Physics · Ch 6 — Optics

Diffraction at Single Slit

6.11.2

Diffraction at Single Slit

A parallel beam of light falling normally on a single slit $AB$ of width $a$ produces, on a distant screen, a diffraction pattern whose intensity at a general point $P$ (making angle $\theta$ with the normal from the slit's centre $C$ ) is found by dividing the slit into pairs of 'corresponding points' and analysing their interference. Dividing the slit into two halves of width $a/2$ each, the path difference between a corresponding pair is $(a/2)\sin\theta$ ; setting this equal to $\lambda/2$ (destructive interference for the pair) gives the condition for the first minimum: $\boxed{a\sin\theta=\lambda}$ . Generalising to $2n$ equal parts (an even number), every corresponding pair cancels when $a\sin\theta=n\lambda$ , giving the $n$ th-order minimum condition $\boxed{a\sin\theta=n\lambda}$ ( $n=1,2,3,\ldots$ ). Dividing the slit instead into an odd number of equal parts leaves one part un-cancelled, producing a much fainter secondary maximum at $a\sin\theta=(2n+1)\lambda/2$ , $n=0,1,2,\ldots$ (the ce …

Figure 6.63Diffraction at single slit

What this figure shows. A parallel beam of light falls normally on a single slit AB of width a, with C marking the slit's centre; a straight line through C perpendicular to the slit meets a distant screen at O. Light reaching a general screen point P, in the geometrically-shadowed region beyond the slit's straight-through projection, arrives from every point across the slit's width along paths that can be treated as parallel, all making the same angle theta with the line CO -- this angle theta is the single variable the entire single-slit diffrac …

Figure 6.64Corresponding points

What this figure shows. The slit width a is divided into two equal halves, AC and CB, each of width a/2; a pair of 'corresponding points', one in each half separated by exactly a/2, is marked, and the path difference between the light reaching point P from this pair of corresponding points is shown to be (a/2) sin(theta). Setting this path difference equal to lambda/2 -- the condition for the pair to cancel by destructive interference -- and noting every other such corresponding pair across the slit cancels in exactly the same way, gives the …

Misc Example 6.31Angular spread and second-minimum distance for a single slit

Worked out. Light of wavelength 500 nm passes through a 0.2 mm wide slit, forming a diffraction pattern on a screen 60 cm away. (i) The angular spread of the central maximum, up to the first minimum (n=1), is sin(theta) = lambda/a = 500e-9/0.2e-3 = 2.5e-3, i.e. theta approximately 0.0025 rad. (ii) Using y = n(lambda)D/a for small angles, the first minimum sits at y1 = lambda D/a = 500e-9 times 0.6/0.2e-3 = 1.5e-3 m = 1.5 mm, and the second minimum (n=2) sits at y2 = 2 lambda D/a = 3.0e-3 m = 3 mm; the distance between the central maximum and the second minimum is therefore y2 - y1... actually y2 itself measured from the centre, 3 mm, while the gap between the first and second minima is y2 - y1 = 1.5 mm -- demonstrating that away from the extra-wide central maximum, every subsequent si …

Misc Example 6.32Slit width from the angular position of the first minimum

Worked out. A single slit produces a diffraction pattern whose central maximum, for light of wavelength 5000 angstrom, extends out to a first-minimum angle of 30 degrees (sin 30 degrees = 0.5). Rewriting the first-minimum condition a sin(theta) = lambda for the slit width, a = lambda/sin(theta) = 5000e-10/0.5 = 1e-6 m = 0.001 mm -- an extremely narrow slit, only about a micrometre wide, needed to spread the central diffraction maximum out to such a large 30-degree half-angle, consistent with the general rule that a narrower slit produces a wider (more spread …

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