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University of Wisconsin - Milwaukee

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Problem 51 Hard Difficulty

A 15.0 -cm-long grating has $6.00 \times 10^{3}$ slits per centimeter. Can two lines of wavelengths 600.000 $\mathrm{nm}$ and 600.003 $\mathrm{nm}$ be separated with this grating? Explain.

Answer

yes

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Video Transcript

so we have the length of the grating. A viable for the diffraction is 15 centimetre. The number of slips per unit length of the grating is six times 10 to the three lines. Press send emitter, which is equivalent to six times to the five lines perimeter. The creating spacing of the grading is going to be de course one over and which is going to give us 1.67 times 10 to the negative six Peter, which is the spacing between two quitting, Okay. But the value of D is what that reminds the maximum order of diffraction that the grating element can have for a particular weapon and and that particular were blanked. In our case is Lambda I call 600 nada meter. So I'm just going to use Lambda equivalent to Lambda One, which is 600 madam eater, or 600 times 10 to the negative nine meter. All right. Also, when the light is normal to the grating element that is there, I cost 90 degree. We obtain the maximum possible order of diffraction. So what that means is, if you remember for the diffraction d sign David, I am. It's going to be m lambda Share M is the order of diffraction. He's displacing. And that I am is the angle. All right? So from here, we can actually write down I am For the maximum, it's going to be d sign 90 degrees because we're considering the normal incidents over the web land. Okay, so from here, signed nineties one. And these want 167 times to the negative six. Lambda is 600 times 10 to the negative nine. So from here, we're actually going to get 2.78 But this value needs to be an in desert, so we actually keep too. Okay, so this is the valley we're going to keep. Now. The total number of slits in the grading is going to be an because 0.15 which is the length of the two off the creating times six times 10 to the five, which is number off lines per unit length. Okay. And this is going to give us nine times 10 to the for slips. Therefore the maximum available resolving power of the grating is going to be our because and max times and which is two times nine times 10 to the fore. And this is going to be 1.8 times 10 to the five. So this is the viable maximum resolving power. Now we know that were given that lamb, that one. Okay, Lambda one is 600 nanometer, and another world link that is given is 600 0.3 unanimity. Er. So from here, we can see that Delta Lambda, which is going to be Linda to minus Lambda one is going to be 0.3 in an emitter. Okay, so from seer, actually, the resulting power, which is the ratio off Lambda by Delta Lambda, is going to be 600 times 10 to the negative nine over Sierra 0.3 times 10 to the negative nine. And this is going to be actually combat into two times 10 to the five. Okay, so this is the value that the resulting power, which we require and you can see that actually the inviolable reserving power are is actually it less than I required. You can go back to the page. Second, you can see our is 1.8 times 10 to the five and are required is two times to notify. So says are is less than I require. The grating cannot be used for two separate delights. So the answer to this question whether it can separate the lances, no.

University of Wisconsin - Milwaukee
Top Physics 103 Educators
Elyse G.

Cornell University

Farnaz M.

Other Schools

Zachary M.

Hope College

Aspen F.

University of Sheffield