Physics · Ch 6 — Optics

Young's Double Slit Experiment

6.10.4

Young's Double Slit Experiment

In Young's double-slit experiment, a single monochromatic source $S$ illuminates a double slit $S_1,S_2$ (separation $d$ , both equidistant from $S$ , so both are automatically fed in phase), and a screen is placed at distance $D$ from the slits. For a screen point $P$ at height $y$ from the centre $O$ , dropping a perpendicular from $S_1$ onto $S_2P$ gives the path difference $\delta=S_2P-S_1P=S_2M=d\sin\theta$ ; using $\sin\theta\approx\tan\theta=y/D$ for a screen far from the slits gives $\boxed{\delta=dy/D}$ . A bright fringe (constructive interference) forms where $\delta=n\lambda$ ( $n=0,\pm1,\pm2,\ldots$ ), giving fringe position $\boxed{y_n=n\lambda D/d}$ (with $n=0$ the central bright fringe); a dark fringe (destructive interference) forms where $\delta=(2n-1)\lambda/2$ ( $n=1,2,\ldots$ ), giving $\boxed{y_n=(2n-1)\lambda D/(2d)}$ . Broad, clear, well-separated fringes require the screen to be as far from the slits as practical (large $D$ ), light of as large a wavelength as practical, and the two coherent slits as close together as practical (small $d$ ). Consecutive bright (or dark) fringes are separated by a constant bandwidth $\beta=\lambda D/d$ , so all fringes are equally spaced on either side of the …

Figure 6.58Young's double slit experiment (overview)

What this figure shows. A single source S illuminates an opaque screen carrying two narrow, closely spaced slits S1 and S2, equidistant from S; beyond the double slit, the overlapping wavefronts spreading from S1 and S2 fall on a distant viewing screen, where alternate equally-spaced bright and dark interference fringes are formed, with the brightest, central fringe appearing directly opposite the midpoint O between the two slits, where light from S1 and S2 arrives wi …

Figure 6.59Young's double slit experimental setup (path-difference geometry)

What this figure shows. Two coherent slits S1 and S2, separated by distance d, sit a distance D in front of a screen; C marks the midpoint between the slits and O the corresponding midpoint on the screen. For a general screen point P at height y from O, a perpendicular dropped from S1 onto the line S2P marks off the path difference delta = S2P - S1P as the short segment S2M; using the small angle theta that P makes at C, together with the approximation sin(theta) approximately equal to tan(theta) approximately equal to y/D for a screen far from the slits, gives delta = d sin(theta), which reduces to the working formula delta = dy/D used thr …

Figure 6.60Formation of bright and dark fringes

What this figure shows. Two overlapping wave trains, one from each slit, are drawn arriving at a screen point where they are exactly in phase (crest-on-crest), reinforcing each other into a bright band marked 'constructive interference'; a second point is drawn where the same two wave trains arrive exactly out of phase (crest-on-trough), cancelling each other into a dark band marked 'destructive interference' -- the diagram makes visually explicit exactly what a bright versus a dark fringe physically corresponds to in …

Figure 6.61Interference fringe pattern

What this figure shows. A horizontal axis marks screen position y in units of D/d, D lambda/d, 2D lambda/d and so on outward from the centre, lined up against the corresponding path difference (in units of lambda) and phase difference (in units of pi) at each point; below this, an intensity curve is plotted showing sharp, equally-spaced bright peaks of essentially the same height I0 at every integer multiple of D lambda/d, separated by dark troughs falling to zero exactly halfway between -- graphically summarising the regular, evenly-spaced bright-a …

Misc Example 6.28Fringe positions, fringe width, and the effect of moving the screen or immersing the setup in water

Worked out. In Young's double-slit experiment, the slits are 0.15 mm apart (d), illuminated with light of wavelength 450 nm, with the screen 2 m away (D). (i) The second bright fringe sits at y2 = 2 lambda D/d = 2 times 450e-9 times 2/0.15e-3 = 12e-3 m = 12 mm from the centre, and the third dark fringe sits at y3 = 2.5 lambda D/d = 15 mm. (ii) The fringe width is beta = lambda D/d = 450e-9 times 2/0.15e-3 = 6e-3 m = 6 mm. (iii) Moving the screen farther away (increasing D) increases the fringe width proportionally, since beta is directly proportional to D. (iv) Immersing the whole setup in water of refractive index 4/3 shortens the wavelength to lambda' = lambda/n, shrinking the fringe width proportionally to beta' = beta/n = 6/(4/3) = 4.5 mm -- fringes get closer together underwater exactly because the waveleng …

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