Question 11 1 pts Instruction: Input numerical values only;...

Question 11 1 pts Instruction: Input numerical values only; no unit. Keep two decimal points for your numerical answer. Example: -3.14, -314.51, 3.14, 314.51 You pass $633 \times 10^{-9}$ m laser light through a narrow slit and observe the diffraction pattern on a screen. The distance on the screen between the centers of the first minima on either side of the central bright fringe is $\Delta h = 73 \times 10^{-3}$ m. If the width of the slit is $w = 2.5 \times 10^{-4}$ m, what is the distance L between the slit and the screen (unit in meters, m)?

Transcript

0:00 Hi there.

00:01 So for this problem we have a sleet that has a distance wide that is equal to 0 .240 millimeters and is illuminated by parallel light rays of a wavelength that is equal to 540 nanometers.

00:31 The diffraction pattern is observed on a screen that is 3 meters from the lead, we're going to call that this distance alt, that is 3 meters.

00:44 And intensity at the center of the central maximum is going to be equal to theta, that is equal to 0 degrees at the right.

01:02 And at that point, the intensity of 0 is equal to 6 times 10, to the minus 6 watts per centimeters to the square.

01:20 Now for part a of this problem, we are asked about, what is the distance on the screen from the center of the central maximum to the first minimum? so to solve this exercise, it is necessary to apply the concepts on the principle of superposition.

01:44 Specifically constructive interference.

01:47 Now, constructive interference is defined by the distance, the width of the slit, the width of the slit, sign of theta, and this is going to be equal to the order m times the wavelength, where m is called interference order, and is the number of wavelengths by which the two paths differed.

02:11 And lambda, of course, is the wavelength.

02:13 And d is the distance of the wave.

02:17 Now for smaller angles, we know that the sign of theta is approximate to the tangent of theta.

02:30 So we are going to have now that this is the distance d, tangent of theta, is equal to m times the wavelength.

02:41 Now, from the trigonometry properties, it is understood that so the in this case the length measured vertically by the reason of the distance so we're going to have that the tangent is the height divided by the distance separation else so that is going to be equal to m times the wavelength so if we solve for the distance why that is going to be that's the distance on the screen so if we solve for that we are going to obtain that this is m times the wavelength times the distance l divided by d.

03:24 So in here we can substitute all of the values that we have for this problem.

03:29 We're going to have that this is for the first order.

03:33 So m is equal to 1.

03:36 The wavelength is the one given 540 nanometers, which means tens to the minus 9 meters.

03:44 And this times the length that in this case is also given, that length from the sleet to the screen is 3 meters, and this divided by the distance d.

04:04 That in this case, it is also given that is the width of 0 .240 millimeters, that in meters is 10 to the minus 3 meters.

04:17 So using the calculator, we obtain a value of.

04:37 So we obtained that the value of y is equal to 6 .75 times 10 to the minus 3 meters, that we can also write as 6 .75 millimeters.

04:53 So that's a solution for part a of this problem...

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