Consider a single-slit diffraction pattern for λ=589nm, 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?

Question

Physics

Joanna

5 January, 22:45

Consider a single-slit diffraction pattern for λ=589nm, 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?

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Answers (1)

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Douglas · Answer 1

Answer: i) 2.356 * 10^-3 m = 2.356mm, ii) 4.712 * 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R * (mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589*10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 * 589*10^-9 * 1/0.00025

y = 589*10^-9/0.00025

y = 2,356,000 * 10^-9

y = 2.356 * 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 * 2 * 589*10^-9/0.00025

y = 1178 * 10^-9 / 0.00025

y = 4,712,000 * 10^-9

y = 4.712 * 10^-3 m = 4.712mm