# Problems (and solutions)

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Biomedical & X-ray Physics
Kjell Carlsson
Problems
(and solutions)
for courses SK2500 & SK2501,
Physics of Biomedical Microscopy,
Physics of Biomedical Microscopy, Extended Course
by
Kjell Carlsson
2
Note:
This little collection of problems was originally compiled in the academic year
2003/2004 when the course Physics of Biomedical Microscopy was given for the first
time, and therefore no previous written examinations existed. Such previous
examinations are usually of considerable interest to students, and therefore the
following problems were extracted from examinations given in a previous course (that
was similar, but not identical, to the current course). The problems are “mixed,”
meaning that they are not arranged in the chronological order of the course content.
There is also a strong bias toward the parts covered in the compendium “Imaging
Physics” and the confocal part of “Light Microscopy.” In order to cover some of the
new stuff in the present course, such as Köhler illumination and DIC, problems 1 and
2 were added (these are not old examination problems).
3
Problem 1
Köhler illumination is much used in microscopy. Look at Fig. 4 in the compendium “Light
Microscopy” and answer the following questions.
a)
When the luminous field diaphragm is closed down, less light from the lamp will reach the
specimen. Does this mean that the illuminance level in the specimen is reduced? Explain why,
or why not, this is the case.
b) In photography, closing down the diaphragm opening of the camera lens will increase the
depth of field. Which of the two diaphragms in a microscope will affect the depth of field, and
why doesn’t the other one have the same effect?
c) Which of the two microscope diaphragms affects the resolution, and how should it be adjusted
for best results? Why doesn’t the other diaphragm affect the resolution?
d) Sometimes people with eyeglasses cannot get close enough to the eyepiece of the microscope.
What is the effect if the pupil of the eye is placed too far away from the eyepiece, and why?
Problem 2
A DIC microscope is set up in the usual way, i.e. with a phase shift ϕ =
π
between the sheared beams
2
for equal specimen pathlengths, see “Light Microscopy” page 28. The light intensity obtained when the
2⎛ ϕ ⎞
two sheared beams are recombined is given by the function I = I 0 cos ⎜
⎟ . Thus, for the case of
⎝2⎠
I0
, or 50% of maximum intensity. To clearly distinguish a change in
2
intensity level, we require a change of, say, 20% (i.e., I becomes 40% or 60% of maximum level).
equal pathlengths, we get I =
Under these circumstances, what is thickness, d, of the thinnest plane-parallel piece of glass that can be
detected when inserted into a DIC microscope according to the figure? We assume that monochromatic
light with λ = 550 nm is used.
Air, n = 1
Two beams in the DIC
microscope sheared by
Glass, n = 1.5
d
the distance δ.
δ
Problem 3
True or false, and why? (Consider incoherent imaging)
•
•
•
•
If we know what the point spread function of an optical system looks like, we know
everything there is to know about the imaging properties of the system.
If we know what the point spread function of a diffraction-limited optical system looks like,
we know everything about the imaging properties of the system.
If we know the Rayleigh resolution limit of an imaging system, we know everything about the
imaging properties.
If we know the Rayleigh resolution limit of a diffraction-limited imaging system, we know
4
Problem 4
Let us define ”volume resolution” in confocal microscopy as the product of the resolution figures in the
x, y and z directions. (If, for example, the resolution figures are 1.0 μm, 0.5 μm and 1.5 μm, the
volume resolution would be 1.0 × 0.5 × 1.5 = 0.75 μm3.)
a)
What is the mathematical relationship between the wavelength of the light and the volume
resolution in confocal fluorescence microscopy?
b) What is the mathematical relationship between the numerical aperture of the objective and the
volume resolution in confocal fluorescence microscopy? (It’s okay to assume that the
numerical aperture is small in order to simplify the calculations)
For simplicity you can assume that the excitation and fluorescence wavelengths are equal. You are free
to select which measure of resolution you want from the ones presented in the course.
Problem 5
In this problem you are asked to select a suitable dichroic beam-splitter and barrier filter for confocal
fluorescence microscopy. A simplified schematic drawing of the set-up is shown in the figure below.
Objective
Dichroic beam-splitter
Aperture
Barrier filter
Detector
Fluorescent
specimen
λexc = 488 nm
The emission spectrum of the fluorophore used is given in the figure below.
Intensity
400
500
600
700
λ (nm)
5
You have a choice of 4 different dichroic beam-splitters, whose transmission curves are shown below.
Note that in all cases light that is not transmitted is reflected.
Transmission
90%
a
d
c
b
5%
400
500
700
600
λ (nm)
488
You also have a selection of different barrier filters, whose transmission curves are given in the figure
below.
Transmission
95%
2
1
3
4
5
6
0%
400
500
600
700
λ (nm)
488
Since the light intensity is very low, it is important that as many fluorescence photons as possible can
reach the detector. However, it is also important that absolutely no laser light (which is many orders of
magnitude stronger than the fluorescent light!) can reach the detector. Select a suitable dichroic beamsplitter (a-d) and barrier filter (1-6) to give the best results under the circumstances. Explain carefully
6
Problem 6
To avoid aliasing (i.e. false spatial frequencies due to insufficient sampling density) an “anti-aliasing”
filter can be used between the microscope objective and the CCD detector matrix used for recording
digital images of the specimen. Let’s assume that the detector elements in the CCD are arranged as
shown in the figure below.
5 μm
10 μm
4 μm
y
10 μm
x
The MTF curve of the microscope objective is given by:
MTF
1
0
50
100
mm-1
where the spatial frequency (in units of mm-1) refers to the image plane of the objective, i.e. where the
CCD is located. You have a choice of three different anti-aliasing filters, whose MTF curves are
illustrated on next page.
7
MTF
MTF
mm-1
100 mm-1
MTF
mm-1
100 mm-1
mm-1
50 mm-1
100 mm-1
Which anti-aliasing filter do you choose, and why? Can you see any disadvantages with using an antialiasing filter concerning the image quality? Explain!
It is sufficient to treat the one-dimensional problem (for example x-direction only).
Problem 7
It is a well-known fact that the maximum attainable signal-to-noise ratio (SNR) in light measurements
depends on the number of photons detected during the measurement. Higher SNRs require more
photons. The question is whether it is possible to obtain, in a given measurement situation, a higher
SNR by using an image intensifier. An image intensifier is a device that produces multiple output
photons for each incoming photon (this doesn’t violate the principle of energy conservation, because
we input electric power). Imagine that we couple an image intensifier, that produces 1000 secondary
photons for each incoming one, in front of the detector. We then detect many more photons, but will
we get a higher SNR? Answer this question by studying the SNR in terms of average signal level and
standard deviation when performing repeated measurements. Assume that the detector output signal is
directly proportional to the number of detected photons during the exposure time, and that both
detector and image intensifier introduce negligible amounts of noise (i.e. only photon noise is present).
(Formulas for standard deviation etc. can be found in App. 8 of Imaging Physics)
Problem 8
Christmas is the proper time for making wishes. This year we shall wish for microscope optics that are
useful for all wavelengths from X-rays right up to visible wavelengths. And while we’re at it, let’s wish
for diffraction limited performance at all wavelengths. Supposing Santa delivers the goods, we will
build two microscope set-ups for fluorescence microscopy. In both cases we will use soft X-rays (λ =
2.5 nm) for illuminating fluorescent specimens, which then emit light in the visible region (assume
monochromatic light at λ = 550 nm). One set-up is according to Fig. 13 in the compendium Light
Microscopy. In this set-up conventional (i.e. non-confocal) microscopy is used. A large specimen area
is ”illuminated” by an X-ray ”lamp,” and the fluorescent light from the specimen is imaged by the
objective and eyepiece. The other set-up is confocal, and according to Fig. 18 in the compendium. In
this case we assume that the ”laser excitation light” consists of X-rays. What is the highest spatial
frequency (measured in the specimen plane) that could be imaged by these two systems? We will
assume that the confocal detector aperture is very small.
Problem 9
In cases where we have some prior knowledge concerning the specimen, it may be possible to extract
useful information from images even if the sampling theorem has not been fulfilled. Consider the
following case:
We are using an NA = 0.90 microscope objective for imaging a periodic structure, whose spatial
frequency we know is > 1.8 μm-1. The microscope used is non-confocal, and the imaging is incoherent
at a wavelength of 550 nm. The image is digitally recorded with a sampling distance of 0.33 μm
(measured in the specimen plane). In the recorded image we see a periodic, low-contrast pattern with a
spatial frequency of 0.90 μm-1. Is it possible from these data to uniquely determine the spatial
frequency of the periodic specimen structure, and if so what is the frequency?
8
Problem 10
When working with light microscopy at high magnification, the image quality is sometimes adversely
affected by mechanical vibrations in the set-up. Let’s assume that we are recording images with a (nonconfocal) microscope equipped with a high-quality 100/1.4 objective (i.e. 100x magnification and 1.4
numerical aperture). The specimen emits fluorescent light with a wavelength of approximately 580 nm.
The effects of the vibrations can be described by a point spread function, psfvibrations, according to the
figure below, where x is given in units of meters (in the specimen plane). We only consider a onedimensional case, with vibrations perpendicular to the optical axis of the microscope.
psfvibrations
1
1+
0
x2
4.3 × 10 −15
x
a) Determine MTFvibrations and PTFvibrations.
b) Will the vibrations have any decisive influence on the image quality? (Motivation needed)
Problem 11
In a non-confocal fluorescence microscope the specimen is uniformly illuminated with 436 nm light
from a mercury lamp. The irradiance level is 1.0 mW/mm2. The specimen is imaged using a 100x
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objective, and the “efficiency” of the imaging process is such that it takes 1.0 × 10 illumination
photons to produce one fluorescent photon in the image plane of the objective. Images of the specimen
are recorded by a CCD detector with 1024x1024 elements, each with a light-sensitive area of 10x10
μm2. The detector is located in the image plane of the objective. The specimen is living and therefore
moving. As a result, we want to record as many images per second as possible. We demand, however,
that photon quantum noise must not reduce the signal-to-noise ratio below 20. The quantum efficiency
of the detector is 50%. How many images per second can we record?
Problem 12
We are imaging a specimen in reflected light (λ = 633 nm) using a confocal microscope with a very
small detector aperture. As a result, imaging is purely coherent. The microscope is optimally focused
on the reflecting surface. If, during scanning, the specimen vibrates in a direction parallel to the optical
axis, this will cause intensity variations in the images. What is the maximum allowed vibrational
amplitude if we cannot tolerate larger intensity variations than 5% due to vibrations. Scanning is
performed with an N.A. = 0.65 dry objective.
(Hint: If you encounter an equation you cannot solve analytically, use trial-and-error to get an
approximate solution)
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Problem 13
You want to record a specimen volume of 100μm × 100μm × 100μm using confocal microscopy.
The microscope optics are diffraction limited, and the detector aperture is so small that its influence on
the resolution is negligible. You want to store the volume as a stack of digital images in a computer.
Each pixel value in the images is represented by one byte of data. How many megabytes of data do you
need to record if you don’t want to lose any information due to the sampling process?
The microscope objective has 100 × magnification and a numerical aperture of 1.3. Oil with a
refractive index of 1.52 is used as immersion medium. The specimen is fluorescenctly labelled, and
you may use an “average wavelength” of 550 nm in the calculations.
Problem 14
Single molecule detection has been demonstrated in fluorescence microscopy in recent years. Consider
the following experimental situation:
•
The fluorophore has an excited state lifetime of 3.5 ns (i.e. after excitation the molecule
returns to the ground state emitting a fluorescent photon after, on average, 3.5 ns).
The excitation intensity is “high,” which means that the molecule is excited again immediately
after returning to the ground state.
The fluorophore molecule is expected to last for ten thousand excitation/emission cycles
before it is destroyed by the excitation light.
Fluorescence photons are emitted isotropically in all directions.
The fluorophore molecule is located in immersion oil (n = 1.52), and the objective is a planapo 40/1.0 oil immersion type.
50% of the photons that enter the microscope objective will be lost due to
absorption/reflection in the optical system.
The detector quantum efficiency is 40%.
•
•
•
•
•
•
Estimate (i.e. reasonable approximations are allowed) the number of photons that can be recorded from
a single fluorophore molecule before it is destroyed. How long will it take to record this number of
photons?
Problem 15
A fluorescence microscope according to Fig. 13 in the compendium “Light Microscopy” uses a highpressure mercury lamp with an emission spectrum according to the figure below.
Int.
365
436
302
546
405
313
577
334
λ (nm)
300
400
500
600
700
The specimen is labelled with a fluorophore whose excitation and emission spectra are given by the
figure on next page. (The excitation spectrum shows how efficiently the fluorophore is excited by
different wavelengths.)
10
Int.
Excitation
Emission
λ (nm)
200
300
400
500
600
You have a choice of 5 excitation filters that transmit the following wavelength bands in nanometers:
313 ± 5 , 334 ± 5 , 365 ± 5 , 405 ± 5 , 436 ± 5 , 546 ± 5
Outside these wavelength bands the transmission is nil.
Transmission spectra for available dichroic beamsplitters are given below. Light that is not transmitted
is reflected. Note that these beamplitters are “leaky,” which means that there is a small transmission
also for wavelengths that should be reflected.
Transmission
5%
leakage in
reflection
band
d1
d4
d3
d2
λ (nm)
300
400
500
600
700
The transmission curves for the available barrier filters are shown below.
Transmission
100%
b1
b2
b3
b4
λ (nm)
0
300
400
500
600
700
11
Select suitable excitation, barrier and dichroic filters for the given lamp and fluorophore. Remember
that the excitation light is many orders of magnitude stronger than the fluorescence light. The
microscope objective absorbs wavelengths below 350 nm.
Include short motivations for the choice of filters.
Problem 16
Technologicus microscopicus is an interesting little bacterium whose back is covered with green
fluorescent stripes, see figure. The fluorescence intensity profile resembles a sinewave with a period
length of 0.80 μm, and the contrast is high (nearly 100% modulation).
Fluorescence intensity
0.80 μm
x
x
You are recording a digital image of this stripe pattern using an ordinary (i.e. non-confocal)
fluorescence microscope, equipped with a 25/0.65 dry objective (nearly diffraction-limited). In the
image plane of the objective (see Fig. 1 in “Light Microscopy” compendium) a CCD area array sensor
is located. The sensor comprises 1024 × 1024 detector elements, each with an area of 7.5 × 7.5
μm2, see figure blow. The light sensitivity is uniform over the entire 7.5 × 7.5 μm2 area of an
element, and zero outside.
y
x
7.5 μm
7.5 μm
a)
Estimate what the modulation will be in the recorded image data from the CCD (you may use
your own calculations and/or equations and figures in the compendium to get the result).
b) Will the period length of the recorded pattern be correct (i.e. will aliasing occur or not)?
(Motivation needed)
Problem 17
The confocal principle can be used not only in microscopic imaging, but also in “macroscopic”
applications. In this problem we will consider the use of a confocal set-up for imaging of the blood
perfusion in the skin of a patient. After intravenous injection of a fluorescent substance, the skin
fluorescence is recorded with an optical set-up according to the figure on next page. What optical
12
section thickness (measured as full-width-half-maximum) can we expect when using this kind of setup. The detector aperture is so small that it does not affect the depth resolution.
( λ laser ≈ λ fluorescence ≈ 500 nm).
Laser light
Aperture
Detector
Dichroic beamsplitter
Lens diameter 50 mm
0.50 m
Air
Fluorescence light
Patient
Problem 18
Aliasing, i.e. moiré-patterns due to insufficient sampling density, is often a problem when using area
array sensors. In some cases, however, a bit of “detective work” can tell us whether a periodic pattern
has been recorded correctly or not. In the present example we have no data to tell us whether the
sampling criterion was fulfilled or not. An area array sensor with 256 x 256 detector elements was
used, each with a light-sensitive area of 10 x 10 μm2 (see figure 1). The light sensitivity is uniform over
the entire area of each detector element and zero outside, and the response is linear (i.e. pixel values are
proportional to exposure). Using this sensor, a pattern according to fig. 2 was obtained. Explain why
we can be sure that in this particular case no aliasing has occurred, and that therefore the pattern in fig.
2 is recorded with correct period length.
(Hint: consider detector MTF)
y
x
10 μm
10 μm
Fig. 1.
13
Pixel value
255
(white)
0
(black)
x
Fig. 2.
Problem 19
You want to record, with as much detail as possible, a very large specimen volume (approx. 2 mm x 2 mm x 2
mm) in a confocal fluorescence microscope. You are using a (dry) 5x objective with numerical aperture 0.10.
Suggest a suitable sampling density (sampling points/mm in the specimen volume) in all three dimensions. Would
you expect that differences in resolution in the horizontal and vertical directions will be pronounced in this case?
Explain!
(The detector aperture is “very small,” and the objective is nearly diffraction-limited. Both excitation and
fluorescence wavelengths are close to 550 nm.)
Problem 20
In fluorescence measurements the light flux is often quite low, and therefore the signal-to-noise ratio
(SNR) is often rather poor. The situation can be improved by prolonging the measurement time so that
more photons are collected. One problem with using long measurement times is that the fluorescence
intensity often decays gradually over time. The reason for this is that fluorophore molecules are
destroyed by the excitation light (this is referred to as photo-bleaching). Consider the two cases shown
in the figure below. In case 1 no photo-bleaching occurs. Therefore, the detection rate of photons (i.e.
the number of photons detected per second) is I0 = 10 000 s-1 independent of the measurement time. In
case 2 photo-bleaching occurs, so that the detection rate is given by I (t ) = I 0 ⋅ e
− αt
, where
I0 = 10 000 s and α = 100 s .
-1
-1
I0
Case 1
Case 2
t
a)
Calculate for cases 1 and 2 the maximum SNR that is theoretically possible as a function of
measurement time τ.
b) In case 2 it is pointless to use very long measurement times. Where approximately would you
say that this limit is (use a common sense estimation).
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Solutions.
Problem 1
a)
When closing the luminous field diaphragm, the illuminance level in the specimen remains
constant. The illuminated area is, however, reduced. The total luminous flux is thus reduced,
but this smaller flux is distributed over a correspondingly smaller area, leaving the
illuminance level constant.
b) Depth of field is controlled by the aperture diaphragm, which controls the total solid angle
under which the objective collects light. Reducing this angle reduces the blur circle that occurs
when an object is out of focus. Changing the luminous field diaphragm does not change this
solid angle, and therefore does not affect the depth of field.
c) Resolution is affected by the aperture diaphragm. It should be adjusted so that the numerical
aperture of the condenser matches that of the objective, see “Light Microscopy” p. 19.
d) If the eye pupil does not coincide with the exit pupil of the microscope, we will not be able to
see the entire field of view simultaneously. This feels rather like looking through a keyhole.
The reason for this phenomenon is that rays of all different directions (corresponding to
different positions in the specimen plane) form a beam waist at the exit pupil. If the eye is held
there all these rays can enter the eye simultaneously. If not, rays from some directions will
miss the eye pupil.
Problem 2
If the phase angle ϕ changes by 12°, corresponding to an optical path difference of
λ
, the light
30
intensity will change by approx. 20%. The optical path difference is then given by
(n − 1)d = 0.5d =
550
nm ⇒ d = 37 nm. Thus, we can expect that a glass thickness of
30
approximately 40 nm can be detected.
Problem 3
True. Given the psf we can calculate, for an arbitrary object, what the image function will be.
(The image function is the convolution of the object function and the psf, see Imaging Physics
page 22)
True. This is a more limited case than the one above. Both are therefore true.
False. For an arbitrary (i.e. not diffraction-limited) optical system we cannot deduce the psf
from the Rayleigh limit alone. Systems with different psfs can have the same Rayleigh limit.
Knowledge of the psf is necessary to completely characterize the imaging properties.
True. If we know the Rayleigh limit of a diffraction-limited system we know what the psf
looks like (Light Microscopy, p. 13-14). According to case 1 above, we then know exactly
what the imaging properties are.
•
•
•
•
Problem 4
a) If the Rayleigh criterion and optical section thickness are taken as resolution measures, we get:
λ
0.45 λ
8.5 λ
Volume resolution = 0N.45
2 α . This means that the volume resolution is proportional
. A. × N . A. ×
8πn sin ( 2 )
to λ .
3
b) In order to solve this, we have to find out how the optical section thickness depends on the
numerical aperture. If we assume that N.A. is small, we get sin
Therefore the volume resolution is proportional to
(N . A.)
−4
( ) ≈ 14 ⋅ sin 2 (α) = 41n (N . A.)2 .
2 α
2
2
.
Problem 5
A suitable choice would be beam-splitter b and barrier filter 3. In this way most of the laser light will
be reflected towards the specimen, and the fluorescent light will be efficiently transmitted by the beamsplitter (a would reflect very little laser light towards the specimen, whereas c and d would mean that
we waste fluorescent light). Barrier filter 3 is the best choice, given the requirement that we must
15
block all laser light (note that all beam-splitters transmit a few percent of the laser light reflected back
from the specimen) and transmit as much as possible of the fluorescent light.
Problem 6
With the CCD matrix used, we have a sampling frequency of 100 mm-1, meaning that the highest
spatial frequency we can correctly record is 50 mm-1. The objective can reproduce much higher
frequencies than this, and this is true also for the detector MTF, whose first zero value occurs at 200
mm-1 (x direction) and 250 mm-1 (y direction). Total MTF is the product the component MTFs, and
therefore we need to add a component whose limiting frequency is 50 mm-1. We should therefore add
the anti-aliasing filter with the lowest limiting frequency, which is 50 mm-1; the others cannot prevent
aliasing completely. A disadvantage with using an anti-aliasing filter is that the modulation (i.e.
contrast) for frequencies well below 50 mm-1 will also be considerably reduced. Therefore the image
will not appear as sharp as it would otherwise be.
Problem 7
SNR =
mean value
=
standard deviation
n
∑i
1
n
n
∑ (i
k
k =1
, where ik represents the value from an individual
− imean )2
k
k =1
n −1
measurement. When using an image intensifier, all light values will be larger by a factor of F (= 1000
in this case). Thus, all i-values in the above equation will be replaced by Fi. We then get
SNR =
mean value
=
standard deviation
1
n
n
∑ Fi
n
∑ (Fi
k
k =1
k
, which gives exactly the same SNR as in the
− Fimean )
2
k =1
n −1
previous case. Using an image intensifier will produce both a larger output signal and higher noise, but
the SNR will be the same (compare electron multiplication in a photomultiplier tube, which will
produce the same effect).
Problem 8
In the non-confocal case the highest spatial frequency that can be imaged is given by
2 N.A.
(see Fig.
λ
8 in compendium “Light Microscopy”), where λ is the fluorescence wavelength (all imaging is done in
visible light - the X-rays are only used for “flooding” an extended area of the specimen with energy to
excite the fluorophore molecules). In the case of confocal microscopy, however, the X-rays will be
focused to a diffraction-limited (i.e. very small) spot by the objective. The same objective will then
image fluorescent light from the (X-ray) illuminated spot onto the detector aperture. As a result, the
spatial frequecy cut-off for the system will be given by
2 N.A. 2 N.A.
(see compendium “Light
+
λ exc.
λ fluor .
Microscopy” p. 36). In the first case, the spatial frequency cut-off is given by
2 N.A.
= 3.6 × 10 6 ⋅ N.A. m-1, which is equal to 3.6 ⋅ N.A. μm-1. Thus, in this case resolution
−9
550 × 10
is not improved by using X-ray excitation. In the second case, the resolution is determined almost
entirely by the first term
2 N.A.
2 N.A.
=
= 8.0 × 108 ⋅ N.A. m-1, which is equal to
−9
λ exc.
2.5 × 10
800 ⋅ N.A. μm-1. In this case we really get a benefit from the short wavelength of the X-rays.
16
Problem 9
Using an N.A. 0.90 objective, the highest spatial frequency that can be imaged is
2 N . A.
2 ⋅ 0.90
=
m -1 = 3.3 μm-1. Therefore, we know that the true spatial frequency is
−9
λ
550 × 10
somewhere between 1.8 and 3.3 μm-1. The sampling frequency is 3.0 μm-1, meaning that the highest
frequency that can be correctly recorded is 1.5 μm-1. We clearly have a case of aliasing, and frequency
νalias = 0.90 μm-1. Using the formula for aliasing in Appendix 8 of “Imaging Physics,” we get
ν alias = n ⋅ ν sampling − ν real , where n is a positive integer number. n = 1 is the only possibility, since
n > 1 would give νreal > 5 μm-1. For n = 1 we get two possibilities, namely νreal = 2.1 and νreal =
3.9 μm-1. 3.9 is outside the possible range, and therefore the spatial frequency of the structure must be
2.1 μm-1.
Problem 10
a)
The OTF is the Fourier transform of the psf, normalized to unity at zero spatial frequency.
Using the FT table on p. 310 in β, we find that OTFvibrations =
−15
−7
e − 4.3×10 ⋅2 πν = e − 4.12 ×10 ⋅ ν . This means that MTFvibrations = OTFvibrations, and
PTFvibrations = 0.
b) With the objective used (asuming nearly diffraction-limited performance), we get an
MTFobjective that decreases almost linearly, reaching zero at
ν lim it =
2 N.A.
2 ⋅ 1.4
= 4.83 × 106 m-1. We therefore get MTFvibrations(νlimit) =
=
−
9
λ
580 × 10
0.14, and MTFvibrations(νlimit/2) = 0.37. This means that the vibrations will cause a serious
deterioration in image quality.
Problem 11
An irradiance level of 1.0 mW/mm2 means that
1.0 × 10 −3
= 2.19 × 1015 excitation photons reach
hc
λ
−4
the number of photons will
the specimen per second and mm2. Since the efficiency is only 1.0 × 10
be reduced by this factor. Furthermore, because of the magnification, they will be spread out over an
7
area that is 100 × 100 times larger. As a result, 2.19 × 10 fluorescence photons reach the image
−4
3
mm2, 2.19 × 10 photons
plane per second and mm2. With a detector element area of 1.0 × 10
will reach a detector element per second. Only half of the photons will be detected, and therefore
1.10 × 103 photons will be detected per second. In order to get a SNR of 20, we need to detect (on
average) 400 photons in a measurement. With the given photon flow, this will number is obtained in
400
1.10 × 103
per second.
= 0.36 seconds. Therefore, we can record a maximum of approximately three images
17
Problem 12
2
⎛ sin( u / 2) ⎞
⎟ , where
⎝ u/2 ⎠
The light intensity as a function of defocus, z, is given by I( u ) = ⎜
u=
8π
⋅ z ⋅ sin 2 ( α / 2) , and α is obtained from N.A. = sin α. A little trial-and-error qickly shows
λ
that u = 0.78 will give I(u) = 0.95. From this we can calculate that
uλ
0.78 ⋅ 633 × 10 −9
z=
=
= 1.6 × 10 − 7 m = 0.16 μm. This means that when the
2
2
o
8π sin ( α / 2)
8π sin ( 20.3 )
vibrational amplitude is greater than 0.16 μm, we will get a drop in light intensity that exceeds 5%.
Problem 13
From page 39 in the compendium “Light Microscopy” we get the maximum spatial frequency
components that can be imaged in all three dimensions:
4 N.A.
4 ⋅ 1 .3
=
= 9.45 × 10 6 m-1
λ
550 × 10 −9
( N.A.)2
(1.3)2
= 2.02 × 106 m-1
=
=
λn
550 × 10−9 ⋅ 1.52
νmax,xy =
ν max,z
According to the sampling theorem we must sample with the following frequencies to record all
information:
ν s,xy = 2ν max,xy = 1.89 × 10 7 samples/m.
ν s,z = 2ν max,z = 4.04 × 10 6 samples/m.
The total number of samples necessary in a 100μm × 100μm × 100μm volume is
(1.89 × 107 ⋅ 100 × 10−6 )2 ⋅ 4.04 × 106 ⋅ 100 × 10−6 = 1.4 × 109 sampling points. With one byte
of data per sampling point we get 1.4 gigabytes of data (which is no big deal today).
Problem 14
We can expect that altogether 1.0 × 10 photons will be emitted by the fluorophore molecule we are
studying. How many of these can we expect to record?
Since the photons are emitted isotropically we must find the solid angle under which we collect
photons and compare this with the total solid angle of 4π.
4
Spherical surface,
area = A
dφ
φ
α
r
Flat, circular surface,
area = A’
N.A. = n ⋅ sin α
Specimen
18
The microscope objective will collect photons under a solid angle of Ω =
A
. We can calculate Ω
r2
r ⋅ dφ ⋅ 2π ⋅ r ⋅ sin φ
= 2π ⋅ sin φ ⋅ dφ , and integrate over φ from 0 to α. We then get
r2
Ω = 2π(1 − cos α ) , i.e. the same result as eq. 10 in “Light Microscopy.” This is an exact formula.
from dΩ =
Since the task is to estimate the photon number, it is allowed to make a simpler “small angle”
estimation of the solid angle which gives Ω ≈
A′ π(r ⋅ sin α )2
=
= π ⋅ sin 2α .
r2
r2
For an oil immersion objective with N.A. = 1, we get α = 41.1˚. The exact formula then gives Ω = 1.55
steradians (the approximate formula gives 1.36).
1.55
= 0.123 . Because of
4π
losses in the optics and detector, only the fraction 0.50 × 0.40 = 0.20 of these will be detected.
Thus, the fraction of photons that will be collected by the objective is
Therefore, the total number of detected photons is expected to be
1.0 × 104 ⋅ 0.123 ⋅ 0.20 = 2.5 × 102 (using the approximate solid angle, we get 2.2 × 102 ).
How long will it take to record this number of photons?
On average, one photon will be emitted every 3.5 ns. This means that it takes
3.5 × 10−9 ⋅ 1.0 × 104 = 3.5 × 10−5 s = 35 μs for the molecule to emit ten thousand photons (of
Problem 15
The objective transmits λ > 350 nm. For excitation we need λ < 400 nm. The only excitation
wavelength that fulfills these requirements is 365nm. This wavelength is not ideal for excitation of the
fluorophore ( ≈ 310 should be optimal), but the emission is intense and should be quite sufficient. The
excitation filter should therefore be 365 ± 5 .
The dichroic filter should reflect 365 nm efficiently, but transmit as much as possible of the fluorescent
light. d2 fulfills these requirements.
The barrier filter must remove all of the remaining 365 nm radiation that is transmitted by the dichroic
filter. This means that we have to choose b2 even though it will absorb some of the fluorescent light.
Problem 16
a)
Degree of modulation in recorded image = Degree of modulation in object × MTF. We must
determine the MTF-value for the entire imaging chain at the spatial frequency of the
fluorescent pattern. MTFtotal = MTFoptics × MTFdetector . MTFoptics is given by the
diffraction-limited curve in Fig. 8 in the compendium “Light Microscopy.” The spatial
frequency of the pattern is
ν = (0.8 × 10− 6 ) = 1.25 × 106 m-1. The limiting frequency is
−1
2 N.A.
2 ⋅ 0.65
=
= 2.36 × 106 m-1 (we have assumed a wavelength of 550 nm). This
λ
550 × 10−9
means that the spatial frequency of the pattern is 0.53 times the limiting frequency. From the
curve in Fig. 8 in “Light Microscopy” we can measure that MTFoptics ≈ 0.38 at this
frequency.
For
the
detector
we
get
(see
compendium
on
Imaging
Physics)
sin (πνL )
, where ν is the spatial frequency of the magnified image of the
MTFdetector =
πνL
1.25 × 106
= 5.0 × 104 m-1, and L is the width of a detector element, i.e.
pattern, i.e.
25
19
7.5 × 10−6 m. Inserting these numbers, we get MTFdetector = 0.78. As a result,
MTFtotal ≈ 0.38 ⋅ 0.78 = 0.30 . Since the degree of modulation in the object was nearly
100%, we can expect that the degree of modulation in the recorded image will be nearly 30%.
b) In order to correctly record the spatial frequency of the image we must have at least two
sample points per period of the pattern (sampling theorem). The period length in the image
−6
−5
will be 0.80 × 10 ⋅ 25 = 2.0 × 10 m = 20 μm. Since the distance between sample points
is 7.5 μm, we get 2.7 sample points per period, and therefore the sampling theorem is fulfilled.
Yes, the pattern will be recorded with correct period length.
Problem 17
Equation 18 in the compendium “Light Microscopy” gives the FWHM optical sectioning thickness in
⎛ 25 × 10−3 ⎞
⎟⎟ = 2.86o . We get
0
.
50
⎝
⎠
confocal fluorescence microscopy. In the current example α = arctan⎜⎜
FWHM =
8.5 ⋅ 500 × 10−9
= 2.7 × 10− 4 m = 0.3 mm. This is far from microscopic resolution, but
o
2
8π ⋅ sin (1.43 )
can still give valuable information concerning skin structures.
Problem 18
Let us denote the sensor width (10 μm) by D. The sensor then has a sampling frequency in the x
direction of
1
1
. This means that aliasing occurs for frequencies above
(= half the sampling
D
2D
frequency). A sensor with uniform sensitivity over the entire width D, and zero outside, will have an
MTF given by MTF(ν ) =
1
sin (πνD )
, where ν is the spatial frequency. At a frequency of
we
πνD
2D
get an MTF-value of 0.64, and for higher frequencies we get lower values. This means that for spatial
frequencies where aliasing occurs, we always get MTF < 0.64. This in turn means that when aliasing
occurs the degree of modulation of the pattern obtained will always be lower than 0.64 (the degree of
modulation in the original pattern can never be higher than unity). A simple measurement in the figure
gives a degree of modulation slightly above 0.8, and therefore we can conclude that aliasing cannot
have taken place.
Problem 19
Assuming perfect optics, and an infinitely small detector aperture, a confocal microscope can record
spatial frequencies up to
4 N.A.
( N.A.)2
in the horizontal direction and
in the vertical direction (p.
λ
λ
39 in compendium “Light Microscopy”). With current data inserted into these equations, we get the
frequencies 727 mm-1 (horizontal) and 18.2 mm-1 (vertical). In order to correctly record these
maximum frequencies without aliasing, we have to use sampling densities of twice these values, i.e.
1455 mm-1 and 36.4 mm-1. In reality we could probably use somewhat lower sampling densitites,
because the MTF-value becomes very low as we approach the limiting frequency, see Fig. 22 in “Light
Microscopy”).
As we can see, there is a large difference in the maximum frequencies that can be recorded in the
horizontal and vertical directions respectively (a factor of 40!). Therefore we can expect rather serious
problems with different resolution in the horizontal and vertical directions.
20
Problem 20
The maximum SNR possible in a measurement where we can expect N photons to be detected is
N . In case 1 N = 1.0 × 10 4 ⋅ τ , where τ is the measurement time, and therefore we get
τ
SNR = 100 τ . In case 2 we get N = ∫ I 0e − αt dt =
0
−100 τ
I0
(1 − e−ατ ) = 100 (1 − e−100τ ) , and
α
.
therefore SNR = 10 1 − e
In case 2 one can easily calculate that at τ = 0.04 seconds we have reached 99% of maximum SNR, and
certainly it is rather pointless to go on measuring after that.
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