Simple Nature - Light and Matter

Double-slit diffraction of blue and red light example 12Blue light has a shorter wavelength than red. For a given doubleslitspacing d, the smaller value of λ/d for leads to smaller valuesof sin θ, and therefore to a more closely spaced set of diffractionfringes, (g)The correspondence principle example 13Let’s also consider how the equations for double-slit diffractionrelate to the correspondence principle. When the ratio λ/d is verysmall, we should recover the case of simple ray optics. Now if λ/dis small, sin θ must be small as well, and the spacing betweenthe diffraction fringes will be small as well. Although we have notproven it, the central fringe is always the brightest, and the fringesget dimmer and dimmer as we go farther from it. For small valuesof λ/d, the part of the diffraction pattern that is bright enough tobe detectable covers only a small range of angles. This is exactlywhat we would expect from ray optics: the rays passing throughthe two slits would remain parallel, and would continue movingin the θ = 0 direction. (In fact there would be images of the twoseparate slits on the screen, but our analysis was all in terms ofangles, so we should not expect it to address the issue of whetherthere is structure within a set of rays that are all traveling in theθ = 0 direction.)Spacing of the fringes at small angles example 14At small angles, we can use the approximation sin θ ≈ θ, whichis valid if θ is measured in radians. The equation for double-slitdiffraction becomes simplyλd = θ m,which can be solved for θ to giveθ = mλd.The difference in angle between successive fringes is the changein θ that results from changing m by plus or minus one,∆θ = λ dFor example, if we write θ 7 for the angle of the seventh brightfringe on one side of the central maximum and θ 8 for the neighboringone, we haveθ 8 − θ 7 = 8λd − 7λd= λ d.,r / Interpretation of the angularspacing ∆θ in example 14.It can be defined either frommaximum to maximum or fromminimum to minimum. Either way,the result is the same. It does notmake sense to try to interpret ∆θas the width of a fringe; one cansee from the graph and from theimage below that it is not obviouseither that such a thing is welldefined or that it would be thesame for all fringes.and similarly for any other neighboring pair of fringes.Section 12.5 Wave Optics 789