a screen is placed 37 cm from a single slit which is illuminated with 564 nm light. if the distance from the central maximum to the first minimum of the diffraction pattern is 2.7 mm, how wide is the slit in micrometer?

A parallel electron beam is focused on the sample in the same manner as in imaging mode to produce diffraction patterns. The beam width transmitted is controlled by a chosen area aperture, and the intermediate lens concentrates the diffraction pattern onto the screen.

Given data:

separating the single slit from the screen = L = 37cm

λ = 564 nm

distance between the first minimum and the central maximum = Z = 2.7mm

We are aware that the destructive diffraction angle is:

θ = pλ/a

where p is the minimum's order, p = 1 for the first minimum, and a is the slit's width, we get the following:

θ = (564nm/a)

We also know that a triangle rectangle can be created, with the distance between the slit and the screen serving as the adjacent cathetus to this angle and the distance between the first maximum and the first minimum serving as the opposite cathetus:

Tg(θ) = Z/L

Tan(564nm/a) = 2.7/37

The same units must first be used on the right side:

2.7mm = 0.27cm

Tg(636nm/a) = 0.27cm/43cm

636nm/a = Atg( 0.27cm/43cm ) = 0.406

a = 636nm/0.406 = 1,556.50nm

1 μm = 1000nm

then:

a = 1,556.50 nm = (1,556.50/1000) μm = 1.55650 μm

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