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Setler [38]
3 years ago
15

n optician is performing Young's double-slit experiment for her clients. She directs a beam of monochromatic light to a pair of

parallel slits, which are separated by 0.134 mm from each other. The portion of this light that passes through the slits goes on to form an interference pattern upon a screen, which is 4.50 meters distant. The light is characterized by a wavelength of 553 nm. (a) What is the optical path-length difference (in µm) that corresponds to the fourth-order bright fringe on the screen? (This is the fourth fringe, not counting the central bright band, that one encounters moving from the center out to one side.)
Physics
1 answer:
steposvetlana [31]3 years ago
7 0

Answer:

Explanation:

This is a problem based on interference pf light waves.

wavelength of light λ = 553 nm

slit separation d = .134 x 10⁻³ m

screen distance D = 4.5 m

for fourth order bright fringe, path- length  difference = 4 x λ

= 4 x 553

= 2212 nm .

= 2.212 μm

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