Tutorial Solutions 9 Holography 9.1 Efficiency of Amplitude Hologram Show that if the fringe intensity pattern of a hologram is ; µ= · δ ´ ; µ g´x y g0 g x y then the Amplitude transmission of a thin hologram formed from this distribution is 2 ; µ= δ ´ ; µ· ´δ ´ ; µµ Ta ´x y T0 a gˆ x y b gˆ x y where a and b are constants dependent of the film type. and δ ´ ; µ δ g x y ; µ= gˆ´x y g0 When reconstructed this amplitude transmission will result in first and second order recon- structions. Calculate the ratio of power between the first and second order reconstructions, and hence estimate the required ratio of intensities between the object and reference beams used to form the hologram so that the first order reconstruction has 5 times the power of the second. γ : (assume the film has = 1 3). 2 2 2 2 Hint: You will have to assume that r o where r is the reference beam intensity and o the object beam intensity to get to the expression quoted in lectures. : Estimate total efficiency of such a hologram, assuming that the plate is exposed to an OD 0 5. The total efficiency is defined at the intensity in the First Order reconstruction over the total input power. Hint: Make the same assumption regarding r2 and o2. This part is rather difficult. Solution Using the notation as in lectures, we have that the intensity on the holographic plate is ; µ= · δ ´ ; µ g´x y g0 g x y wherewehavethat δ ; µ= ´ ; µ ´ κ θ φ´ ; µµ g´x y 2ro x y cos xsin x y In coherent light the amplitude transmitted by the hologram is γ= D0 =2 2 Ta = 10 E where the exposure E = gt,andt is the exposure time. We can then write this as γ= 2 Ta = Kg where we have collected the constants in K to give γ= D0 =2 2 K = 10 t Department of Physics and Astronomy Revised: August 2000 ; µ If we then substitute for g´x y from above, we get that γ= δ 2 γ= g γ= 2 2 ´ · δ µ = · Ta = K g0 g Kg0 1 g0 We then substitute δ δ g gˆ = g0 so we have that γ= γ= γ=2 2 2 ´ · δ µ = [ · δ ] Ta = K g0 g Kg0 1 gˆ δ j jδ j If we have that j g g0,then gˆ 1, so we can expand in terms of a power series to get, to second order that γ · µ γ γ´ 2 γ= 2 2 δ µ = · ´δ µ ´1 gˆ 1 δgˆ gˆ 2 8 so we get that 2 δ · ´δ µ Ta = T0 a gˆ b gˆ wherewehavethat γ= 2 T0 = Kg0 γ γ= 2 a = K g 2 0 γ · µ γ´ γ= 2 2 b = K g 8 0 This is a exactly as shown in lectures On reconstruction we apply a reconstruction wave of the κ θµ r exp ´ı xsin So in the first order reconstruction we get, δg r2 κ θµ = ´ ; µ ´ φµ r exp ´ı xsin a a o x y exp ı g0 g0 r2 ´ ; µ ´ ´κ θ φµµ · a o x y exp ı xsin 2 g0 so that the amplitude of the reconstruction is given by 2 ; µ ar o´x y g0 In the second order we get, (as shown in lectures), we get terms of the form, 3 2 ; µ r o ´x y φ´ ; µµ ´ κ θµ b 2 exp ´ı2 x y exp xsin g0 so each of these terms have amplitude given by 3 2 ; µ br o ´x y 2 g0 Department of Physics and Astronomy Revised: August 2000 However what we measure is the intensity ratio, which is given by 2 2 a g0 R = 2 2 2 ; µ b r o ´x y 2 2 · now noting that g0 = r o , and then substituting for a and b, we (hopefully!), get that ¡ 2 2 2 16 r · o R = 2 2 2 γ · µ ´ 2 r o 2 2 Now if we assume that r o , then we can take the approximation that ¡ 2 2 2 4 2 2 · r · o r 2r o then 16 r2 · R 2 2 2 γ · µ ´ 2 o which is the expression quoted in lectures. = γ = : If we want the first order intensity to be ¢5 of the second order, then R 5, so if 1 3, then we have that 2 2 : r 5 4o : so we require the intensity of the reference beam to be approximately ¢5 4 that of the object beam. This is approximately the ratio noted in lectures, and “just about” justifies the approxi- 2 2 mation that r o ! : Efficiency: If the plate is exposed to an optical density of D = 0 5, then the intensity transmit- tance D = : T = 10 0 316 so that the amplitude transmittance, Ô = : Ta = T 0 562 Noting that, in the above analysis, T0 is the average amplitude transmittance, then we have that = : T0 = Ta 0 562 We have, from above, than the intensity of the first order is given by a2r4o2 2 g0 while the intensity of the straight through beam is 2 µ ´T0r so the ratio of the intensities of the first order to the straight through beam is a2r2o2 R = 2 2 T0 g0 Department of Physics and Astronomy Revised: August 2000 now, if we substitute for T0 and a, then after some manipulation, we get that γ2 r2o2 R = 2 4 g0 2 2 2 4 i): if we assume that r o ,theng0 r ,sowehavethat γ2 o2 R 4 r2 which was the expression quoted in lectures. 2 2 : ii): in this case we know that r = 5 4o , so we get that γ2 r2o2 = : R = 2 2 2 0 056 · µ 4 ´r o This says that about 5.6% of the intensity transmitted through the hologram is diffracted into : the reconstruction. But the intensity transmittance of the hologram is T = 0 316, so the total efficiency is = : = : 0:056T 0 016 1 6% so about 1.6% of the intensity in the reconstruction beam is diffracted into the image recon- struction. This calculation does not include reflections from the holographic plate and scattering losses in the gelatine, which will typically result in the loss of another 30%, which gives a typical, overall efficiency of about 1%, which was the figure quoted in lectures. 9.2 Bleached Holograms If an amplitude hologram is bleached to form a phase distribution of Φ´ ; µ Ta = exp ı x y where ; µ Φ g´x y ; µ= π ´x y 2 g0 ; µ where g´x y and g0 are as defined in the previous question. Show that this hologram will reconstruct in a similar manner to the amplitude hologram. Calculate the ratio of power between the first and second order reconstructions, and hence estimate the required ratio of intensities between the object and reference beams used to form the hologram so that the first order reconstruction has 5 times the power of the second. Compare with with the result for the question above, and comment. Solution As for the amplitude case the incident intensity is ; µ= · δ ´ ; µ g´x y g0 g x y Department of Physics and Astronomy Revised: August 2000 wherewehavethat δ ; µ= ´ ; µ ´κ θ ´ ; µµ g´x y 2ro x y cos xsin phi x y after bleaching the phase is given by: ; Φ g´x y ; µ= π = π´ · δ µ ´x y 2 2 1 gˆ g0 so the amplitude transmittance is given by ´ Φ´ ; µµ = ´ πδ ´ ; µµ Ta = exp ı x y exp ı2 gˆ x y δ j j ´ ; µj If, as in the amplitude case, we assume that j g g0,then gˆ x y 1, so we can expand the exponential to give, 2 2 · πδ π ´ µ Ta = 1 ı2 gˆ 2 gˆ This is exactly the same as the amplitude case, and again we have 2 δ · ´δ µ Ta = T0 a gˆ b gˆ wherewehavethat T0 = 1 π a = ı2 2 π b = 2 Since this in exactly the same expression as for the amplitude case we will get exactly the same set of reconstructed terms for first and second order. (See lectures for details of the working). In the above question the expression for the ratio between the intensity of the first and second order reconstruction was calculated to be 2 2 a g0 R = 2 2 2 ; µ b r o ´x y 2 2 · We can substitute g0 = r o , and for the above values of a and b to get 2 2 2 ´ µ r · o R = π2r2o2 2 2 so again of we assume that r o , we get get that 1 r2 · R 2 π2 o2 so if we want R = 5, then we have that r2 47 o2 so, with this system, in order that the first order is ¢5 the intensity of the second, the reference ¢ beam must be approximately ¢50 brighter than the object. This is approximately 10 more than for the amplitude hologram, showing that the phase hologram is much more non-linear, and is likely to have much larger higher order terms. The main advantage in bleaching holograms its that the overall effiency increases since. As shown above amplitude hologram the actual hologram absorbs about 90% of the light, so a factor of in overall brightness is often obtained at the expense of higher order terms.