KTH Applied Physics Examination, TEN1, in courses SK2500/SK2501, Physics of Biomedical Microscopy, 2009-06-05, 8-13, FB51 Allowed aids: Compendium Imaging Physics (handed out) Compendium Light Microscopy (handed out) Extracts from β (handed out) Pocket calculator Note: Write your name on all papers. Only one problem solution on each paper. Give motivations to all your answers, and explain all symbols that are introduced. You are encouraged to use simple figures to clarify explanations. You may answer in English or Swedish. Each problem can give points. (You need 30p to pass.) Problem 1 The angular resolution of the eye is approximately one minute of arc (1/60 of a degree). This means that two point sources must be separated by this angular distance if we are to detect that they are two separate objects. In this problem you will investigate what kind of optical instrument you need to see and count bacteria growing on a cultivation dish. The individual bacteria are approximately 2 μm in diameter. They repel each other, so the center-to-center distance between bacteria is never less than 5 μm. You can order the following equipment (in order of rising cost): 8x magnifying glass 15x magnifying glass Microscope with 5x eyepiece and 5x, x and 20x objectives Microscope with 8x eyepiece and x, 20x and 0x objectives (Continued on next page.)
Which of these instruments do you have to order to carry out the task? 2 You may assume that the bacteria are seen with good contrast against the background, and that the optical equipment is so good that the resolution limit is set by the eye. When you do the calculations, assume that the center-to-center distance between the bacteria must exceed by a factor of two the resolution limit of the eye. You can then view the bacteria comfortably without straining the eye. Assume that all optical equipment is used in such a way that you see a virtual image at infinite distance. Problem 2 When a photon hits a photogate-type detector, it may release an electron that is collected and stored in the detector. This process occurs repeatedly during the exposure time, after which the total charge is read out and measured. Assume that we have a detector with a quantum efficiency of 73%, and a well capacity of.2 electrons. (Well capacity is the maximum number of electrons that can be stored. If we continue to expose the detector to light we get saturation, i.e. the measurement will be ruined.) Due to thermal effects in the detector we get a certain dark value. This means that, on avarage, 36 electrons will be created per second when the detector is in complete darkness. To compensate for this, a dark value obtained with the same exposure time is subtracted in all measurements. The only noise sources of importance are photon quantum noise and dark signal noise, both of which are described by Poisson statistics. Assume that we have a measurement situation where, on average, 127 photons are incident on the detector per second. a) What is the maximum exposure time that can be used without saturating the detector? (2p) b) What is the maximum signal-to-noise ratio we can get when using this detector? (8p) Problem 3 You want to study a fluorescently labeled specimen in a microscope. A mercury lamp is used for illumination, see emission spectrum on next page. Also shown on next page are the excitation and emission spectra for the fluorophore used.
3 Int. 365 Mercury lamp spectrum 36 302 313 33 05 56 577 λ (nm) 300 00 500 600 700 Fluorophore Excitation Emission 00 50 500 550 600 Wavelength (nm) When viewing the specimen you can choose between two different filter sets, each including excitation filter (E), dichroic beam splitter (D) and barrier filter (B). The transmission curves for each filter set is shown on next page. Both filter sets produce images with similar brightness, so the specimen can bee seen equally clearly with either filter set. The fluorophore photobleaches easily, however, and you want to avoid this as much as possible. Does it matter which filter set you use? (Explain why or why not.) Photobleaching is caused by chemical reactions when a fluorophore molecule is in the excited state just prior to emitting a photon.
Transm. Transm. 0% 0% E D B E D B λ (nm) λ (nm) 00 500 600 700 00 500 600 700 Filter set #1 Filter set #2 Problem A microscope is used with a 25/0.65 objective. To record images of the specimen, a CCD sensor is mounted in the image plane of the microscope objective. Below is the MTF curve for the objective. The MTF for the CCD is seen on next page. Please note that the scale for the spatial frequency refers to the specimen plane for the objective, and the image plane for the sensor. MTF Objective Spatial frequency in specimen plane (μm -1 )
5 MTF CCD Spatial frequency in image plane (mm -1 ) a) Is the amount of detail (image sharpness) in the recorded images limited mainly by the objective or by the CCD? (6p) b) Does it make sense at all to improve the MTF of the better of the two components (objective or CCD)? Or is it only the weakest link in the imaging chain that determines the quality (i.e. does the strength of the chain depend on its weakest link alone)? Motivation needed. (p) Problem 5 The optical sectioning capability in confocal microscopy is very important. To test this, a flat mirror can be scanned vertically and the intensity as a function of focus position be recorded. The curve below shows the result for a dry 25/0.65 objective. A laser wavelength of 88 nm was used in the experiment. Would you say that the performance of this objective is very good (aberrations negligible), acceptable or quite bad? Intensity Focus position (μm)
Problem 6 6 When looking through the eyepieces of a fluorescence microscope you can see a dense line pattern (but it is rather difficult to see, because the contrast is low). When the same pattern is recorded with a CCD camera attached to the microscope, the pattern looks coarser, i.e. it has a lower spatial frequency (moiré effect). What conclusion can you draw concerning the density (spatial frequency) of the pattern in the specimen? A 20/0.60 dry objective is used, and the CCD sensor is located in the image plane of the objective. The CCD sensor has 5 megapixels uniformly distributed over an area of 1.1 mm x 1.1 mm. The illumination wavelength is 36 nm, and the fluorescence wavelength is approximately 530 nm. Good Luck! Kjell Carlsson
7 Solutions to examination in course SK2500/SK2501, 2009-06-05. (Also other reasonable solutions may be acceptable) Problem 1 When looking at the bacteria with a naked eye, at a distance of 25 cm, the angular separation of the 6 5.0 5 bacteria is approximately = 2.0 rad. 0.25 1 When looking through the optical instrument, we want the angular distance to be not less than 30 degree = 5.8 rad. This means that we need an angular magnification of at least 5.8 M = 30. Therefore the magnifying glasses will not be sufficient, we need the 5 2.0 microscope with 5x eyepiece and objectives 5-20x. Problem 2 a) On average, 127 0.73 = 92. 7 electrons/sec are created by incoming photons. In addition, 36 electrons/sec are (on average) contributed by the dark signal. This gives a total of 92.7 + 36 = 128.7 electrons/sec. This means that we reach the well capacity after.2 = 326 sec. = 5. minutes. 128.7 b) Maximum SNR is obtained for the maximum allowed measurement time = 326 sec. (In reality we would like to be a bit below saturation, but let s use the saturation values anyway because they represent the theoretical limit.) The signal will then be 326 92.7 = 3.02 electrons. The standard deviation will be.2 = 20. 9 electrons. (All created electrons, whether they are caused by photons or thermal effects, will contribute to the noise.) 3.02 2 Thus, the maximum SNR max = = 1.5 = 150. 20.9 Problem 3 Looking at the figures, we can see that with filter set 1 we excite the fluorophore rather poorly, but we collect the fluorescent light very efficiently. With filter set 2 on the other hand we excite the fluorophore much more efficiently, but we collect very little of the fluorescent light. To minimize photobleaching we must collect the emitted photons as efficiently as possible, because each time a molecule is excited there is a risk that it will be destroyed. Exciting molecules without efficiently collecting the emitted light is therefore very bad concerning photobleaching. On the other hand, it doesn t matter much if the excitation is not so efficient. Wasted excitation photons will not contribute to the fluorescence, but on the other hand they will not contribute any photobleaching either. If the lamp intensity is sufficient that s quite OK. So, the conclusion is that filter set #1 is best if we want to minimize photobleaching.
Problem 8 a) Let s look at the MTF-values for some spatial frequencies, for example 0.5, 1.0 and 2.0 μm -1 in the specimen plane. The objective then has MTF-values of approx. 0.73, 0.8 and 0.07. In 6 0.5 the image plane the corresponding frequencies are = 2.0 m -1 = 20 mm -1, 0 25 mm -1 and 80 mm -1. At these frequencies the sensor has MTF-values of approx. 0.95, 0.80 and 0.39. In all cases the MTF for the sensor is considerably higher than the MTF for the objective. Therefore the objective is the main limitation. b) The total MTF is given by MTF total = MTFobjective MTFsensor. At the above-mentioned frequencies we get MTF total values of approximately 0.69, 0.38 and 0.03. These values are substantially lower than for the objective alone, especially at higher spatial frequencies. The sensor (better component) therefore significantly influences the image sharpness. It does pay off to improve the better of the two components. But, of course, the pay-off is even higher if we can improve the objective (but that is often difficult due to fundamental limitations like diffraction). So, the analogy with the weakest link of a chain is not valid here. Problem 5 According to the compendium, the FWHM for the optical section is given by 5.6λ FWHM = for a diffraction-limited objective. Using this equation in the present case 2 α 8πnsin 2 9 5.6 88 7 we get FWHM = = 9.1 m = 0.91 μm. 2 o 8πsin 20.3 ( ) From the recorded profile we can see that the FWHM 1. 5 μm. This means that we can label the objective as acceptable. It is clearly worse than a diffraction-limited objective, but it can still produce reasonably thin optical sections and the depth profile is rather symmetric with distinct zeros on each side of the maximum. (What one considers acceptable or bad is, of course, a matter of personal opinion. This is not critical as long as the calculations are correct.) Problem 6 Let s find the upper and lower limits for the spatial frequency in this case. We can see the pattern when looking through the microscope, so the spatial frequency must be lower than 2N. A. 2 0.60 6 νlimit = = = 2.26 m -1 = 2.26 μm -1, which is the frequency limit (where MTF 9 λ 530 = 0) for a perfect diffraction-limited objective. In reality we probably need an MTF-value of something like 0.1 to see the pattern, and then the highest frequency is 0.8 νlimit = 1. 8 μm -1. When the pattern is recorded by the CCD, the sampling criterion is not fulfilled because we see aliasing. Therefore, the frequency must be higher than half the sampling frequency. Let s find out what the sampling frequency is in this case. We have a square CCD with 5 Mpixels. This means we have 5.0 6 = 2236 pixels over a side-length of 1.1 mm. This gives a center-to-center distance betwen pixels of 3 1.1 6 2236 = 6.31 m = 6.31 μm. Thus, the sampling frequency is
9 1 = 0.159 μm -1 6.31. Since we get aliasing the pattern frequency must be 0.159 > = 0. 2 079 μm -1 in the image plane, corresponding to 0.079 20 = 1. 59 μm -1 in the specimen plane. The conclusion therefore is that the pattern frequency must be between 1.6 and 2.3 μm -1, and most likely it is between 1.6 and 1.8 μm -1.