A double slit produces a diffraction pattern that is a combination of single and double slit interference. Find the ratio of the width of the slits (D) to the separation between them (d), if the first minimum of the single slit pattern falls on the seventh-order maximum of the double slit pattern. (This will greatly reduce the intensity of the seventh-order maximum. Assume the central maxima of the single slit and double slit interference patterns are located at the same place on the screen.)

d = 5D

Explanation:

The destructive interference for a single slit is given by the formula;

Dsin∅ = nλ  1

where;

D = slit width

n = order of the minima

θ = Angle to the original direction

λ= wavelength of light

For first minima, n = 1 and θ = θ₁

Substituting into equation 1, we have

Dsinθ₁ = λ 2

The destructive interference for a double slit is given by the formula;

dsin∅ = mλ  3

where;

d = distance between the slit

∅  = Angle between the path

m = order of interference

λ  = wavelength of light

For the fifth maxima, m = 5 and ∅ = ∅₅

Equation 3 becomes;

dsin∅₅ = 5λ   4

Dividing equation 2 by equation 4, we have

Dsinθ₁/dsin∅₅  = λ/5λ

Since the angles and the wavelength are the same, the equation reduces to;

D/d =1/5

d = 5D

Therefore, the split separation is 5 times the slit width

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beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge