1. Thin Lenses f
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2. Thin Lenses: Converging Lensf r2 C2 F1 F2 C1
1/s1+n/s’=(n-1)/r1
-n/s’+1/s2=(1-n)/r2 (here r2<0)
1/s1+1/s2=(n-1)(1/r1-1/r2)
Parallel rays refract twice
Converge at F2 a distance f from center of lens
F2 is a real focal pt because rays pass through
f > 0 for real focal points

3. Thin Lenses: Diverging Lens
f < 0 for virtual focal points C1 F2 F1 C2 r2 f r1
Extension
Rays diverge, never pass through a common point
F2 at a distance f
F2 is virtual focal point

4. Images from Thin LensesC2 F1 C1 O I s’ s
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5. Images from Thin Lenses
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6. Images from Thin LensesC1 F2 C2 r2 s’ r1 O s
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7. Locating Images by Drawing Rays1 3 2 F2 F1 O I
Ray initially parallel to central axis will pass through F2.
Ray passing through F1 will emerge parallel to the central axis.
Ray passing through center of lens will emerge with no change in direction because the ray encounters the two sides of the lens where they are almost parallel.
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8. Locating Images by Drawing RaysF2 F1 O 2 1 3 I
Ray initially parallel to central axis will pass through F2.
Backward extension of ray 2 passes through F1
Ray 3 passes through center of lens will emerge with no change in direction.
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9. Locating Images by Drawing RaysF2 F1 3 I 2 1 O
Backward extension of ray 1 passes through F2
Extension of ray 2 passes through F1
Ray 3 passes through center of lens will emerge with no change in direction.
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10. Two Lens System
Lens 1
Lens 2 O s1
Note: If image 1 is located beyond lens 2, s2 for lens 2 is negative.
1. Let s1 represent distance from object, O, to lens 1. Find s1’ using:
2. Ignore lens 1. Treat Image 1 as O for lens 2.
3. Overall magnification:
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11. Example: Two Lens System
A seed is placed in front of two thin symmetrical coaxial lenses (lens 1 & lens 2) with focal lengths f1=+24 cm & f2=+9.0 cm, with a lens separation of L=10.0 cm. The seed is 6.0 cm from lens 1. Where is the image of the seed?
Lens 1:
Image 1 is virtual.
Lens 2: Treat image 1 as O2 for lens 2. O2 is outside the focal point of lens 2. So, image 2 will be real & inverted on the other side of lens 2.
Image 2 is real.
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12. Example Figure
Lens 1
Lens 2 O s1 f1 f2 L
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13. Table for Lenses Lens Type Object Location Image Location Image Type
Image Orientation
Sign of
f, s’, m
Converging
Inside F
Same side as object
Virtual
Same as object
+, -, +
Outside F
Side of lens opposite the object
Real
Inverted
+, +, -
Diverging
Anywhere
-, -, +
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14. Quiz lecture 25 The radius r in the formula is: Positive Negative
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15. Quiz lecture 25 The radius r in the formula is: Positive Negative O r
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16. Quiz lecture 25 The focal length f in the formula is: Positive
Negative
Positive
Negative
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17. LECTURE 26: Interference
Light diffracts…..multiple path lengths ….causes multiple waves to interference producing perfectly destructive interference when it straddles the pit. Wavelength of red light is roughly 700 nanometers

18. Interference
When two waves with the same frequency f and wavelength combine, the resultant is a wave whose amplitude depends on the phase different, .
Easiest way to achieve the same frequency and wavelength is to use the same source. Lasers are an excellent source of this kind of light. Splitting a laser beam in two beams and making them meet again to combine can result in constructive & destructive interference.
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19. Interference: Phase Differences
 = 0 or 1
Constructive Interference
 = 0.5 
Destructive Interference
 = 0.3 
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20. Coherence Only coherent waves can produce interference!
*If the difference in phase between two (or more) waves remains constant ( i.e., time-independent), the waves are said to be perfectly coherent.
*A single light wave is said to be coherent if any two points along the propagation path maintains a constant phase difference.
In reality, there are no completely coherent light waves, but partial coherence can be achieved. For instance, sunlight remains coherent only over short distances.
*Coherence length: the spatial extent over which a light wave remains coherent.
Only coherent waves can produce interference!
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21. Coherence Single Photon Infinite coherence length
If the duration of the E-field oscillation for a photon is one nano-sceond & it travels with the speed of light, the coherence length would be about 30 cm or 1 foot. Coherence length is important because is allows you to determine the length over which interference occurs.
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22. Intensity of Two Interfering Waves
*Two light waves not in phase:
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23. Intensity of Two Interfering Waves
*Maxima occur for:
for m = 0, 1, 2, 3….
*Minima occur for:
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24. Interference
*Three ways in which the phase difference between two waves can change:
By traveling though media of different indexes of refraction
By traveling along paths of different lengths
By reflection from a boundary
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25. Quiz lecture 25 top layer bottom layer ntop = nbottom
The light waves of the rays in the figure have the same wavelength and are initially in phase. If 7.60 wavelengths fit within the length of the top layer and 5.50 wavelengths fit within that of the bottom layer, which layer has the greater index of reflection?
Top layer
top layer
bottom layer
ntop = nbottom
ntop
nbottom L
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26. Interference: Different Indexes of Refraction
*The phase difference between two waves can change if the waves travel through different material having different indexes of refraction. n1 n2 L
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27. Interference: Different Path Lengths
*The phase difference between two waves can change if the waves travel paths of different lengths.
Thomas Young experiment
(1801)
Remember Huygen’s
principle.
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28. Interference: Different Path Lengths
*The phase difference between two waves can change if the waves travel paths of different lengths.
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29. Interference: Different Path Lengths
*The phase difference between two waves can change if the waves travel paths of different lengths.
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30. Interference: Different Path Lengths
*The phase difference between two waves can change if the waves travel paths of different lengths.
Spacing between fringes:
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