1224 Flashcards

Question

What happens if string is not terminated with correct impedance?

Answer 1

Reflection and transmission will occur according to their coefficients

Answer 2

Send wave with frequency ω down 1st string then all points on string will oscillate harmonically with frequency ω =\> excite waves of same frequency in 2nd string

Answer 3

* Pressure is fixed by surrounds =\> pressure is a node * Change in pressure (p) is given by the picture * Pressure is pi/2 out of phase with displacement * Ψ (x=0) = Ψ (x=L) = 0 for both closed ends

Answer 4

* Group velocity (envelope) gives energy and info

Answer 5

* Group velocity = phase velocity * ω = ck * shallow water when ω=sqrt(ghk) since speed of water decreases as water gets shallower

Answer 6

* Hard to compress * Low viscosity

Answer 7

* Deep water kh \>\> 1 (then tanh(kx) → 1) * Shallow water kh \<\< 1 (then have to expand tanh(kx))

Answer 8

* relationship between group and phase velocity

Answer 9

* Gravity waves * group velocity \< phase velocity * dispersion does not contain density hence will hold for any non-viscous incompressible fluid

Answer 10

* expression from tanh(kh) expansion around 0 is simplified further to eqn in picture * these waves are dispersionless

Answer 11

* depth of water (h) decreases so group and phase velocity decreases * amplitude increases to conserve E (since E = KE + PE) * so PE increases due to increasing amplitude to keep E constant * e.g. waves breaking at sea-shore or tsunami

Answer 12

* small λs =\> surface tension γ dominates * large λs =\> gravity g dominates

Answer 13

* large number of multiple reflected rays (could be infnite but amplitude between successive wave decreases) between 2 partially silvered plates [higher number of interfering light sources = better resolution] * which are out of phase by a constant increment - constant path diff. of mλ (2tcosθ=mλ) [pi phase change at each mirror hence no overall phase change] * increases sharpness of inference max. =\> bright fringes * interference fringes focussed by convering lens =\> concentric circles

Answer 14

* light approaching diffracting object is parallel & monochromatic (interfering light rays are II to each other) * screen is at distance much larger compared to size of diffracting object * dsinθ=mλ

Answer 15

* 2 wavelengths are just resolved when max. of one lies at 1st min. of the other * θr=λ/a * for circular aperature: θr = 1.22λ/D (D=diameter of circle)

Answer 16

* interference pattern of recombined beams that have travelled a different distance d=Lf-Lm * hence altering posⁿ of movable mirror =\> control path diff. * path diff. = 2Δ which equals 2ndcosθ (so Δ=ndcosθ) * pi shift in beam that strikes fixed mirror (reflection from back of beam splitter) & approximate n_air = 1 hence bright fringes when 2dcosθ = (m+0.5)pi * [light strikes mirrors at angle θ to normal)

Answer 17

* to add a λ-dependent phase diff. since path diff. in glass for each beam may not be the same if use light of different λs and n_glass is slightly λ dependent * compensator plate have same thickness & material as beam splitter * also to make sure beams travel through beam splitter twice each

Answer 18

* use (kx-wt) where x is distance travelled by light * for movable mirror, light travels 2Lm * for fixed mirror, light travels 2Lf= 2Lm+2Δ * take out exponentials as common factor since exp(ix)e(-ix)=1

Answer 19

* d increases =\> cosθ decreases =\> θ for same fringe (same m) increases * expand outwards & new fringes appear at centre =\> ring appear from centre and expand

Answer 20

The plane of constant phase (and amplitude?) on a wave. The normal to the wavefront.

Answer 21

refractive index, n =c/v Optical path length (OPL) = nΔx

Answer 22

Fermat - Light, in going between two points, traverses the path with the smallest optical path length.

Answer 23

Huygen - Every point on a wavefront acts as a point source for secondary wavelts propagating radially outward, such that at a later time, the primary wavefront is the envelope of the secondary wavelets.

Answer 24

![Basic illustration of Snell's Law.](http://www.math.ubc.ca/~cass/courses/m309-01a/chu/Fundamentals/snell01.gif)

Answer 25

For light travelling from high to low refractive index, the angle of refraction must be greater than the angle of incidence. The largest possible value for refraction is pi/2, which corresponds to the maximum angle of incidence for refraction to occur. Beyond this angle, there are no real angles that satisfy the equation: ![\theta_c = \theta_i = \arcsin \left( \frac{n_2}{n_1} \right),](http://upload.wikimedia.org/math/d/a/0/da0a2cae5df97d6d7e34fa0bad52e009.png). Beyond this angle, total internal reflection occurs (TIR).

Answer 26

Light whose plane of vibration has a fixed orientation. The electric field vector is confined to a given plane along the direction of propagation.

Answer 27

If we have plane polarised light incident at an angle ![\theta](http://upload.wikimedia.org/math/5/0/d/50d91f80cbb8feda1d10e167107ad1ff.png) to the plane of polarisation of a perfect polariser, onl the componet of the field along the direction of the transmision axis is transmitted. ![](https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcTw7EN3kVGFcxC4WmMznqzH9WtNiYsXFHG8t4qyx3e1HiJf6TTG) Half the intensity of natural light is transmitted since light has components travelling in all directions.

Answer 28

Light with constant amplitude whose plane of vibration rotates in time. If rotation is clockwise - right-circularly polarised light, if anticlockwise - left-circularly polarised light. It can be produced by obtaining linearly polarised light whose orthogonal components (i.e. x and y) have equal amplitudes and either advancing or delaying one component by pi/2. E = E0 cos(kz − ωt)i + E0 sin(kz − ωt)j.

Answer 29

Circular polarisers are used in 3D cinemas where the ﬁlm is shown in the two polarisations and glasses receive one type of circular polarisation in one eye and the other type in the other eye. If glasses with linear polarisers were used the wearer would have to keep their head at exactly the right orientation so that one eye would continually receive just one polarisation and the other eye the other. Tilting would cause a mixture to be observed in each eye

Answer 30

Similar to cicular polarisation, except the orthogonal components (x and y) do not have equal amplitudes and they do not have to be pi/2 out of phase. For this reason, linear and circular polarisations are special cases of elliptical polarisation with linear having theta = 0 and circular having the amplitudes equal and theta = 90.

Answer 31

Fresnel's Equations: ![r_s = \frac{n_1 \cos \theta_\text{i} - n_2 \cos \theta_\text{t}}{n_1 \cos \theta_\text{i} + n_2 \cos \theta_\text{t}}](http://upload.wikimedia.org/math/8/6/4/864251cabf65ac7455025532f7e83652.png) ![t_s = \frac{2 n_1 \cos \theta_\text{i}}{n_1 \cos \theta_\text{i} + n_2 \cos \theta_\text{t}}](http://upload.wikimedia.org/math/a/e/9/ae9ad83ce1f0befb4a0fa239b448f6ca.png) ![r_p = \frac{n_2 \cos \theta_\text{i} - n_1 \cos \theta_\text{t}}{n_1 \cos \theta_\text{t} + n_2 \cos \theta_\text{i}}](http://upload.wikimedia.org/math/2/b/c/2bc3b5034c610b74e5de6fcdbc49490f.png) ![t_p = \frac{2 n_1\cos \theta_\text{i}}{n_1 \cos \theta_\text{t} + n_2 \cos \theta_\text{i}}](http://upload.wikimedia.org/math/9/3/3/933e12192b1ffd8ea7b7adf503c24794.png)

Answer 32

At one particular angle of incidence, the component of the field parallel to the plane of incidence is entirely transmitted with its reflection coefficient going to zero. This possible because: ![](http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/brewster.gif)

Answer 33

![](http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/refce.gif)

Answer 34

![](http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/refci.gif)

Answer 35

![Fresnel reflection.svg](http://upload.wikimedia.org/wikipedia/commons/thumb/a/a4/Fresnel_reflection.svg/800px-Fresnel_reflection.svg.png)

Answer 36

![Fresnel reflection.svg](http://upload.wikimedia.org/wikipedia/commons/thumb/a/a4/Fresnel_reflection.svg/800px-Fresnel_reflection.svg.png) ## Footnote Notice that brewster's angle is still there but no critical angle.

Answer 37

1. Reflection - There is a preferential reflection of light polarised perpendicular to the plane. This is entirely the case at brewster's angle. 2. Absorption - Some crystalline materials absorb more light in one direction than another. This anisotropy is called dichroism. 3. Birefringence - Crystalline materials may have diﬀerent indices of refraction associated with diﬀerent directions. 4. Scattering - Light scattering of molecules causes oscillations perpendicular to the direction of travel of light. Light travelling along z causes oscillations in x and y, therefore the nuclei/charges oscillating in x/y will radiate in y/x.

Answer 38

* increase path diff. by 2t(n - n_air) since light travels in both directions in each arm hence goes through object twice * correponds to 2t(n - n_air)λ wavelengths =\> phase shift that corresponds to wavelength shift as well * pattern changes by 2t(n - n_air)λ fringes * (imagine moving along sin curve, object introduced shift is past pi/2 so add pi/2 to get overall intensity)

Answer 39

* consider amplitude of each successie transmitted beam * I is given by I=I_max(1/1+Fsin^2γ) where F=4R/(1-R)^2 * max. amplitude when I=I_max i.e. when sinγ=0 so this happens for γ=m\*pi

Answer 40

* m=1 * since at m=0, no fringe because it's at right angles to etalon where no light is transmitted

Answer 41

* condition: max. of one coincides with HALF max. of the other (since intensity never becomes 0) * hence need distance between 1/2 I_max and I _max for one wavelength (consider γ at 1/2 and let γ=γmax +Δγ =\> use double angle formulae and the fact that you know what sinγmax and cosγmax =\> use small angle approximation) * then find 2 formulas for Δcosθ by when λ & t are fixed and when m & t are fixed

Answer 42

* dsinθ=mλ where θ=angle normal to interfering light rays * consider distance travelled by r1 and r2 * OPD = r2-r1 =\> use Pythagoras to find them * take out L as a common factor so eqn is in form for binomial expansion since ASSUMED that L \>\> (y+d/2) or (y-d/2) so (y+d/2)/L \<\< 1 or (y-d/2)/L \<\< 1 * y/L=tanθ =\> use small angle approximation so tanθ=sinθ

Answer 43

* consider distances travelled by r1 and r2 =\> in the case of r2, it travels r1+dsinθ (note: δ is kdsinθ) * put out a factor of exp(ikr1 +iδ/2) * max. when δ/2=m\*pi * 0 when δ/2=(m+0.5)\*pi * 1/2 when δ/2=(m+0.75)\*pi or (m+0.25)\*pi

Answer 44

* d→nd and λ→λ/n * adds path diff. (n-1)t =\> fringe shifted * IF (n-1)t=λ/2 =\> corresponds to phase shift of pi between 2 beams hence points of bright fringes become dark fringes

Answer 45

* path. diff becomes large ( m\>\>1) =\> coherence lost so fringe pattern starts to degrade * y becoming larger =\> also affects pattern

Answer 46

Coherence: A measure of how well a real wave approximates an infinite plane wave. Spatial - The distance over which the relationship Δφ=kΔx is valid. Temporal - The time over which the relationship Δφ=ωΔt is valid.

Answer 47

* same with equal amplitude, take out factor of exp(ikr1+iδ/2) * Then write exponentials in terms of cos and sin * Then group together terms with cos and sin * max. when δ/2=m\*pi [I is proportional to (E1+E2)^2)] * 0 when δ/2=(m+0.5)\*pi [I is proportional to (E1-E2)^2)] * i.e. same conditions for equal amplitude waves

Answer 48

* since I_max and I_min depends on amplitude hence if sufficiently different fringes may be hard to see * V needs to be greater than some value for reasonable clarity

Answer 49

![OPD = 2n_2d\cos\big(\theta_2)](http://upload.wikimedia.org/math/0/1/f/01f85c8c7b664f700bed79f0c3681c90.png)

Answer 50

* array of N slits (equally spaced) =\> path diff. between successive rays is dsinθ * calculate total amplitude using same method as double slit except you have a geometric series * but still take out common factors like exp[iNδ/2] and exp[iδ/2]

Answer 51

- High to low refractive index - 0 phase change - Low to high refractive index - π phase change

Answer 52

* x→0 =\> sinx = x (small angle approximation) hence I → I₀N² * primary max. when sin²((π/λ)dsinθ) = 0 =\> π/λ)dsinθ=mπ i.e. dsinθ=mλ * min. when sin²((Nπ/λ)dsinθ)=0 i.e. (Nπ/λ)dsinθ=kπ (k = another integer) EXCEPT in cases where k/N=m =\> primary max. (since denominator is also 0) hence k≠Nm for zeros * subsidiary max. when sin²((Nπ/λ)dsinθ)=1 i.e. (Nπ/λ)dsinθ=(m+0.5)π

Answer 53

![](http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/soapfi2.gif)

Answer 54

![](http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/oilfilm.gif)

Answer 55

![](http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/tfil2.gif)

Answer 56

* dsinθ=mλ (condition for primary max.) * λ decreases → m increases for fixed theta * =\> λblue will have more maxima than λred =\> can use diffraction grating to resolve spectral lines like Fabry Perot

Answer 57

* condition: resolved if primary max. of one coincides with min. of other * consider change in angle from primary max. to 1st zero and how maxima posⁿ changes as λ changes * primary max when dsinθ=mλ * zero when dsinθ=(k/N)/λ EXCEPT when k/N=m =\> so let k=mN+1 for 1st zero hence dsinθ=(m+1/N)/λ * so for fixed λ change from max. to 1st zero is dΔsinθ=λ/N =\> Δsinθ=λ/Nd * how max. posn changes as λ changes (fixed m): dΔsinθ=mΔλ =\> Δsinθ=mΔλ/d * equate two expressions for Δsinθ =\> λ/Δλ=mN

Answer 58

Remember that the distance d is almost the same in the case of a very small angle, and that the small angle approximation for d = x tan α ≈ αx is used.

Answer 59

Light travels from air to a material with some refractive index and is reflected from the top and bottom surfaces. Assuming the medium below the wedge is the same as that above, i.e. air, there must be one π phase change, and this is due to reflection off the top or bottom surface depending on whether the material or air has the higher refractive index. Therefore, the conditions are: ## Footnote constructive interference - 2nαx = (p + 1/2 )λ destructive interference - 2nαx = p λ,

Answer 60

* single slit of width (a) = very large number N of equally spaced narrow slits (d=separation of slits) i.e. Nd=a (as N→infinity, d or a/N→0)

Answer 61

* replace d with a/N in intensity of an array of N slits (diffraction grating) * use small angle approximation for denominator =\> get factor of N² * but as N→infinity =\> amount of light transmitted from each region shrinks by 1/N hence effective amplitude should be divided by N to reflect this (since intensity is proportional to amplitude squared - divide by N²) * I→I₀ as x→0 * min. when (π/λ)asinθ=mπ * max. when π/λ)asinθ=(m+0.5)π

Answer 62

* using θr=1.22λ/D

Answer 63

* Product of that for grating & single slit * d\>a =\> 1st min. of single slit at greater angle than 1st primary max. of diffracton grating away from centre * primary max. of diffraction grating decreases until 1st single slit min. then rise and fall again but at smaller peaks * primary max. of diffraction grating when dsinθ=mλ * min. of single slit when asinθ=kλ * hence when d/a=m/k =\> coincide so order m grating maxima missing * good gratings when d\>\>a since slower fall of primary max.