Single slit diffraction

Transcription

1 Single slit diffraction Book page

2 Review double slit Core Assume paths of the two rays are parallel This is a good assumption if D >>> d PD = R 2 R 1 = dsin θ since sin θ = PD d Constructive interference occurs, when PD = nλ nλ = dsin θ n = 0, 1, 2, 3,. Destructive interference occurs when PD = n λ n λ = dsin θ n = 0, 1, 2, 3,

3 Rays are not parallel but converge We can still use the same mathematical approach sin θ = PD d tan θ = y D If θ is small, sin θ ~ tanθ PD d = y D PD = yd D where y = distance between adjacent nodal lines, d = slit separation, D = cgrahamphysics.com distance between 2016 slit and screen

4 PD = nλ For constructive interference nλ = y nd or y D n = nλd d Fringe separation: y = X n+1 X n = λd d d = λd y

5 Requirements Single slit diffraction occurs when a wave is incident on a slit of about the same size as the wavelength The diffraction pattern consists of a series of bright and dark fringes Note Central maxima has twice the angular width of the secondary maxima, each of which have same width Intensity falls significantly intensity of 1 st secondary maxima ~ 5% of I 0 2 ~ 2% 3 ~ 1% No light reaches the center of a minima, but going towards a maxima, the intensity gradually increases I A 2 The narrower the slit width, the cgrahamphysics.com greater the amount 2016 of diffraction

6 The single slit equation A slit maybe considered as a row of n point sources, close together and vibrating in phase, each source producing secondary circular wavelets At each point on the screen, resulting interference will be the sum of the contributions from each of the n point sources, taking into account the phase of each contribution

7 Direction of incident waves When light passes through a narrow gap the range of point sources is restricted to the width of the gap Interference between wavelets causes cancellation in certain directions Consider wavelets arriving straight at the screen Arrive in phase Bright central maximum Constructive interference d D

8 Consider waves arriving at angle θ 1 with PD λ The waves arrive in phase at the slit, but the wavelets leaving the slit at an angle will no longer be in phase This means the top ray travels PD further than bottom ray of λ, the second ray travels PD of λ 2 and so on The difference in path between the top and middle ray is λ 2 destructive interference occurs In this way all wavelets in the top half of the slit will interfere destructively with wavelets from the lower part of the slit When the wavelets reach the screen, they will interfere destructively

9 The first nodal line θ 1 For n = 1 sin θ 1 = λ 2 d 2 = λ d λ = d sinθ 1 equation for minimum at P 1 where d = slit width and θ = angle of PD

10 The second nodal line Consider a wider angle θ 2 with PD 2λ sin θ 2 = 2λ d 2λ = d sin θ 2 mλ = d sin θ m single slit equation for destructive interference

11 Constructive interference Consider waves arriving at an angle θ with PD 3λ 2 Pair rays from bottom third will cancel with rays from middle third, each pair canceling out because PD between them is λ 2

12 Constructive interference PD = λ Rays passing through top third have no matching pairs left to cancel with These reach the screen as maximum: 3λ 2 λ 2 = λ sin θ 1 = 3λ 2 3λ = d sin θ d 2 1 for 1 st order maximum, excluding central maximum

13 Higher order maximum Increasing the angle θ we can create a PD of 5λ 2, 7λ 2,.. up to m λ, m I When the PD is an odd number multiple of λ, higher 2 order maxima are produced m + 1 λ = dsin θ 2 m constructive interference

14 Remember The single slit and the double slit have opposite maxima and minima Double slit: mλ = d sin θ m This is maximum / constructive interference Single slit: mλ = d sin θ m This is minimum / destructive interference

15 Example A slit of width m is illuminated by red light of wavelength 620nm. At what angle does the third order minimum occur? Solution mλ = d sin θ m sin θ = mλ = = d θ = sin =

16 Basic observations d(slit opening)~λ Geometrical shadow Diffraction pattern a) Straight edge b) Single long slit d~3λ c) Circular aperture d) Single long slit d~5λ

17 Intensity of maxima Determined by slit width d Areas of min intensity at sin θ = λ d, 2λ d, 3λ d.. Areas of max intensity at sin θ = 0, 3λ, 5λ, 7λ 2d 2d 2d..

18 Intensity Intensity of light decreases for each successive bright area since more of the n light sources interfere destructively To find the distance of the center point of the diffraction pattern d D tan θ 1 = y 1 D

19 Finding an equation for interference d PD=λ sin θ 1 = PD d = λ d For small angle sin θ~ tan θ = y 1 D For D >>> y y 1 = λ λ= dy 1 D d D where d = slit width, y = distance from central point, D = distance from slit

20 Dark lines - sin θ = nλ d General equation Bright lines - sin θ = n+1 2 λ d for n = 1, 2, 3.. For θ = 0 there is a bright central maximum that is twice as wide as the others The equation λ= dy 1 can be used to D find λ predict dimensions of pattern predict positions of nodal lines

21 Example Light with wavelength 670nm passes through a single slit with width 12μm. Viewed on screen, 30 cm away, find a) how wide is the central maximum in i)degree and ii)cm b) what is the separation of adjacent minima (excluding the pair on either side of the central maximum) Solution A)i) sin θ = nλ 6 = θ = sin = width = 2 x 3.2 = = d ii) sin θ = y D y = D sin θ= 0.30 x sin 3.2 = m ~ 3.4cm B) λ= dy 1 D Dλ y= = d = 1.7cm

22 Note: effect of variations of d and λ Central maxima is exactly twice the width of the separation of adjacent nodal lines Diffraction pattern becomes more noticeable the narrower the slit becomes Narrower slit lots of maxima of nearly same intensity Maxima very closely together Width of pattern in proportion to λ Wavelength inversely with width of slit If slit width is >>> λ, then central maxima is very small

23 Note θ = nλ d minima where n = 1 is the position of first Minima are not actually separated by equal distances However, for values θ < 10 0 it is a good approximation to consider minima to be equally spaced

24 Summary Narrower slit effect more noticeable lots of maxima same intensity closer together To be noticeable λ~d y as λ maxima increases as d y as 1 as d d If d >>> λ than y is very small as D y

25 Circular slits If the slit is not rectangular but circular, we can find the diffraction pattern by using the formula θ = 1.22λ d where d = diameter of aperture θ = in radians The angle θ locates the first dark fringe relative to the central bright region

26 Single slit diffraction with white light A single slit illuminated by white light each component color has a specific λ max and min for each λ will be located at a different angle For a given slit width colors with long λ (red, orange, etc) will diffract more than color with short λ (blue, violet, etc) Resulting diffraction pattern will show all colors of rainbow with blue and violet nearer to central position Red will appear at greater angles