In a double-slit interference experiment, the slit separation is 2.39 pm, the li

Question

In a double-slit interference experiment, the slit separation is 2.39 pm, the light wavelength is 532 nm, and the separation between the slits and the screen is 4.25 m. (a) What is the angle between the center and the third side bright fringe? If we decrease the light frequency to 94.9% of its initial value, (b) does the third side bright fringe move along the screen toward or away from the pattern's center and (c) how far does it move? (a) Number Units (b) (c) Number Units Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

Here ,

slit seperation ,d = 2.39 um

d = 2.39 *10^-6 m

wavelength = 532 *10^-9 m

screen distance , D = 4.25 m

a)

let the angle is theta

as d * sin(theta) = m * wavelength

2.39 *10^-6 * sin(theta) = 3 * 532 *10^-9

theta = 41.9 degree

the angle will be 41.9 degree

b)

for the light frequency to be 94.9% of initial frequency

wavelength = 532 nm/(0.949)

wavelength = 560.6 nm

as the wavelength is increased

the third bright fringe will move away from the centre.

c)

let the angle is theta

change in distance = 3 * (560.6 - 532) *10^-9 * 4.25/( 2.39 *10^-6)

change in distance = 0.152 m

the fringe will move 0.152 m

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