In a two-slit interference experiment the distance between the slits is 0.400 mm and the angle of the line from the slits to the m = 4 bright region on the screen is 0.407. Find the wavelength of the light used. Express your answer with the appropriate units. HA ? Value Units %D Submit Request Answer Part H Find the angle of the line from the slits to the m = 2 dark region on the screen. Express your answer in degrees.

In a two-slit interference experiment the distance between the slits is 0.400 mm and the angle of the line from the slits to the m = 4 bright region on the screen is 0.407. Find the wavelength of the light used. Express your answer with the appropriate units. HA ? Value Units %D Submit Request Answer Part H Find the angle of the line from the slits to the m = 2 dark region on the screen. Express your answer in degrees.

University Physics Volume 3

17th Edition

ISBN:9781938168185

Author:William Moebs, Jeff Sanny

Publisher:William Moebs, Jeff Sanny

Chapter4: Diffraction

Section: Chapter Questions

Problem 32P: The width of the central peak in a single-slit diffraction pattern is 5.0 mm. The wavelength of the...

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(a) Sodium vapor light averaging 589 nm in wavelength falls on a single slit of width 7.50 m. At what angle does it produces its second minimum? (b) What is the highest-order minimum produced?
What are the angular positions of the first and second minima in a diffraction pattern produced by a slit of width 0.20 mm that is illuminated by 400 nm light? What is the angular width of the central peak?
The width of the central peak in a single-slit diffraction pattern is 5.0 mm. The wavelength of the light is 600 nm, and the screen is 2.0 m from the slit. (a) What is the width of the slit? (b) Determine the ratio of the intensity at 4.5 mm from the center of the pattern to the intensity at the center.
For a three-slit interference pattern, find the ratio of the peak intensities of a secondary maximum to a principal maximum.
How far would you place a screen from the slit of the previous problem so that the second minimum is a distance of 2.5 mm from the center of the diffraction pattern?
In our study of diffraction by a single slit, we assume that the length of the slit is much larger than the width. What happens to the diffraction pattern if these two dimensions were comparable?
For a double-slit configuration where the slit separation is four times the slit width, how many interference fringes lie in the central peak of the diffraction pattern?
The thickness of an aluminum foil is measured using a Michelson interferometer that has its movable mirror mounted on a micrometer. There is a difference of 27 fringes in the observed interference pattern when the micrometer clamps down on the foil compared to when the micrometer is empty. Calculate the thickness of the foil?
Two slits of width 2 m, each in an opaque material, are separated by a center-to-center distance of 6 m. A monochromatic light of wavelength 450 nm is incident on the double-slit. One finds a combined interference and diffraction pattern on the screen. (a) How many peaks of the interference will be observed in the central maximum of the diffraction pattern? (b) How many peaks of the interference will be observed if the slit width is doubled while keeping the distance between the slits same? (c) How many peaks of interference will be observed if the slits are separated by twice the distance, that is, 12 m, while keeping the widths of the slits same? (d) What will happen in (a) if instead of 450-nm light another light of wavelength 680 nm is used? (e) What is the value of the ratio of the intensity of the central peak to the intensity of the next bright peak in (a)? (f) Does this ratio depend on the wavelength of the light? (g) Does this ratio depend on the width or separation of the slits?
Consider a single-slit diffraction pattern for =589 nm, projected on a screen that is 1.00 m from a slit of width 0.25 mm. How far from the center of the pattern are the centers of the first and second dark fringes?
An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a nonreflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?
Suppose you have a lens system that is to be used primarily for 700-nm red light. What is the second thinnest coating of fluorite (magnesium fluoride) that would be nonreflective for this wavelength?