Problem set 6 for the course Theoretical Optics Sample Solutions

Size: px

Start display at page:

Download "Problem set 6 for the course Theoretical Optics Sample Solutions"

Daisy Short
5 years ago
Views:

1 Karlsruher Institut für Tehnologie KIT) Institut für theoretishe Festkörperphysik SS01 Prof. Dr. G. Shön, Dr. R. Frank Tutorial: Group 1, Name: Group, Group 3. Problem set 6 for the ourse Theoretial Optis Sample Solutions 1 Hologram Of A Point Satterer A point satterer resies on the z-axis in istane to a photographi plate at z = 0. A plane referene wave Ψ ref r) = A ref e ik 0x sin θ interferes on the sreen with the objet wave Ψ obj x, y) = A obj x +y + e ik 0 between objet an sreen. x +y +, where A ref, A obj R. We onsier a large istane point satterer plane wave photographi plate a) We want to alulate the interferene effets on the sreen in the following. Therefore, fin an approximate expression of the objet wave Ψ obj for large istane between objet an sreen. Explain, what exatly this property means mathematially. Explain, why this allows You to approximate the square roots by Taylor polynomials then an why it suffies to expan the amplitue up to zeroth orer in x an y means that the approximation is inepenent of x an y), while for the argument of the exponential the next non-vanishing orer of the variables x an y has to be inlue. [ Points) ] b) The photographi sreen will show the interferene pattern of the referene an objet wave after eveloping, thus ating as an aperture with amplitue transmission

2 oeffiient T x, y) = Ψ obj + Ψ ref. Show that T x, y) = A ref + A obj [ 3 Points) ] + A refa obj os k 0 k 0 sin θ + k 0 x sin θ) + y )). 1) ) To reonstrut the image of the point satterer, we illuminate the evelope sreen with an inoming plane wave Ψ in = Ψ ref of the same frequeny as the referene wave. Diretly behin the sreen, the transmitte fiel istribution is then given by Ψ trans = Ψ in T x, y). Fin the fiel Ψ trans x, y) on the sreen for general θ in terms of propagating waves express sin an os by omplex exponential funtions). Whih of these partial waves reonstruts the image of the point satterer? What is the physial meaning of the remaining partial waves?[ 3 Points) ] a) The sreen oorinates are given by x an y, where the sreen is suppose to have finite extents along these oorinates, e.g., h an w not given in the problem text). A large istane then means, that is always muh larger than any values for x an y that we use impliitly to evaluate any of the wave expressions, i.e. x, y [ 0.5 Points) ]. Now we write the square root as x + y + = 1 + x + y, ) } {{ } 1 whih is basially the funtion ft) = 1 + t evaluate for small positive t 1 only. Hene, the Taylor expansion aroun the point 0 is a goo approximation to the values of this funtion f, whih up to first orer in the argument gives 1 + t t + Ot ). Taylor aroun t 0 = 0 for t 1) 3) Thus, at t = x +y, we obtain the expansion In the expression x + y + + x + y Ψ obj x, y) = ) 1 + O. 4) 3 A obj x + y + e ik 0 x +y +, 5) we nee to expan the square roots now. As isusse in the leture, it is usually suffiient to only keep the first term in the amplitue while the phase is muh more sensitive, so there we shoul also keep the seon term as well Fresnel approximation). This is beause the amplitue variations are not as important as the phase variations for obtaining an interferene pattern if we approximate the phase to be inepenent of x an y, we oul not stuy the interferene effets, beause we woul have approximate it to eath ). Thus, we write Ψ obj x, y) A obj e ik 0 e i k 0 x +y ), 6)

3 whih is the esire asymptoti behavior of the objet wave on the far away sreen that allows to stuy the most prominent interferene ontributions. b) Starting from the original expression, we fin So, we nee to evaluate Re ) A Ψ ref Ψ obj A ref obj = Re T x, y) = Ψ obj + Ψ ref 7) = A obja ref = A obja ref = Ψ obj + Ψ ref + Ψ refψ obj + Ψ ref Ψ obj }{{} ReΨ ref Ψ obj) e ik 0 x sin θ x +y Aing a 0 allows to reast the expression as = A obja ref rearranging terms gives = A obja ref = A obja ref «) 8) entere given wave expressions) 9) x os k + y ) ) 0 x sin θ took real part) 10) x os k + y 0 + x sin θ )). os θ) = osθ)) 11) x + y os k 0 + k 0 k 0 x sin θ + k 0 sin θ k 0 sin θ 1) }{{} =0 os k 0 k 0 sin θ + k 0 y + x x sin θ sin θ }{{} 13) =x sin θ) x sin θ) + y )). 14) os k 0 k 0 sin θ + k 0 We therefore fin that the interferene pattern is entere aroun x sin θ, as sugeste by looking at the sketh. Finally, the full expression for T reas T x, y) = Ψ obj + Ψ ref 15) = Ψ obj + Ψ ref + Re ) Ψ ref Ψ obj 16) = A obj ) + A ref + A obja ref os k 0 k 0 sin θ + k 0 x sin θ) + y )). 17)

4 ) So, we write the transmitte fiel iretly behin the sreen) as: Ψ trans = Ψ ref T x, y) 18) = Ψ ref Ψ obj + Ψ ref 19) = Ψ ref Ψobj + Ψ ref + Ψ ref Ψ obj + Ψ refψ obj ) 0) = Ψ obj + Ψ ref ) Ψ ref ) + Ψ }{{} refψ obj + Ψ ref Ψ }{{}}{{ obj } term 1 term term 3 From this expression, we fin alreay, that the first two partial wave in term 1 simply reonstrut the referene beam with a moifie amplitue). Similarly, term 3 orrespons to the objet wave outgoing spherial wave) with a moifie amplitue. To interpret the remaining term we nee to investigate a little bit further. Inserting the full expressions we have = A obj ) + A ref + A obja ref Ψ trans = Ψ in T x, y) = A ref e ik 0x sin θ T x, y) with T from above here we use the simpler form, not the final result) as T x, y) = A obj ) + A ref + A )) obja ref os k 0 + x + y x sin θ ««k 0 + x +y x sin θ k + e i 0 we fin: e +i Ψ trans = A ) obj ) + A ref A ref e ik 0x sin θ = + A ref A obj e ik0 e ik 0 x +y e ik 0x sin θ e ik 0x sin θ + A ref A obj e ik0 e ik 0 x +y A ) obj ) + A ref A ref e ik 0x sin θ e ik 0x sin θ + x +y x sin θ 1) ) ««) 3) 4) 5) 6) term 1 7) + A ref A obj e ik0 e ik 0 x +y term 3 8) + A ref A obj Summary of the various terms: e ik0 e ik 0 x sin θ) +y e ik 0 sin θ 1. The image of the referene beam plane wave pattern on the sreen) term 9). Inoming spherial wave from position z = + in front of the sreen) at x = sin θ, y = 0). This is a real image of the satterer. In this image, epth information is reverse, i.e. bumps are ents an vie versa. That s why it is also alle the pseuosopi image the name enotes the epth reversion). See Fig.??. 3. Outgoing spherial wave of the point satterer. This orrespons to a virtual image of our satterer at its original position x = 0, y = 0) in istane z = behin the sreen). This is the hologram.,

5 Thus, Gabor) holograms simultaneously generate both a virtual an a real image, also alle twin images. point satterer plane wave photographi plate enter of inwars prop. spherial wave, the pseuosopi/real image Figure 1: The hologram of a point an its pseuosopi image. 13 Optial Coherene Tomography Optial oherene tomography is a sanning optial mirosope with improve vertial resolution. From a soure, light with amplitue E 0 t) emerges an is split into a referene ray an a measurement ray. This is fouse by a onventional mirosope into a sample with mean refrative inex n. E r t) is the amplitue of the reflete referene ray at the position of the oupler; E s z; t) is the amplitue of the light sattere insie the sample at the position of the oupler. The light is sattere at some partile loate at epth z with the real value sattering effiieny a s. a) Formulate E r t) an E s z; t) in terms of E 0 t). Express the etete intensity I et = Er t)+e s t) in terms of the autoorrelation funtion Γτ) = E0 t)e 0t τ).[ 4 Points) ] b) Assume that E 0 has a Gaussian spetral power ensity with mean angular frequeny ω 0 an with ω: E0 ω) = i 0 exp ω ω ) 0) 30) ω Fin Γτ). [ 3 Points) ]

6 ) Consier a single point satterer at z = z s. Calulate I et as a funtion of s := l r l s using the results of a) an b). Sketh I et s) an annotate the main features. [ 3 Points) ] ) Whih features of the interferometer an of I et s) have to be known to etermine the quantities z s an a s? How oes the light soure influene the vertial resolution z-iretion)? What etermines the lateral resolution x-iretion)?[ Points) ] OCT is use in eye iagnostis an more general in biology an meiine as an imaging metho. It allows for goo z-resolution epth) in the images. See Wikipeia for more etails. a) E r t) is the light that was reflete at the referene mirror. Hene, by onservation of energy intensity is half) it s elaye by lr with a relative amplitue of 1/ : E r t) = 1 E 0 t τ r ), 31) τ r = l r. 3) Analogously, the light sattere insie the sample is elaye by ls aounting for the free spae until it enters the sample an z n aounting for the propagation insie the sample: E s t) = 1 a s E 0 t τ s ), 33) τ s = l s + z n. 34) The sattere amplitue is furthermore multiplie with the sattering effiieny. The intensity at the etetor is given by I et = E r t) + E s t) 35) = E r t) + E s t) ) E r t) + E st) ) 36) = E r t)e r t) + E s t)e st) + Re { E r t)e st) } 37) = E r t)er t) + E s t)est) + Re { E r t)est) } 38) [ = 1 E0 t τ r )E 0t τ r ) + a s E 0 t τ s )a s E0t τ s ) + Re { 39) }] E 0 t τ r )a s E0t τ s ) = 1 [Γ0) + a sγ0)] + Re { a s Γτ s τ r ) } 40) The esire result is I et = 1 { 1 + a s)γ0) + a s Re Γ ls l r + nz )} 41)

7 b) Aoring to the Wiener-Khinthine theorem isusse in sript an leture, the autoorrelation funtion is the Fourier transform of the spetral ensity Sω) = Eω) : Γτ) = ωsω) exp iωτ) 4) The spetral ensity was given in the problem to be a Gaussian with with ω, entere aroun ω 0 : Sω) = i 0 exp ω ω 0) ) 43) ω We an use the Fourier table of Bronstein there is a typo in eition 4, this is the orret version): fω) = 1 e ω /4a e aτ = τ fω)e iωt, 44) 4πa or, equivalent but more onvenient for us here an I heke it), english Wikipeia: fω) = e αω π α e τ /4α = τ fω)e iωτ. 45) We enter Sω) into the Fourier integral an get Γτ) = ω i 0 exp ω ω 0) ) exp iωτ) 46) ω Using the substitution of variables ω = ω ω 0 ω = ω we obtain ) Γτ) = i 0 exp iω 0 τ) ω exp ω exp iω τ) 47) ω In this form, we an apply 45) an ientify α = 1 ω, wih finally gives Γτ) = πi 0 ω exp iω 0 τ) exp 1 ω τ ). 48) The final esire result is Γτ) = πi 0 ω exp τ ω ). 49) ) In terms of s, we fin that τ = τ r τ l = s nz)/, hene I et = 1 { s nz )} 1 + a s)γ0) + a s Re Γ Inserting the expliit form for Γτ) from b) yiels I et = [ 1 { πi 0 ω 1 + a s) + a s Re exp s nz) ω ) ) s nz) )}] exp iω 0 51) = [ 1 ) πi 0 ω 1 + a s) + a s exp ω s ω0 )] nz) os s nz) 5) 50)

8 1.8 / ω) normalize Is) a a^)/ s 0.5 nz Figure : Fringe pattern orresponing to part ), with normalize Is) = I et s)/ πi 0 ω. The sketh f. Fig. ) of I et as a funtion of s shows a onstant intensity of πi0 ω1 + a s) for s 0. Aroun the position s = nz, there is a fringe pattern with amplitue πi 0 ωa s on top of the onstant intensity. The fringe pattern has a Gaussian envelope an eays with a sprea of. ω ) The epth z s of the satterer is enoe as the maximum of the Gaussian fringe envelope in the I et /s plot. By s max = nz, we nee the mean refrative inex n of the sample, the istane between oupler an sample surfae l s an the length of the referene l r arm an the intensity amplitue i 0 in orer to etermine z s an a s quantitatively. The z-resolution is illustrate in Fig. 3. The sattering effiieny a s enters the etete intensity as part of the bakgroun intensity an as the amplitue of the fringe pattern. The first ourene oes not provie any information beause the ontributions of satterers at ifferent vertial positions in the sample are inistinguishable. Thus, the fringe amplitue efines the sattering amplitue as long as the fringe patterns of ajaent satterers o not overlap in this ase, the problem beomes muh harer). A narrow fringe pattern is esirable beause then satteres an be loser together without having overlapping fringe patterns. The with of the fringe pattern is the with of the Gaussian envelope an is reiproal to the spetral with. This means

9 3 *a 3 *a.5 I for z 1 =1 I for z =1.5 I et.5 I for z 1 =1 I for z =1.5 I et normalize Is) 1.5 normalize Is) ?m*1+a^)? 0.5?m*1+a^)? s s Figure 3: Naive shemati of superpose intensities of two satterers at z 1 = 1 an z = 1.5 interferene effets have been neglete, but the basi behavior is visible). Left: small ω, orrespons to high egree oherent light. Right: large ω, orrespons to low egree of oherene. The zero-line of the intensity of height m1 + a s)/ is no goo measure for a s, sine an unkown number m of satterers ontribute to this value. The Gaussian envelope however is always of with a s, no matter how many satterers ontribute to the signal. Left: High oherene auses reflete signals to overlap an reue the z-resolution. Right: Low oherene narrows the fringes an z-positions of iniviual satterers are better isernible. This allows better z-resolution than onventional mirosopy. that a reue oherene of the light soure improves the vertial resolution. In the lateral iretion, the spetrometer behaves as a onventional mirosope an, thus, it is iffration-limite. In partiular, there is no epenene on the oherene properties of the light soure. Han in solutions in leture on

l. For adjacent fringes, m dsin m

l. For adjacent fringes, m dsin m Test 3 Pratie Problems Ch 4 Wave Nature of Light ) Double Slit A parallel beam of light from a He-Ne laser, with a wavelength of 656 nm, falls on two very narrow slits that are 0.050 mm apart. How far