Diffraction Review Today Single-slit diffraction review Multiple slit diffraction review Diffraction intensities Diffraction grating and spectroscopy Suary of single-slit diffraction Given light of wavelength passing through a slit of width a. There are dark fringes (diffraction inia at angles given by a sin where is an integer. Note this exactly the condition for constructive interference between the rays fro the top and botto of the slit. Also note the pattern gets wider as the slit gets narrower. The bright fringes are roughly half-way between the dark fringes. (Not exactly but close enough. Double-slit diffraction slits of zero width 1 slit of width a 5 Two-slit and one-slit patterns Actual photograph: (a two slits (bone list covered slits of width a 5 (Figure 36-15 fro text page 1003. Scaling of diffraction patterns Notice a coon feature of interference and diffraction patterns: The large-scale features of the pattern are deterined by the sall-scale regularities of the object, and vice-versa. Hologras and X-ray diffraction patterns are exaples. Single-slit Intensity We know where to find the dark fringes in the single-slit pattern. But can we calculate the actual intensity at a general point? Yes, using the phasor ethod. Book gives result on page 998: I sinα α Here I is the intensity at angle on the screen. is the intensity at the central axiu. The angle α φ /, and φ is the phase difference between the rays fro top and botto of slit.
Phasors for Single Slit Break up the slit into any tiny zones, giving any rays of light, which coe together on the screen. φ φ Phase difference between adjacent rays Phase difference between top and botto rays Aplitude at center Su of all phasors Aplitude at angle, get fro diagra First Maxiu and Miniu Reeber of course the relation between phase difference and path difference π φ ( asin φ 0 0 φ (π asin Intensity for Single Slit Rφ Rsin( φ / I sin( φ / φ 4sin ( φ / φ Which gives the textbook result: I sinα α Q.36-3 Light of wavelength 600 n is incident on a slit of width 30 µ. What is the diffraction angle for the first diffraction iniu (dark fringe? (1.01 rad (.0 rad (3.03 rad (4.04 rad Q.36-3 Light of wavelength 600 n is incident on a slit of width 30 µ. What is the diffraction angle for the first diffraction iniu (dark fringe? Q.36-4 In the sae diffraction experient, where the angle for the first iniu is.0 rad, consider a point on the screen where the phasor diagra for the interfering waves is as shown. Solution: asin 600 10 sin a 30 10 9 sin.0 rad 1.15 6 0 10 3 Roughly what diffraction angle is this point found at? (1.01 rad (.0 rad (3.03 rad (4.04 rad (1.01 rad (.0 rad (3.03 rad (4.04 rad
Q.36-4 In the sae diffraction experient, where the angle for the first iniu is.0 rad, consider a point on the screen where the phasor diagra for the interfering waves is as shown. Roughly what is this diffraction angle? Five Slits Solution: For.0 rad, we have a closed circle. Thus for.04 rad, we would have the second closed circle. So this point is about idway between:.03 rad. Note we still have ax / d But ore slits akes the peaks sharper. ƒ (1.01 rad (.0 rad (3.03 rad (4.04 rad For any slits, we get a diffraction grating. Diffraction Grating Wavelength Resolution of a Grating Width of sharp lines Resolving Power Positions of axius: d sin Sall angles: / d / d Widths of sharp axius: Wavelengths just resolved: / Nd / d / Nd / N Resolving power definition: R So we get: R N
Spectroscopy Resolving Power The grating spectroeter Many grooves gives sharp lines High resolution: Measure wavelengths precisely Distinguish between alost-equal wavelengths Many uses in science and technology Diffraction grating G Very sharp axiu when all rays are in phase. d sin For N rulings, the resolving power is R N Spectru of a Hot Gas Using a grating spectroeter to look at light fro a hot gas we see spectru lines. Only certain specific wavelengths (colors are eitted by any particular gas. This characteristic spectru can be used as an identification tag for specific atos and olecules. The figure shows four bright lines of the hydrogen ato. xaple: Yellow sodiu vapor lines Proble 36-50 The strong yellow lines in the sodiu spectru are at wavelengths 589.0 n and 589.6 n. How any rulings are needed in a diffraction grating to resolve these lines in second order? We need R R N 589 0.6 n n 98 98 But so N 491 R Diffraction grating recap Position of lines is deterined by separation of rulings. d sin One last exaple: Proble 35-91 Sharpness of lines is deterined by nuber of rulings. / Nd Resolving power is deterined by nuber of rulings and order of line. R N A icrowave transitter sends waves across a lake to a receiver on the other side. D>>a and >a. Find x to ake the signal a axiu. (Hint: reflection fro surface gives phase change.
xaple continued Direct ray: ( ( a x L d D + a x D + D Reflected ray: ( ( a x L r D + a + x D + + D ( a + x ( a x ax Path difference: L Lr Ld D D D After accounting for phase change, condition ax D L so x for axiu is: D 4a