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frutty [35]
3 years ago
6

A pair of narrow, parallel slits separated by 0.254 mm are illuminated by green light of wavelength 559 nm. An interference patt

ern is observed on a screen 1.71 m away from the plane of the slits. Calculate the distance from the central maximum to the first bright region on either side of the central maximum. Answer in mm. Answer in units of mm.
Physics
1 answer:
UkoKoshka [18]3 years ago
4 0

Answer: 3.76 mm

Explanation: The formulae that defines the position of a nth fringe from the center of a double slit interference experiment is given below as

y =R×mλ/d

Where y = distance between the nth fringe and the center

R = distance between slits and screen = 1.71 m

m = position of fringe = 1

d = distance between slits = 0.254 mm = 0.000254m

λ = wavelength of light = 559 nm = 559×10^-9m

By substituting parameters, we have that

y = 1.71×1×559×10^-9/0.000254

y = 0.00000095589 / 0.000254

y = 0.00376m = 3.76 mm

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<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

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