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Red laser light ( = 650 nm) is incident on & diffraction grating patlern is formned on grating charectcnzedb} JOleta screen 2.,0 m behind the grating Adlnt (2 ...

Question

Red laser light ( = 650 nm) is incident on & diffraction grating patlern is formned on grating charectcnzedb} JOleta screen 2.,0 m behind the grating Adlnt (2 pts) Determine the grating constant &. Express _ Jour ansuer in melen(4 pts) Determine and 0z. the angles of the first-ordcr and answer in degrees: sccond-ordcr maxima obecned Extesjoapts) Determine the positions and Yz of the obscrvcd maxima on the scrccn;

Red laser light ( = 650 nm) is incident on & diffraction grating patlern is formned on grating charectcnzedb} JOleta screen 2.,0 m behind the grating Adlnt (2 pts) Determine the grating constant &. Express _ Jour ansuer in melen (4 pts) Determine and 0z. the angles of the first-ordcr and answer in degrees: sccond-ordcr maxima obecned Extesjoa pts) Determine the positions and Yz of the obscrvcd maxima on the scrccn;



Answers

Red light falls normally on a diffraction grating ruled 4000 lines/cm, and the second-order image is diffracted $34.0^{\circ}$ from the normal. Compute the wavelength of the light. From the grating equation $m \lambda=a \sin \theta_{\mathrm{m}}$ $$ \lambda=\frac{a \sin \theta_{2}}{2}=\frac{\left(\frac{1}{4000} \mathrm{~cm}\right)(0.559)}{2}=6.99 \times 10^{-5} \mathrm{~cm}=699 \mathrm{~nm} $$

Okay during his 35 problem. 45 years. So we have monochromatic light, which is these red lines and their incoming falling on a transmission to fraction grading at an angle by. So you through these are all entering at an angle Phi um, because there are parallel and we went to show that equation 13 for diffraction. Maximum must be replaced by this equation here of the sign I plus sign later in this sequel, plus or minus for positive ends. Okay, so So let's start doing some geometry here. So let's look at this first path. We have this first top top Redpath coming in, hitting it at angles, respect to the horizontal of fei and transmitting off with the respect cherie angle of data. So we want to figure out the differential path links here. So this one is also gonna come off that something go out here such that this angle is also data. So let's figure out how much extra path was traveled. So we extend these this rectangle out, we should see that. Um um um um Mark, that said years of this section in this section are extra path that has traveled then the next long. So we want to figure out we're trying to find the path Link Jane is gonna be this distance here. We're gonna call. Don't, uh, fell, too. In this distance up here, which we call, don't l one. So now let's figure out how to find his. Well, if we look at this triangle, it's no sorry. This is don't tell one. So if we look at the triangle Mead Um, sorry. I'm just trying to figure out this triangle we want to draw All. So it's this triangle right here. Her as an angle phi here. So what this tells us is that Delta l, too was given by the sign five. And we can do the same thing over here with the struggle. So one triangle there. One triangle here. And from that we can see that belts L one is given by the sign data. Okay, So if passing through the grating at the higher rate travels and extra distance out l one plus felt out to which we're gonna define dealt out. So each ray above travels and extra distance delta l from the re below it. So the total pathway difference we just set this equal to ginger times the Webley, and we're gonna add a plus or minus. And this is the same as what we just added. He signed by, plus the Earth sign data. So this is exactly what we wanted to show because in lander pulsar minus. Okay, well, so let's go back and do that again. So this right here, girl, a sentry green. So this right here, this by in this along here is D. This year is data. So we can find this value in this value respectively, by using science. That's that's what we did. And we set that equal to endanger times wave links to get the interference pattern. How we like. Okay, so moving on the part b, it says, What's up with this? Plus a minus sign. Right? So the poster minus sign just allows for the incident angle and attracted angle toe have either positive or negative values, and we can also think of it as positive or negative integers right on either side of coming in. That's all that is. Is a louse positive or negative? They flash five cool and lastly, Part C says green light with a wavelength. 5 50 millimeters from Adam is incident at an angle of 15 degrees to normal on the traction grading. And it's a 5000 lines, percent lines per centimeter. So we want to set find first angles that wish the first maximum occur. Okay, well, we know that sign Data is in named of D. So data is the inverse sine of in the over D. And we said the first maximum So was take the inverse sign here of Yeah, you could. So this is for the normal diffraction grating, right? But we just figured out that we need to replace it with this, so I'm actually going to erase what I have here already. Tried it. Okay, so now we have the sign. Hi. Plus the sign later equals and possibly remember. Okay, so from miss, we can figure out science data is negative. Sci fi plus or minus in named over D. So, like I am. So now we can take the under sign, and we get data. What we want to figure out is the inverse sine off negative sign. Bye. And we're using positive one for our ever pulsar minus one and equals one. So this is just Lambda over D. So it's progressing. So now we got inverse sine a view tive sign of 15 degrees plus or minus 400 groups. Not 400 mrs 5 50 5 59 centimeters over. What's our D? Well, that is 0.1 meters, divided by 5000 lines. So if we plug this end, our calculator for both plus in likeness data's of 0.93 degrees and negative 32 degrees.

Hello, friends. Here it is given, like off a violent 5 20 nanometer that is 5 20 in tow, 10 to the power minus 9 m passes through regretting having 4000 lines per centimeter regretting having 4000 lines percent emitter and false on the screen located 1.6 m from the grating. So l is given 1.6 m in first part. We have to draw the diagram. So we is schematically represent the process and they will all known distances on angle, right? Be part angular Deflection for second bright went you tough work Second bright went See part right distance your separation off Second order right from central right French No to see it. This is the grating having 4000 lines and one cent emitted These are the wavefront off the light scrape at the distance off 1.6 m diffraction grating d to be one upon 4000 scientific er that is 2.5 micrometers. So with off the grating is 2.5 micrometers right wavelength is given 5 20 nanometer. This is central. Right? Finch first started second daughter and third order. This is why tree she the separation between the slits DS in the distance off one centimeter, there are 4000 lines. So it is to be 0.25 and tow 10 to the power ministries and Demeter, or 2.5 into 10 to the power when a 6 m and Guler separation for I met order. Bright French sign in Worse off, I am land up on the for second part. Angular separation off. Second order by French theater too. Sign in Worse off. Two in tow. 5. 20 into 10 to the power nine upon 2.5 in tow. 10 to the power minus six. It will be 25 degree the separation between second order Maxima and Central right French. Why, you will be, um lambda l upon the ed is Yeah. 1.6 m. Andi is 125 10 to the par minus six. So you will get by two to be 0.73 m. That's all for it. Thanks for watching it

And this problem we have to calculate the number of maximums formed in the friction patron and the corresponding angles. So let's start by writing an expression for the number of maximum as uh and maximum is less than or equal to D. Divided by lambda. Here this is the space creating and lambda is the wavelength of the light. We can wind this A. D. As it is equals to one divided by capital and where this capital and is the number of uh lions. But unit length by setting the expression into this world. And we can. Why? This one is ah one divide by 6000 multiple laboratories. Power to per meter, Divide by the Celinda, which is equals to 632 pointed Multiply witnessed for -9 m. So from here we can write the value for this uh and maximum As in maximum should be equals to two. Now we can ride the total number of uh next month is uh total number of maxima will be equals to two times of the two because uh the number of two represents the number of maximus from the one side from from the central maxima to the one side. So we have two sides from the central maximum. So we have to multiply it by two plus the central maxima. So from here we will get the total number of maximum says five mm. We can sign the order for these number of maximums. As for the central maximum we will have uh and as energy equals to one and four. Uh The others as a plus one is one and the plus one is two. We need to calculate the angles for these. Uh Makes more so for the central maximum for central maximum our maximum we can wipe the angle as a theater is equals to sign universe of and linda divide by D. So for central make somehow we can wipe this question is three to not as equals two signing orders of zero multiplied by lambda divided by T. So from here we will get hit A zero Z equals to zero degree. Now for first daughter maxima on either side of the central maximum we can white and is equals two plus one is one and for this case we can ride the theater is plus one is one is equals to sine inverse into. Ah here we have one Into Lambda which is equals to 632.8 multiply bitterness bar minus nine m divided by D. Which is equals two. One divide by the number of lines which is equals to 6000 multiply bitterness power to permit. Er So from here we will get to the angle for this uh Uh maximum is 22.3 degree. Similarly, we can wait for any Z equals two plus minus two. For the second order maximum, the angle will be plus minus three to plus minus two and this is equal to sine inverse of To into the violent, which is equals two, 632.8 multiple eyewitness par -9 m, divide by one, divide by One, divide by into 606,000. Sorry 6000 multiply waiters for two per meter. So square work clothes from here we will get the value for this. 30 plus one is two As stated plus ministries equals two 49.4 Degree. Thank you.


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