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Light of 581 nm is incident on a pair of slits. A diffraction pattern appears on a projection screen 3 distance 1.2 m away from the slits. If the distance between t...

Question

Light of 581 nm is incident on a pair of slits. A diffraction pattern appears on a projection screen 3 distance 1.2 m away from the slits. If the distance between the central maximum to the second maximum is 42 mm; what is the spacing between the slits in htn?

Light of 581 nm is incident on a pair of slits. A diffraction pattern appears on a projection screen 3 distance 1.2 m away from the slits. If the distance between the central maximum to the second maximum is 42 mm; what is the spacing between the slits in htn?



Answers

Light of wavelength 633 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. What is the distance between
the two dark fringes on either side of the central bright fringe?

Hi in this given problem there is a double slit experiment, interference experiment in which the we've land of light that is given as lambda Is equal to 650 nanometer sleep separation That is given as D. is equal to 0.500 millimeter. And distance of screen from double slits that is B is equal to 3.00 meter. No, in the first part of the problem we have to find the distance of first bright fringe from the central bright so using an expression for the distance of and it bright fringe from the center and this expression is Y. N. Is equal to N. D. Lambda by D. So for the first bright range by one That is equal to one. Multiplied by D means 3.00 m multiplied by violent 650 Into a 10 for -9 m divided by 0.500 into standish four minus three m, cancelling this meter. Finally, the distance of first bright fringe from the central bright fringe comes out to be 3.9 into 10 days par -3 m. Or we can say this is why one is going to 3.9 millimeter. Which is the answer for the first part of this given problem here. No. In the second part of the problem we have to find the distance of 2nd dark fringe from the center and distance of and net dark fringe from the center is given as Why and dash that is two and -1 Into the lamb to buy two d. So for the second dark fringe Y two dash is equal to two times soft to -1 in today Means 3.00 m equivalent, 650 into 10 for minus nine m, divided by two times of small D, which is 0.500 Into a 10 for -3 m. Finally, the distance of second dark fringe from the central bright fringe comes out to be 5.8 five milli meter, which is the answer for the second part of this problem. Thank you.

The expression for distance of the first order Dark Prince. First order Dark means distance and very tennis. Why equals two and minus one developer too multiplied with linda l divided where? Mm. So let us substitute the value given in our question. So from here we can write we can right linda equals two. Whitey divided by and minus one day wherever to multiplied with. And so we will substitute the value given in our question. So why is given us 3.4 multiplied with 10 to the power 10 to the power of minus three m. Multiplied with the value of these. Given us 0.12 multiplied, would tend to the power of minus three m divided by and and is given as one minus one divided by two multiplayer. Where L. L. Is given as one point double five m. But resolving we will get the equivalent linda will come out to be 526 multiplied by 10 to the power of minus nine m. Or it will be equal to 526 nanometer. This is the required wavelength, which has been asked in our question, so the required length is 526 nanometer.

In the given problem, we have been given a light off. Waveland 4 78 nanometer. The virulent off the light is 4 78 nanometer, or for 70 it into tenders for minus nine meter. It is falling. One notable streaks. The first order bright band is appearing three millimeter away from the central bright man, so the stream limiter will behave like the fringe. Wait so we can take it as a fringe bit. Beat up 3.0 millimeter or we can try tips 3.0 in tow. Turn dish for minus tree meter the scene if 0.91 meter away from the slips. So this is the distance off screen from the cellars. We have to find the distance between us alerts for rich. The former use will be better for French with it's given us the Lambda by D. So the phone for slit separation will become delenda by beetle or weaken. Saves. The open nine millimeter into 4 78 into 10 takes a par minus nine meter and divided by 3100 into tenders for minus three meter. Yeah, calculate on the pains. The final answer comes out to be one point for five into tender for minus four meter. What we can see. 0.145 milli meter. This is a separation between the sticks. Thank you.


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