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Ch 36: Diffraction

Chapter 35, Problem 36

Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens, the distance from the central maximum to the first minimum is 8.65 mm. What is the width of the slit?

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Hello, fellow physicists today, we're going to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. So monochromatic light of wavelength nanometers falls on a narrow slit and then passes through a lens with a focal length of 70.0 centimeters. Determine the slit width if the distance between the first order maxima from the center of the screen is 7.55 millimeters. OK. So first off, we need to recall the equation for the distance from the central maxima or it's also called Mima. OK. So let's call it equation one. So equation one for the distance between the distance from the central max and up is Y subscript M is equal to the focal length multiplied by the mimma value which is some integer M represented by M multiplied by the wavelength which is represented by lambda divided by the slit width, which is represented by A. So let's make a note that M in this case, the minimum value is one because we're dealing with a first order M so we can rewrite equation one to say that why subscript one since M equals one, so Y subscript one equals the focal length multiplied by one multiplied by wavelength divided by the slit width. OK. So now we need to rearrange equation one to solve for a, the slit. So A equals X multiplied by LAMBDA divided by Y one. OK. And let's remember that Y one is the distance between the Mima from the center of the screen and that X is the focal link and LAMBDA is the wavelength just to be clear. So let's plug in all of our known values. So note that we, the focal length is given to us in centimeters, but we need to convert that to meters. And when we convert it to meters, we get 0.700 meters multiplied by the wavelength which we need to also convert the wavelength from nanometers to meters. So all we have to do is take 678 multiplied by 10 to the power of negative nine meters divided by Y one, which is the distance between the minimum from the center of the screen, which was given to us as 7.55 millimeters. But we need to convert that to meters. So when we convert that to meters, we get 7.55 multiplied by 10 to the power of negative 3 m. So when you plug that into a calculator you should get that the slit width equals 6.29 multiplied by 10 to the power of negative 5 m. But we need to convert this to micrometers because all of our multiple choice answers are micrometers. So in order to do that, all we have to do is multiply by 10 to the power of six. So our focal or I should say correction. So our slit width her A should equal 62.9 micrometers. So A equals slit width equals 62.9 micrometers. OK. So that means that our final answer has to be a 62.9 micrometers. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.
Related Practice
Textbook Question
Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum 1u = 0°2 is I0. What is the distance on the screen from the center of the central maximum (a) to the first minimum
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Textbook Question
Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum 1u = 0°2 is I0. What is the distance on the screen from the center of the central maximum (b) to the point where the intensity has fallen to I0>2?
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Textbook Question
If the planes of a crystal are 3.50 Å (1 Å = 10-10 m = 1 Ångstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0°, and in what part of the electromagnetic spectrum do these waves lie?
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Textbook Question
Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin u can be? What does this tell you is the largest that m can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?
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Textbook Question
Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point O in Fig. 36.5a . The microphone is then moved in a direction perpendicular to the line from the center of the slit to point O. At what distances from O will the intensity detected by the microphone be zero?
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Textbook Question
A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at +-61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier?
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