FIGURE EX33.17 shows the interference pattern on a screen 1.0 m b...

Question

FIGURE EX33.17 shows the interference pattern on a screen 1.0 m behind an 800 lines/mm diffraction grating. What is the wavelength (in nm) of the light?

Accepted Answer

Hey, everyone in this problem, a 750 line per millimeter diffraction grating produces an interference pattern that's shown in the figure we'll get to in just a minute. The pattern is projected onto a screen located 1.1 m away from the grating. And we are asked to determine the wavelength of the light causing this pattern. We're given four answer choices all in micrometers. Option A 0.6 option B 0.41 option C 0.32 and option D 0.2. So let's take a look at this diagram we're given and what we have is the position X and centimeters along the X axis and the intensity along the Y axis. And we're shown that the distance from the central peak or that central Maxima to the next peaks on either side. So to the first order fringe is 35.2 centimeters, then going from the first order fringe to the second order fringe. So the next peaks as we move outward from that central maxima is a distance of 50.7 centimeters. We wanna find the wavelength of light here and we're gonna start just by writing out some of the information we're given in the problem. So we're told that we have 750 lines per millimeter. What that means is that we can write d the distance between the slits in our dra action grade as one divided by this 750 value, one divided by 750. This is gonna be in millimeters and we wanna write this in our standard unit of meters. So we can write this as one divided by 750,000 m and just dividing by 1000. Yeah, we're also told the distance from the screen to our grading. That's gonna be L and that's 1.1 m. What we're looking for is Lambda that wavelength. All right. So we have some information about the distance to the fringe. We have D we have L we're looking for LAMBDA. So let's recall when we're dealing with the diffraction grading that we have a couple of different equations that we can use. The first tells us that D multiplied by sine as theta M is going to be equal to M multiplied by lambda. OK. So this is, this is an attractive equation at this point for this problem because it has that wavelength lambda in it that we want to find a Emma's gonna indicate the fringe. So we can figure that out using our diagram, we know D but this, the value we don't know yet. So we're gonna need another equation to use in order to figure out what that the value is gonna be. And recall, we have another equation that says ym is gonna be equal to L tangent of theta M. And now why this is the value that we can get from our diagram, that's that distance that we're given, we know L so that'll allow us to find the first, then we can use that other equation to find the wavelength. So let's start with the first order range. So we're gonna be starting with the peaks that are 35.2 centimeters away from that central maximum. We're gonna use the equation to find the first. OK. So we have that Y one is gonna be equal to L multiplied by a tangent of theta one. We're using M equals one here because we're talking about the first order of French. Now, why one that distance from the central maximum we know is 35.3 centimeters. We can convert to meters and write this as 0.352 m. This is going to be equal to the L value 1.1 m multiplied by a tangent of theta one. This tells us that tangent of theta one will be equal to 0.352 m divided by 1.1 m. The unit of meters will divide out. We can take the inverse tangent and we get that theta one it's gonna be approximately 17.74 degrees. So we have our, the value. Now, now we can get to calculating the wavelength when we do that, we are gonna find the sign of theta one is equal to one multiplied by Lambda. Again, using that equation with M is equal to one. We are gonna want to solve for Lambda. OK. So we're gonna have Lambda is equal to D sine of theta one and we know d we know theta one. So we have Lambda is going to be equal to one divided by 750,000 m multiplied bys of 17.74 degrees. When we work all of this out, we get that Lambda is gonna be equal to about 4.06 264 multiplied by 10 to the exponent negative 7 m. Our answer choices are in micrometer. So we wanna convert, we're gonna convert by multiplying by 10 to the exponent six. And this is gonna give us a Lambda value of approximately 0.41 micrometers when we consider two significant digits. And that is the correct number of significant digits for the answer choices because in the problem, the values we are given have two significant digits. OK? So we can't go any more than that. We're gonna need to stick to that two significant digits. Now we have our lambda value. This is enough to answer this as a multiple choice problem. If this was on a test and you were under a time crunch, you could go ahead and answer this question and move on. But what we're gonna do is we're actually gonna consider the second order fringe as well. Maybe because we were given this diagram and make sure that the wavelength we found matches what it would for the second order fringe because it should be the same. OK. If it's not, maybe we've done something wrong or maybe this is a trick question. So now we're gonna consider this second order fringe where M is equal to two and we're gonna perform those same calculations. So now we have Y two is equal to L tangent of theta two substituting in our values. Now, we have to be a little bit careful about Y two. It is not 50.7 centimeters. And remember that, that distance Y is always measured from that central maximum. So it has to be the distance from the central Maxima out to this point. So it's gonna be 35.2 centimeters plus 50.7 centimeters which gives us 85.9 centimeters converting to meters. We're gonna have Y two is 0.859 m. This is gonna be equal to that L value 1.1 m multiplied by a tangent of the two dividing by 1.1 m on both sides, tangent of the two will be equal to 0.859 m divided by 1.1 m. The unit of meters will divide out. We can take the inverse tangent on both sides. And we're gonna find that theta two is approximately equal to 37.99 degrees. We have the, now we go on to our calculation for the wavelength lambda. Again, we have D sine. This time of theta two is equal to that M value two multiplied by lamp. OK. M is two because we're dealing with the second or fridge. So if we think about calculating lambda, we're gonna have to divide both sides by two this time. So we're gonna have DS sign of beta two divided by two. OK. So LAMBDA will be equal to one half multiplied by one divided by 700,000 m multiplied by sign of 37.99 degrees. If we work all of this out on our calculator, we get that Lambda, it's approximately equal to 4.10349 multiplied by 10 to the exponent negative 7 m converting to micrometers and using the appropriate number of significant digits, which we said was two. We're gonna get that this wavelength Lambda is 0.41 micrometers. OK. So we are getting the same answer whether we look at the first order fringe or the second order fringe. So that's great. That's what it should be. If we go up to our answer choices, we can compare what we found and we see that the correct answer here is going to be option B that is the wavelength of light that is producing this pattern. Thanks everyone for watching. I hope this video helped see you in the next one.

FIGURE EX33.17 shows the interference pattern on a screen 1.0 m behind an 800 lines/mm diffraction grating. What is the wavelength (in nm) of the light?

Key Concepts

Diffraction Grating

Interference Pattern

Wavelength Calculation

Watch next