A screen is placed 43 cm from a single slit which is illuminated with 636 nm light. If the distance from the central maximum to the first minimum of the diffraction pattern is 3.8 mm, how wide is the slit in micrometer

The width is 1.25692 μm

Explanation:

The data that we have here is:

Distance between the single slit to the screen = L = 43cm

λ = 636 nm

Distance from the central maximum to the first minimum = Z =  3.8mm

We know that the angle for the destructive diffraction is:

θ = pλ/a

where p is the order of the minimum, for the first minimum we have p = 1, and a is the width of the slit,

then we have:

θ = (636nm/a)

And we also know that we can construct a triangle rectangle, where the adjacent cathetus to this angle is the distance between the slit and the screen, and the opposite cathetus is the distance between the first maximum and the first minimum:

Tg(θ) =  Z/L

Tan(636nm/a) = 3.8cm/43cm

First, we need to use the same units in the right side:

3.8mm = 0.38cm

Tg(636nm/a)  = 0.38cm/43cm

636nm/a = Atg(  0.38cm/43cm ) = 0.506

a = 636nm/0.506 = 1,256.92 nm

1 μm = 1000nm

then:

a = 1,256.92 nm = (1,256.92/1000) μm = 1.25692 μm

v=ir = 3x15 =45 v