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TRANSCRIPT
Physics 202Professor P. Q. Hung
311B, Physics Building
Physics 202 – p. 1/26
Interference and Diffraction
Interference
Light is electromagnetic waves. Shine twolight beams on a given spot and observe thesuperposition of these two beams.
Superposition of two waves can lead toconstructive interference or destructiveinterference.
Physics 202 – p. 2/26
Interference and Diffraction
Interference
Light is electromagnetic waves. Shine twolight beams on a given spot and observe thesuperposition of these two beams.
Superposition of two waves can lead toconstructive interference or destructiveinterference.
Physics 202 – p. 2/26
Interference and Diffraction
Interference
Physics 202 – p. 3/26
Interference and Diffraction
Interference
Physics 202 – p. 4/26
Interference and Diffraction
Interference
Constructive interference:Start with two identical sources in phase, i.e.oscillating in the same way. If they continue tooscillate in phase when they reach a givenpoint, the interference is constructive ⇒
Brightest or most intense spot. The conditionfor this to be so is that their difference in pathlengths should be a integral multiple of thewavelength.l2 − l1 = mλm = 0,±1,±2, ...
Physics 202 – p. 5/26
Interference and Diffraction
Interference
Destructive interference: Start with twoidentical sources in phase, i.e. oscillating inthe same way. If they are out of phase(crest-to-trough) when they reach a givenpoint, the interference is destructive ⇒ Darkspot. The condition for this to be so is thattheir difference in path lengths should be ahalf-integral multiple of the wavelength.l2 − l1 = (m −
12)λ
m = 0,±1,±2, ...
Physics 202 – p. 6/26
Interference and Diffraction
Young’s double-slit experimentYoung’s experiment demonstrated the wavenature of light by showing that two coherent lightbeams can interfere with each other.
Physics 202 – p. 7/26
Interference and Diffraction
Young’s double-slit experiment
If light were just “corpuscules”, the screenwould be illuminated just behind these twoslits.
Observation: Light is spread throughout thescreen with bright and dark fringes. ⇒ Wavenature of light.
Physics 202 – p. 8/26
Interference and Diffraction
Young’s double-slit experiment
If light were just “corpuscules”, the screenwould be illuminated just behind these twoslits.
Observation: Light is spread throughout thescreen with bright and dark fringes. ⇒ Wavenature of light.
Physics 202 – p. 8/26
Interference and Diffraction
Young’s double-slit experimentHuygens’s principle
Physics 202 – p. 9/26
Interference and Diffraction
Young’s double-slit experimentPath Difference
Physics 202 – p. 10/26
Interference and Diffraction
Young’s double-slit experiment
Bright Fringes:∆l = d sin θ = mλm = 0,±1,±2, ...
Dark Fringes:∆l = d sin θ = (m −
12)λ
m = 0,±1,±2, ...
Physics 202 – p. 11/26
Interference and Diffraction
Young’s double-slit experiment
Bright Fringes:∆l = d sin θ = mλm = 0,±1,±2, ...
Dark Fringes:∆l = d sin θ = (m −
12)λ
m = 0,±1,±2, ...
Physics 202 – p. 11/26
Interference and Diffraction
Young’s double-slit experiment
Physics 202 – p. 12/26
Interference and Diffraction
Young’s double-slit experiment: ExampleRed light is used in a double-slit experiment withthe slits separated by a distanced = 1.20 × 10−4 m. Here λ = 664 nm in vacuum.The screen is located at a distance L = 2.75 mfrom the slits. Find the distance y on the screenbetween the central bright fringe and thethird-order bright fringe.
Physics 202 – p. 13/26
Interference and Diffraction
Young’s double-slit experiment: Example
The angle of the 3rd order bright fringe isgiven bysin θ3 = (3)(664×10−9
m)1.20×10−4 m
= 0.0166.
y = L tan θ3 ≈ 4.56 cm
Physics 202 – p. 14/26
Interference and Diffraction
Young’s double-slit experiment: Example
The angle of the 3rd order bright fringe isgiven bysin θ3 = (3)(664×10−9
m)1.20×10−4 m
= 0.0166.
y = L tan θ3 ≈ 4.56 cm
Physics 202 – p. 14/26
Interference and Diffraction
Reflection of an interface
When light travels from a medium of lowerindex of refraction to a region of higher indexof refraction, the reflected wave suffers a 1800
phase change.
When light travels from a medium of higherindex of refraction to a region of lower indexof refraction, the reflected wave suffers a nophase change.
Physics 202 – p. 15/26
Interference and Diffraction
Reflection of an interface
When light travels from a medium of lowerindex of refraction to a region of higher indexof refraction, the reflected wave suffers a 1800
phase change.
When light travels from a medium of higherindex of refraction to a region of lower indexof refraction, the reflected wave suffers a nophase change.
Physics 202 – p. 15/26
Interference and Diffraction
Reflection of an interface
Physics 202 – p. 16/26
Interference and Diffraction
Reflection of an interface: Air Wedge
Physics 202 – p. 17/26
Interference and Diffraction
Reflection of an interface: Air Wedge
Physics 202 – p. 18/26
Interference and Diffraction
Reflection of an interface: Air Wedge
Constructive interference:Let d be the thickness of the air gap at anypoint.2 d = (m −
12)λ
m = 1, 2, 3, ..
Destructive interference:2 d = mλm = 1, 2, 3, ..
Physics 202 – p. 19/26
Interference and Diffraction
Reflection of an interface: Air Wedge
Constructive interference:Let d be the thickness of the air gap at anypoint.2 d = (m −
12)λ
m = 1, 2, 3, ..
Destructive interference:2 d = mλm = 1, 2, 3, ..
Physics 202 – p. 19/26
Interference and Diffraction
Reflection of an interface: Air Wedge exampleA very fine wire 7.33 × 10−3mm in diameter isplaced between two flat glass plates (see Fig.28-9). Light whose wavelength in air is 600 nmfalls perpendicular to the plates, and a series ofbright and dark bands is seen. How many brightand dark bands will there be? Will the area nextto the wire bright or dark?
(2)(7.33 × 10−3mm)/(6 × 10−7m) = 24.5.Half-integer area next to the wire will bebright.
Total of 25 dark lines and 25 bright lines.
Physics 202 – p. 20/26
Interference and Diffraction
Reflection of an interface: Air Wedge exampleA very fine wire 7.33 × 10−3mm in diameter isplaced between two flat glass plates (see Fig.28-9). Light whose wavelength in air is 600 nmfalls perpendicular to the plates, and a series ofbright and dark bands is seen. How many brightand dark bands will there be? Will the area nextto the wire bright or dark?
(2)(7.33 × 10−3mm)/(6 × 10−7m) = 24.5.Half-integer area next to the wire will bebright.
Total of 25 dark lines and 25 bright lines. Physics 202 – p. 20/26
Interference and Diffraction
Reflection of an interface: Newton’s ringLike an air wedge
Physics 202 – p. 21/26
Interference and Diffraction
Reflection of an interface: Thin Film
Physics 202 – p. 22/26
Interference and Diffraction
Reflection of an interface: Thin FilmLet n > 1 be the index of refraction inside the thinfilm of thickness t.
Constructive interference:2 t = (m + 1
2)λvacuum
n
m = 0, 1, 2, ..
Destructive interference:2 t = m λvacuum
n
m = 0, 1, 2, ..
Physics 202 – p. 23/26
Interference and Diffraction
Reflection of an interface: Thin FilmLet n > 1 be the index of refraction inside the thinfilm of thickness t.
Constructive interference:2 t = (m + 1
2)λvacuum
n
m = 0, 1, 2, ..
Destructive interference:2 t = m λvacuum
n
m = 0, 1, 2, ..
Physics 202 – p. 23/26
Interference and Diffraction
Reflection of an interface: Thin Film and color
Physics 202 – p. 24/26
Interference and Diffraction
Reflection of an interface: Thin Film with onephase change
Physics 202 – p. 25/26
Interference and Diffraction
Reflection of an interface: Thin Film exampleA soap bubble appears green (λ = 540 nm) at apoint on its front surface nearest the viewer.What is its minimum thickness? Assume n = 1.35
Minimum thickness ⇒ m = 0. Here we haveconstructive interference.
t = λ
4 n= 5.4×10−7
m
4×1.35 = 100 nm.
Physics 202 – p. 26/26
Interference and Diffraction
Reflection of an interface: Thin Film exampleA soap bubble appears green (λ = 540 nm) at apoint on its front surface nearest the viewer.What is its minimum thickness? Assume n = 1.35
Minimum thickness ⇒ m = 0. Here we haveconstructive interference.
t = λ
4 n= 5.4×10−7
m
4×1.35 = 100 nm.
Physics 202 – p. 26/26