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A series of slides from a university physics lecture on the topic of interference. The slides cover the concepts of superposition, constructive and destructive interference, and the requirements for interference to occur. The lecture also discusses the application of interference to sound and light. The slides include diagrams and examples to help illustrate the concepts.
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Physics 102: Lecture 20, Slide 2
t +1 -
t +1 -
t
In Phase
Different f
Constructive^
Destructive^
Neither 1.5^1 0.5^0 -0.5-1 -1.5 1.5^1 0.5^0 -0.5-1 -1.
(i.e. waves must have definite phase relation)
Two differentpaths Interferencepossible here
Screen a distanceL from slits Single source ofmonochromaticlight^ λ
d 2 slits-separatedby d
Screen a distanceL from slits Single source ofmonochromaticlight^ λ
d L 2 slits-separatedby d
At points where thedifference in pathlength is 0,^ λ,2λ
the screen is bright.(constructive)At points where thedifference in pathlength isthe screen is dark.(destructive)
L
d Physics 102: Lecture 20, Slide 13
L
y θ θ When this Young’s double slit experiment is placedunder water. The separation y between minima andmaxima1) increases^
same^
decreases
Physics 102: Lecture 20, Slide 14
In the Young double slit experiment, is it possible to seeinterference maxima when the distance between slits issmaller than the wavelength of light?1) Yes
n^1 n^2
Reflected wave
n = 1.0 (air)^ n^ (thin film)^1 n^2 Determine^ δ,^ number of extra wavelengths for each ray.^1 2 t Ray 1:^ δ= 0 or ½^1 If |(δ–^ δ)| = ½ , 1 ½, 2 ½ ….^2
(m + ½)^ destructive If |(δ–^ δ)| = 0, 1, 2, 3 ….^2
(m)^ constructive
Note: this iswavelength infilm! (λ=^ λ/n^ )film^ o^1 Reflection^ Distance+ 2 t/^ λfilm This isimportant!Ray 2:^ δ= 0 or ½^2
Physics 102: Lecture 20, Slide 19
Physics 102: Lecture 20, Slide 20
n=1 (air) n =1.5glass^ n =1.8plastic^ (^1 2) t Blue light^ λ^ = 500 nmincident on a thin film(t = 167 nm) of glasson top of plastic. Theinterference is:(1) constructive(2) destructive(3) neither^ δ=^1 δ=^2 Phase shift =^ δ–^2
δ=^1