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Beam 660-nm light passes through closely spaced glass plates close norma Incidence For what Minimum nonzero value of the plate separation tne transmitted light brig...

Question

Beam 660-nm light passes through closely spaced glass plates close norma Incidence For what Minimum nonzero value of the plate separation tne transmitted light bright?showin the figure below-

beam 660-nm light passes through closely spaced glass plates close norma Incidence For what Minimum nonzero value of the plate separation tne transmitted light bright? showin the figure below-



Answers

A beam of 580 -nm light passes through two closely spaced glass plates as shown in Figure $\mathrm{P} 37.30 .$ For what minimum nonzero value of the plate separation $d$ is the transmitted light bright?

A theme of 580 nm light passes through too closely split closely spaced glass plates. As shown in the figure in your textbook. For what minimum non zero value of the plate separation is the transmitted light bright, So the past differences Delta here it's equal to lambda. So if this is true, the transmitted light will be bright. So since Delta is equal to two D, which is equal to lambda, We have the minimum distance is equal to λ over two, which is equal to 580 nm, So the minimum distance is equal to 200 and 90 nanometers.

So the transmitted light path difference to decide data. And so after minimum, you have to demon times, Santana's one. So for minimum D data will be a pirate, too. That will be quarter surgical to the wavelengths. The demon will be 1/2 the weight. That's a 1/2 of 5 95 80 nanometers, eyes equal to 290 nanometers.

In this problem, we are going to find the minimum amount of value off the expression D For with the transmitted light, Thio will appear bright. So in this case, we are going to calculate the value off the minimum. Now, if the party friends our path length of your friends which is a delta, is equals to lambda then the transmitted light will be bright. So we can say that this lambda is equals to R two D. Now we can write discretion as two days equals toe lambda and for minimum value off the we can write this equation as a de minimum is equals to land are divided by two by inserting the value for this lambda into the square and we can write 580 Multiply bitterness for minus 9 m, divided by two. So this is equals to 290 nanometer. So this is the minimum video off the separation of the blacks. Thank you.

This problem covers the concept of the interference due to the thin film. Okay so for the wave one and wave to the we've won the changing face is zero. Yeah. Okay. Now in way too uh I mean it is reflected back from a glass to hear. There will be a face to friends of mine and this way and when again it is reflected back. And the same a medium here from the class there will be again a change of change of angle Phase, angle off. So the total changing face angle is two points. So there will be. Uh huh. Zero. It feels different. So for constructive interference, delta which is the Cuban too uh reflective index of air into two times deep. Okay, that must be reverent. Mm times linda. Okay, so from this we can write B. Must be volunteer. I'm linda up on two times the refractive index off air of the minimum value. D. For the non zero value. When will make you were into one or one into the violence them tiles 580 nanometer upon to. So the minimum separation between the please is uh 219. No, no meat. Yeah.


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