phys 102 – lecture 23
TRANSCRIPT
![Page 1: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/1.jpg)
Phys 102 – Lecture 23Phys 102 Lecture 23Diffraction
1
![Page 2: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/2.jpg)
Today we will...
• Learn about diffraction – bending of light by objectsLearn about diffraction bending of light by objectsSingle slit interference
Circular aperture interference
• Apply these conceptsResolution of optical instruments
X‐ray crystallography
Phys. 102, Lecture 23, Slide 2
![Page 3: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/3.jpg)
Single slit interference?
What happens when light passes through a small slit?
Without diffraction With diffraction
This is NOT what happens!
Diff ti th t b di f li ht d bj tDiffraction – the apparent bending of light around an object or aperture
Phys. 102, Lecture 23, Slide 3
![Page 4: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/4.jpg)
Diffraction & Huygens’ principleCoherent, monochromatic light passes through one narrow slit
Wavelets through slit travel different distances and can interfere!and can interfere!
Phys. 102, Lecture 23, Slide 4Where are the interference maxima and minima?
![Page 5: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/5.jpg)
Single slit diffraction
Consider waves from top and bottom ½ of slit of width aAt this angle, every wave from top ½
21
g , y pof slit interferes destructively with corresponding wave from bottom ½
r1
r1'aa/2 θ
2
2’1’N
1a
sin2 2a λθ =
N’
2 2
1sina θ λ=1' 1r r− =Destructive:
1st minimum
2 ' 2r r− = 'N Nr r= − =Phys. 102, Lecture 23, Slide 5
![Page 6: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/6.jpg)
Single slit diffraction
Consider waves from ¼’s of slit
11’2
2’
At this angle, every wave from top ¼ of slit interferes destructively with
r1
r1'a
a/4N
N’
2corresponding wave from next ¼
a
sin4 2a λθ =
2sin 2a θ λ=1' 1r r− =Destructive:
2nd minimumPhys. 102, Lecture 23, Slide 6
2 ' 2r r− = 'N Nr r= − =
![Page 7: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/7.jpg)
Single slit diffraction minima
Condition for sixths of slit to interfere destructively
3sin 3a θ λ=
1
m = +2
3
m = +1
m = 1
a
m = –1
m = –2
THIS FORMULA LOCATES MINIMA!!
1, 2...m = ± ±sin ma θ mλ=In general,
THIS FORMULA LOCATES MINIMA!!
Note the maximum at m = 0, why? Phys. 102, Lecture 23, Slide 7
![Page 8: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/8.jpg)
ACT: CheckPoint 1
The width a of the slit in the screen is decreased
1
m = +2
m = +1
m = 1sina θ λ=
a
m = –1
m = –2
1sina θ λ=
What happens to the light pattern on the screen?
A It gets wider B Stays the same C It gets narrower
Phys. 102, Lecture 23, Slide 8
A. It gets wider B. Stays the same C. It gets narrower
![Page 9: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/9.jpg)
ACT: Wide double slitsConsider a double slit where the slit width a is not negligible. The separation d between slits is such that d = 3a. p
a
d θ
a
At an angle θ where d sin θ = 3λ, what do you see on the screen?
Phys. 102, Lecture 23, Slide 9
A. A maximum B. A minimum C. In between
![Page 10: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/10.jpg)
Diffraction in 2D
Single slit Circular apertureSquare aperture Hexagonal aperture
Phys. 102, Lecture 23, Slide 10
CheckPoint 2
![Page 11: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/11.jpg)
Diffraction from circular aperture
Central maximum1st diffraction minimum
1sin 1.22D θ λ=θ1
D
Maxima and minima will be a series of bright and dark irings on screen
Phys. 102, Lecture 23, Slide 11
![Page 12: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/12.jpg)
Demo: Resolving PowerResolving powerLight through aperture (of eye, camera, microscope, telescope, etc.) creates diffraction pattern
cing
) p
θobj
Larger spa
θ bj
θmin
Not resolved
θobjθobj θmin
θmin
θobj
Just resolved“Diffraction limit”
Two objects are resolved only when:
obj minθ θ≥ Phys. 102, Lecture 23, Slide 12
θmin
![Page 13: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/13.jpg)
Calculation: microscope resolutionA microscope objective has an aperture size D = 6.8 mm, and a focal length f = 4 mm. What is the closest distance two green g f glight sources (λ = 530 nm) can be to resolve them?
Use small angle approximation:θ
1.22minλθD
≈ objobj
dθ
f≈
Use small angle approximation:θmin
1 22 λfd ≥ 4mm1 22 0 7λ λ⎛ ⎞
≈ ≈⎜ ⎟D
θobjobj minθ θ≥Want:
D
f
1.22objdD
≥ 1.22 0.76.8mm
λ λ≈ ≈⎜ ⎟⎝ ⎠
D
380nm≥
Ultimate limit to resolution: 0.5objd λ≈
Phys. 102, Lecture 23, Slide 13dobj
![Page 14: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/14.jpg)
ACT: CheckPoint 3You are on a distant planet with binary suns. You decide to view them by building a pinhole camera. Light from both suns shines through the hole, but you can only see one spot on a screen.
You should make the pinhole
Phys. 102, Lecture 23, Slide 14A. Larger B. Smaller
You should make the pinhole ________
![Page 15: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/15.jpg)
ACT: Rectangular slit
A goat has a rectangular shaped pupil, with the long axis l h h i lalong the horizontal.
In principle in which direction should a goat’s eye have
A Horizontal B Vertical
In principle, in which direction should a goat s eye have higher resolution?
Phys. 102, Lecture 23, Slide 15
A. Horizontal B. Vertical
![Page 16: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/16.jpg)
Diffraction from a crystalEM waves of short wavelength scatters off of atoms
If there is more than one atom, scattered waves can interfere
dd
Phys. 102, Lecture 23, Slide 16
Crystals – periodic arrangements of atoms – create same interference pattern as diffraction grating!
![Page 17: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/17.jpg)
ACT: Crystal diffraction
In a NaCl crystal, the spacing between atoms is 0.282 nm. Whi h f h f ll i l h ld b dWhich of the following wavelengths could be used to see a clear diffraction pattern?
0.282 nm
A. λ = 0.1 nm B. λ = 1 nm C. λ = 10 nm
Phys. 102, Lecture 23, Slide 17
![Page 18: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/18.jpg)
X‐ray crystallography
Given X‐ray wavelength λ, diffraction angles θ provide information about distance d between atoms in crystalinformation about distance d between atoms in crystal
sin mλθ md
=
θ
d
“Photo 51” Rosalind Franklin
Phys. 102, Lecture 23, Slide 18
Crystalline fiber of DNAAs long as λ < d, small features lead to large θ. BUT need regular ordering of atoms – i.e. a crystal!
![Page 19: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/19.jpg)
Diffraction pattern of DNA
Features of pattern:p1) Layer lines2) Outer diamond3) Cross pattern1
2
33) Cross pattern4) Missing 4th layer line
1
4
Phys. 102, Lecture 23, Slide 19
“Photograph 51” – Rosalind Franklin
![Page 20: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/20.jpg)
Layer lines & diamond patternWhat features in DNA generate layer lines and diamond pattern?
P/10
DNA bases every P/10
λθP/10 = 0.34 nm
+1+2+3
m = +5
λ
/10baseλθ
PΔ ≈
θ
10 sinP θ mλ=
Helix repeats every P
–5
–3–2–1
helixλθP
Δ ≈
sinP θ mλ=
P = 3.4 nm
every P
Note that large (small) distances
sinP θ mλ
Phys. 102, Lecture 23, Slide 20
Note that large (small) distances diffract to small (large) angles
![Page 21: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/21.jpg)
Cross patternWhy does DNA diffraction generate cross pattern?
Slit pattern
InterferenceStrands are tilted at an Interference
patterntilted at an angle
Key to discovery of helix structureP = 3.4 nm
Phys. 102, Lecture 23, Slide 21
Key to discovery of helix structure
Light diffracted ⊥ to slit
![Page 22: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/22.jpg)
ACT: DNA cross pattern
You discover a new structure of DNA in which the diffraction pattern is the same as the “normal” DNA in every respectpattern is the same as the normal DNA in every respect EXCEPT that the cross makes a more acute angle α
“Normal” DNA New form of DNA
α α
Δθ Δθ
Which statement regarding the new DNA structure must be true?A. It cannot be a helix
B Th h li di P b diff
Phys. 102, Lecture 23, Slide 22
B. The helix repeat distance Pmust be different
C. The helix tilt angle must be different
![Page 23: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/23.jpg)
Missing 4th layerWhy is there no interference maximum at m = 4?
“Missing order”
θ
4sin 4P θ λ=Two DNA helices
m = +4
θ4
Two DNA helices shifted by 3P/8 (“minor groove”) m = –4
3 sinP θ3P/8 = 1.3 nm
3 4 38 2P λP λ= =
K t di f d bl h li
31 2 48 sinPr r θ− =
Phys. 102, Lecture 23, Slide 23
Key to discovery of double helix Waves from helix 1 and 2 interfere destructively!
![Page 24: Phys 102 – Lecture 23](https://reader031.vdocuments.net/reader031/viewer/2022020622/61eb7e269bcd2a34c77e311b/html5/thumbnails/24.jpg)
Summary of today’s lecture
• Single‐slit diffractiongInterference minima:
• Circular aperture diffractionsin ma θ mλ= 1, 2...m = ± ±
pFirst interference minimum:
• Resolution in optical instruments1sin 1.22D θ λ=
Angle subtended by objects ≥ angle of first diffraction minimum
• X‐ray diffractionSmall distances ‐> large diffraction angles
Phys. 102, Lecture 23, Slide 24