 0.50. So 0.50 multiplied by 10 to the power of negative 3 m. Fantastic. And then don't forget your parenthesis in your bracket. So let me plug that into our calculator. We will find that delta Y is approximately equal to when I'm gonna not write our final answer in meters. Because initially, when you plug this into your calculator, you'll get a value in meters, but we want our final answer in millimeters. So you'll know that you've gotten your final answer correct when you've done your conversion and you get when you round to one decimal place, 3.4 millimeters as your final answer. So delta Y is approximately equal to 3.4 millimeters. And that's it. That is our final answer. For part. I, I hooray, we did it. So the separation between the 150th and the 151st order maximum is 3.4 millimeters. And these answers that we found together corresponds with multiple choice answer letter B, which once again is I is 3.3 millimeters and I I is 3.4 millimeters. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Two narrow slits 0.070 mm apart are illuminated by a very bright 488-nm light source forming an interference pattern on a screen 4.0 m away. Calculate (a) the distance between the m = 0 and m = 1 lines in the pattern and (b) the distance between the m = 100 and m = 101 lines.

Key Concepts

Young's Double Slit Experiment

Interference Pattern

Path Difference

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