For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ=680nm) through the slit onto a screen 5.5 m away, a diffraction pattern appears. The bright band in the center of the pattern is 8.0 cm wide. What is the width of the slit?

4.675×10^-6 m

Explanation: this is a question under the pattern gotten from an interference experiment.

The formulae below defines the experiment given

y = Rmλ/d

Where y = distance between the center fringe and any other fringe in the pattern = 8cm = 0.08m

m = position of fringe = 1 ( this is because we have only one central band)

λ = wavelength of light used for experiment = 680nm = 6.8×10^-7m

d = width of slit =?

R = distance between slits and screen = 5.5 cm = 0.055m

By substituting the parameters, we have that

0.08 = 0.055 × 1 × 6.8×10^-7/d

By cross multiplying, we have that

0.08 × d = 0.055 × 1 × 6.8×10^-7

d = 0.055 × 1 × 6.8×10^-7/ 0.08

d = 3.74 × 10^-7/ 0.08

d = 46.75×10^-7

d = 4.675×10^-6 m

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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