Transcript
KTH Applied Physics
Examination, TEN1, in courses SK2500/SK2501, Physics of Biomedical Microscopy, 2015-01-13, 8-13, FA31 Allowed aids:
Notes:
Compendium ”Imaging Physics” (handed out) Compendium ”Light Microscopy” (handed out) Extracts from (handed out) Pocket calculator
Write your name on all papers. Write on one side of the paper only. (Papers will be scanned) Only one problem solution on each paper. Give motivations to all your answers, and explain all symbols that are introduced. Data may be given that are not needed for solving the problems. You are encouraged to use simple figures to clarify explanations. You may answer in English or Swedish. Each problem can give 10 points. (You need 30p to pass.)
2 Problem 1.
A friend of yours is a biology teacher, and she plans to buy a new set of student microscopes for laboratory sessions. One of her requirements is that it should be possible to get a good look at red blood cells (diameter approximately 7 m). Each blood cell should then appear reasonably large, something like the full moon in the sky, she says. Then she delegates the task of finding a suitable microscope to you. When searching the web you find a promising candidate, giving good value for money (the school budget is tight). Your candidate has 10x eyepieces, and objectives labelled 10/0.25, 20/0.45 and 40/0.60. But will it fulfill the teacher’s requirement? The moon has a diameter of 3.5 10 m and is at a distance of 3.8 10 m from Earth.
Problem 2.
Today traditional light bulbs are more and more replaced by LED lamps. Let’s assume that you have found an LED lamp to replace your broken halogen lamp in a conventional transmittedlight microscope with Köhler illumination. When you install the new LED lamp, which has approximately the same luminous flow as the old lamp, you notice that the light emitting surface is very much smaller than the filament of the halogen lamp. In fact the new LED lamp can be viewed almost as a point source, emitting light into a half sphere in space. Describe how this change of light source will affect the following properties of the microscope: a) The maximum size of the illuminated field in the specimen plane. b) The maximum spatial frequency that can be imaged. c) Will the answers to questions a) and b) be different if the point source of light is offcenter, i.e. not centered on the optical axis of the microscope? If so, what will the difference be? Problem 3.
Matrix sensors with square elements are often used for image recording. This means that they will give somewhat different responses depending on the orientation of the image. Let’s assume that we are projecting a line pattern onto the sensor shown to the right. The spatial frequency of the pattern is equal to the Nyquist frequency1 for vertical pattern orientation (lines parallel to the columns in the sensor). Calculate how much the MTF of the sensor will change if we rotate this pattern from vertical to 45 degree orientation. Is this sensitivity to orientation large enough so that you think it will influence the appearance of the recorded images?
1.5 m
5.2 m
Detector element (pixel) with uniform sensitivity Detail of sensor matrix over the entire pixel area (shaded)
1
Nyquist frequency = Highest frequency that can be sampled correctly according to the sampling criterion.
5.2 m 1.5 m
3
Problem 4.
Confocal fluorescence microscopy is often used to study living specimens. The scanning rate is then an important parameter, because the specimen should not change too much during image recording. Let’s assume that we want to study a specimen where the maximum allowed recording time is 30 seconds, and that we want to record a volume of 50 m x 50 m x 50 m. A 60/1.2 water immersion objective is used because the specimen refractive index is close to that of water (1.33). What is the maximum allowed recording time per pixel in this case, assuming that the time needed for refocusing, data transfer etc. is negligible (the entire 30 seconds is available for recording light)? The sampling should be carried out in such a way that the sampling criterion is fulfilled in all three dimensions. We will assume that the detector aperture size is close to that given in eq. 21 in the microscopy compendium, and thus the MTF perpendicular to the optical axis is similar to that in a non-confocal microscope. In the depth dimension, however, performance is close to what is theoretically possible in confocal microscopy. You may assume that the excitation and emission wavelengths are similar, and around 500 nm.
Problem 5.
The previous problem highlights the fact that the time available for collecting photons is often very limited for each pixel. Consider the following case:
We have the same specimen and objective as in the previous problem. The number of fluorophore molecules that is recorded in each pixel is only about 10. The average lifetime of the excited state in a fluorophore molecule is 1.5 ns, after which it returns to the ground state by emitting a photon. The laser light intensity is high, so all fluorophore molecules returning from the excited state to the ground state will be immediately excited again. The recording time per pixel is 1.0 s (this is not the same as the answer to the previous problem, but it is of the same order of magnitude). You must take into account that the fluorescent light is emitted isotropically in all directions, and that only a certain percentage will enter the objective. The light loss between the first lens of the objective and the detector (including detector aperture) is 80% (i.e. only 20% gets through). The quantum efficiency of the photomultiplier tube (PMT) detector is 15%. When the PMT is in darkness it emits, on average, 29 electrons from the cathode during 1.0 s (corresponding to as many photons).
a) Approximately how many photons are, on average, detected for each pixel? b) What signal-to-noise ratio, SNR, can we expect in the measurements? Note that the “dark electrons” are not considered as signal when calculating the SNR.
4 Problem 6. You want to record a fluorescently labelled specimen with a confocal microscope. The excitation and emission spectra of the fluorophore are shown in the figure below, as well as the available laser wavelengths. Excitation (dashed lines) Emission (solid lines)
Wavelength (nm)
450
Laser wavelengths:
550
500 488
514
650
600
540
You want to get the maximum possible light intensity from the specimen so that the scanning time can be made as short as possible. The fluorophore is very resistant to photobleaching. Choose a suitable combination of laser wavelength, dichroic beam splitter and barrier filter (see figures below and on next page). The laser power is similar at all three wavelengths. You must give a good motivation for your choice.
100 %
Transmission
1
2
3
4
5
Transmission curves for dichroic beam splitters (light that is not transmitted is reflected).
4%
400
500
600
700
(nm)
5
100 %
Transmission
1
2
3
4
5
Transmission curves for barrier filters (light that is not transmitted is absorbed).
0
400
600
500
Good Luck! Kjell Carlsson
700
(nm)
6
Solutions to examination in course SK2500/SK2501, 2015-01-13. (Also other “reasonable” solutions may be acceptable) (Sometimes extra comments are given in the solutions that are not necessary to get maximum score at the examination.)
Problem 1. .
The angle under which the moon is seen in the sky is approximately 9.2 10 radians. . The angle under which a red blood cell is seen by the naked eye at a distance of 25 cm is approximately .
2.8 10 radians. Thus the moon appears 329 times larger. Therefore the . microscope needs to have a magnification of at least 330, and therefore your candidate is quite suitable (max. magnification 400). .
Problem 2. When answering these questions, it is useful to look at Fig. 4 in the light microscopy compendium. a) The luminous field diaphragm will be filled with light since the point source emits in a half sphere. The diaphragm is imaged by the condenser, forming a sharp real image in the specimen plane. Therefore the size of the illuminated field is determined only by the diaphragm size. No change compared with the halogen lamp. b) If the lamp is a point source on the optical axis, then its image in the aperture diaphragm plane will also be point-like. This means that the aperture diaphragm has no function any more. It can control neither illumination level nor solid angle for the illumination. All light rays illuminating the specimen will be parallel to the optical axis. Therefore we have the case shown in Fig. 9 in . . . the microscopy compendium. The maximum spatial frequency imaged will then be . The halogen lamp, on the other hand, is an extended source as shown in Fig. 4. Opening up the aperture diaphragm means that we illuminate the specimen also with oblique light rays, and we then have the situation in Fig. 10 in the compendium. With a proper setting for the aperture . . . diaphragm we then get a maximum spatial frequency of . This means that we can see higher frequencies (get sharper images) with the halogen lamp compared with the LED lamp. c) If the point source is off-center, it will not change the size of the illuminated field (the luminous field aperture will still be filled with light). But the light rays illuminating the specimen will now be oblique. The more off-center the point source, the more oblique the illumination will be. This means that we have the situation in Fig. 10, and therefore the maximum spatial frequency is higher than it was with the point source centered. If the point source is sufficiently . . . off-center the maximum spatial frequency is .
Problem 3. The center-to-center spacing between the pixels is 6.7 m. This means that we have a sampling 1.49 10 m-1 in the horizontal direction. The Nyquist frequency for a frequency of .
7.46
vertical pattern is then
10 m-1. With this pattern orientation the MTF of the sensor
will be (see Imaging Physics compendium, page 42) get
sin
. .
. .
0.770.
sin
, because
and
0. We
7 .
If we rotate the same line pattern to a 45 degree angle, we get sin
sin
we get
sin
.
.
10 . Now
0.776.
.
.
5.28
√
The MTF difference for the two cases is less than 1%, and therefore the influence of image orientation should be negligible.
Problem 4. The maximum spatial frequency that can be imaged in a direction perpendicular to the optical axis is . . (Fig. 8 in microscopy compendium). According to the sampling criterion we must sample with . .
twice this frequency, i.e.
.
9.6
10 m-1. This means 9.6
10
50
10
480
samples (= pixels) over a distance of 50 m perpendicular to the optical axis (xy-direction). . .
In the depth (z) direction the maximum spatial frequency that can be imaged is . .
sampling frequency in the z direction must be
. .
4.3
. Therefore the
10 m-1. This means
4.3 10 50 10 217 samples (= optical sections) over a distance of 50 m in the z-direction. This means that we need to collect altogether 480 217 5.0 10 pixels to cover the volume while obeying the sampling criterion. Since the maximum total recording is 30 seconds, this means a maximum recording time per pixel of 6.0 10 s = 0.60 s. .
Problem 5. a) With an average lifetime of 1.5 ns, each molecule will emit (on average)
. .
666.7
photons during the recording time of 1.0 s. Consequently 10 molecules are expected to yield 6667 photons. The percentage entering the objective can be calculated using eq. 10 in the microscopy compendium,. giving 0.5
1
1
.
0.28. (Other reasonable
.
estimates are also okay.) This means 6667 0.28 1896 photons will enter the objective during 1.0 s. Taking into account the light loss and quantum efficiency, we get in the end 1896 0.20 0.15 57 detected photons per pixel (on average).
b)
. As we have seen in a), we can expect an average value of 57. This
gives a photon quantum noise (standard deviation) of √57 7.5. In addition we have a dark signal noise of √29 5.4. The total noise level is given by √57
29
9.3. This gives:
.
6.1.
Problem 6 First of all we can rule out the 540 nm laser wavelength. Excitation will be very poor at this wavelength, and also the number 5 barrier filter will be needed to absorb all reflected laser light. Using the number 5 filter means that we will sacrifice a lot of the fluorescent light (wavelengths below approximately 580 nm). Choosing between 488 and 514 nm, we see that 514 will excite the fluorophore more efficiently. From the excitation spectrum it appears that 514 nm will be 60-70 % more efficient. But then we also need to consider how much of the fluorescent light we can collect. That depends on the properties of the
8 dichroics and barrier filters chosen. The dichroic should reflect the laser light well, but some leakage will occur and is acceptable. The barrier filter, on the other hand, must not transmit any laser light at all. With 488 nm excitation a suitable dichroic will be number 3 (or 4), and a suitable barrier filter will be number 3. With this choice of dichroic and barrier filters practically all of the fluorescence spectrum will be transmitted, which means that we have optimum detection efficiency. With 514 nm excitation a suitable dichroic will be number 4, and a suitable barrier filter will be number 4. From the filter curves we see that these filters will transmit nearly all of the fluorescence spectrum (above approximately 560 nm), and will therefore provide nearly optimal detection efficiency. Taking into account that the excitation is clearly more efficient than for 488 nm, this seems to be the best choice. (All solutions discussing the choice of laser wavelength and filters in an intelligent way will give full, or nearly full, credits.)
