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MLS Review Harr - Clinical
Chemistry
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,Which formula correctly describes the relationship D. All of these options
between absorbance and %T ?
D Absorbance is proportional to the inverse log of transmittance.
A. A = 2 - log %T A = -log T = log 1/T
B. A = log 1/T Multiplying the numerator and denominator
C. A = -log T by 100 gives:
D. All of these options A = log (100/100 X T)
100 X T = %T,
substituting %T for 100 X T gives:
A = log 100/%T
A = log 100 - log %T
A = 2.0 - log %T
For example, if %T = 10.0, then:
A = 2.0 - log 10.0
log 10.0 = 1.0
A = 2.0-1.0 = 1.0
A solution that has a transmittance of 1.0 %T would have B. 2.0
an absorbance of:
B
A. 1.0 A = 2.0 - log %T
B. 2.0 A = 2.0 - log 1.0
C. 1% The log of 1.0 = 0
D. 99% A = 2.0
In absorption spectrophotometry: D. Absorbance is directly proportional to concentration
A. Absorbance is directly proportional to D Beer's law states that A = a × b × c, where a is the absorptivity coefficient (a
transmittance constant), b is the path length, and c is concentration. Absorbance is directly
B. Percent transmittance is directly proportional to proportional to both b and c. Doubling the path length results in incident light
concentration contacting twice the number of molecules in solution. This causes absorbance to
C. Percent transmittance is directly proportional to the double, the same effect as doubling the concentration of molecules.
light path length
D. Absorbance is directly proportional to concentration
Which wavelength would be absorbed strongly by a red- A. 450 nm
colored solution?
A A solution transmits light corresponding in wavelength to its color, and usually
A. 450 nm absorbs light of wavelengths complementary to its color. A red solution transmits
B. 585 nm light of 600-650 nm and strongly absorbs 400-500 nm light.
C. 600 nm
D. 650 nm
A green-colored solution would show highest B. 525 nm
transmittance at:
B Green light consists of wavelengths from 500-550 nm. A green-colored solution
A. 475 nm with a transmittance maximum of 525 nm and a 50-nm bandpass transmits light
B. 525 nm of 525 nm and absorbs light below 475 nm and above 575 nm. A solution that is
C. 585 nm green would be quantitated using a wavelength that it absorbs strongly, such as
D. 620 nm 450 nm.
SITUATION: A technologist is performing an enzyme A. Replace the source lamp
assay at 340 nm using a visible-range
spectrophotometer. After setting the wavelength and A Visible spectrophotometers are usually supplied with a tungsten or quartz
adjusting the readout to zero %T with the light path halogen source lamp. Tungsten lamps produce a continuous range of
blocked, a cuvette with deionized water is inserted. With wavelengths from about 320-2,000 nm. Output increases as wavelength
the light path fully open and the 100%T control at becomes longer peaking at around 1,000 nm, and is poor below 400 nm. As the
maximum, the instrument readout will not rise above lamp envelope darkens with age, the amount of light reaching the photodetector
90%T. What is the most appropriate first course of at 340 nm becomes insufficient to set the blank reading to 100%T. Quartz
action? halogen lamps produce light from 300 nm through the infrared region. Deuterium
or hydrogen lamps produce ultraviolet-rich spectra optimal for ultraviolet (UV)
A. Replace the source lamp work. Mercury vapor lamps produce a discontinuous spectrum that includes a
B. Insert a wider cuvette into the light path high output at around 365 nm that is useful for fluorescent applications. Xenon
C. Measure the voltage across the lamp terminals lamps generate a continuous spectrum of fairly uniform intensity from 300-2,000
D. Replace the instrument fuse nm, making them useful for both visible and UV applications.
Which type of monochromator produces the purest D. A prism and a variable exit slit
monochromatic light in the UV range?
D Diffraction gratings and prisms both produce a continuous range of
A. A diffraction grating and a fixed exit slit wavelengths. A diffraction grating produces a uniform separation of wavelengths.
B. A sharp cutoff filter and a variable exit slit A prism produces much better separation of high-frequency light because
C. Interference filters and a variable exit slit refraction is greater for higher-energy wavelengths. Instruments using a prism
D. A prism and a variable exit slit and a variable exit slit can produce UV light of a very narrow bandpass. The
adjustable slit is required in order to allow sufficient light to reach the detector to
set 100%T.
,Which monochromator specification is required in order D. 5-nm bandpass
to measure the true absorbance of a compound having a
natural absorption bandwidth of 30 nm? D Bandpass refers to the range of wavelengths passing through the sample. The
narrower the bandpass, the greater the photometric resolution. Bandpass can be
A. 50-nm bandpass made smaller by reducing the width of the exit slit. Accurate absorbance
B. 25-nm bandpass measurements require a bandpass less than one-fifth the natural bandpass of
C. 15-nm bandpass the chromophore
D. 5-nm bandpass
Which photodetector is most sensitive to low levels of D. Photomultiplier tube
light?
D The photomultiplier tube uses dynodes of increasing
A. Barrier layer cell voltage to amplify the current produced by the photosensitive cathode. It is
B. Photodiode 10,000 times as sensitive as a barrier layer cell, which has no amplification. A
C. Diode array photomultiplier tube requires a DC-regulated lamp because it responds to light
D. Photomultiplier tube fluctuations caused by the AC cycle.
Which condition is a common cause of stray light? C. Dispersion from second-order spectra
A. Unstable source lamp voltage C Stray light is caused by the presence of any light other than the wavelength of
B. Improper wavelength calibration measurement reaching the detector. It is most often caused by second-order
C. Dispersion from second-order spectra spectra, deteriorated optics, light dispersed by a darkened lamp envelope, and
D. Misaligned source lamp extraneous room light.
A linearity study is performed on a visible D. Stray light
spectrophotometer at 650 nm and the following
absorbance readings are obtained: D Stray light is the most common cause of loss of linearity at high-analyte
concentrations. Light transmitted through the cuvette is lowest when absorption
Concentration of Standard is highest. Therefore, stray light is a greater percentage of the detector response
10.0 mg/dL when sample concentration is high. Stray light is usually most
20.0 mg/dL significant when measurements are made at the extremes of the visible spectrum
30.0 mg/dL because lamp output and detector response are low.
40.0 mg/dL
50.0 mg/dL
Absorbance
0.20
0.41
0.62
0.79
0.92
The study was repeated using freshly prepared
standards and reagents, but results were identical to
those shown. What is the most likely cause of
these results?
A. Wrong wavelength used
B. Insufficient chromophore concentration
C. Matrix interference
D. Stray light
Which type of filter is best for measuring stray light? C. Sharp cutoff
A. Wratten C Sharp cutoff filters transmit almost all incident light until the cutoff wavelength
B. Didymium is reached. At that point, they cease to transmit light. Because they give an "all or
C. Sharp cutoff none effect," only stray light reaches the detector when the selected wavelength
D. Neutral density is beyond the cutoff.
Which of the following materials is best suited for D. Holmium oxide glass
verifying the wavelength calibration of a
spectrophotometer? D Wavelength accuracy is verified by determining the wavelength reading that
gives the highest absorbance (or transmittance) when a substance with a narrow
A. Neutral density filters natural bandpass (sharp absorbance or transmittance peak) is scanned. For
B. Potassium dichromate solutions traceable to the example, didymium glass has a sharp absorbance peak at 585 nm. Therefore, an
National Bureau of Standards reference instrument should give its highest absorbance reading when the wavelength dial
C. Wratten filters is set at 585 nm. Holmium oxide produces a very narrow absorbance peak at
D. Holmium oxide glass 361 nm; likewise, the hydrogen lamp of a UV spectrophotometer produces a
656-nm emission line that can be used to verify wavelength. Neutral density
filters and dichromate solutions are used to verify absorbance accuracy or
linearity. A Wratten filter is a widebandpass filter made by placing a thin layer of
colored gelatin between two glass plates and is unsuitable for
spectrophotometric calibration
, Why do many optical systems in chemistry analyzers B. To minimize error caused by source lamp fluctuation
utilize a reference light path?
B A reference beam is used to produce an electrical signal at the detector to
A. To increase the sensitivity of the measurement which the measurement of light absorption by the sample is compared. This
B. To minimize error caused by source lamp fluctuation safeguards against measurement errors caused power fluctuations that change
C. To obviate the need for wavelength adjustment the source lamp intensity. Although reference beams increase the accuracy of
D. To reduce stray light effects measurements, they do so at the expense of optical sensitivity since some of the
incident light must be used to produce the reference beam
Which component is required in a spectrophotometer in C. Photodiode array
order to produce a spectral absorbance curve?
C There are two ways to perform spectral scanning for compound identification.
A. Multiple monochromators One is to use a stepping motor that continuously turns the monochromator so
B. A reference optical beam that the wavelength aligned with the exit slit changes at a constant rate. A more
C. Photodiode array practical method is to use a diode array detector. This consists of a chip
D. Laser light source embedded with as many as several hundred photodiodes. Each photodiode is
aligned with a narrow part of the spectrum produced by a diffraction grating, and
produces current proportional to the intensity of the band of light striking it
(usually 1-2 nm in range). The diode signals are processed by a computer to
create a spectral absorbance or transmittance curve.
The half-band width of a monochromator is defined by: A. The range of wavelengths passed at 50% maximum transmittance
A. The range of wavelengths passed at 50% maximum A Half-band width is a measure of bandpass made using a solution or filter
transmittance having a narrow natural bandpass (transmittance peak). The wavelength giving
B. One-half the lowest wavelength of optical purity maximum transmittance is set to 100%T (or 0 A). Then, the wavelength dial is
C. The wavelength of peak transmittance adjusted downward, until a readout of 50%T (0.301 A) is obtained. Next, the
D. One-half the wavelength of peak absorbance wavelength is adjusted upward
until 50%T is obtained. The wavelength difference is the half-band width. The
narrower the half-band width, the better the photometric resolution of the
instrument.
The reagent blank corrects for absorbance caused by: A. The color of reagents
A. The color of reagents A When a spectrophotometer is set to 100%T with the reagent blank instead of
B. Sample turbidity water, the absorbance of reagents is automatically subtracted from each
C. Bilirubin and hemolysis unknown reading. The reagent blank does not correct for absorbance caused by
D. All of these options interfering chromogens in the sample such as bilirubin, hemolysis, or turbidity.
A plasma sample is hemolyzed and turbid. What is C. Substitute saline for the reagent
required to perform a sample blank in order to correct the
measurement for the intrinsic absorbance of the sample C A sample blank is used to subtract the intrinsic absorbance of the sample
when performing a spectrophotometric assay? usually caused by hemolysis, icterus, turbidity, or drug interference. On
automated analyzers, this is accomplished by measuring the absorbance after
A. Substitute deionized water for the sample the addition of sample and a first reagent, usually a diluent. For tests using a
B. Dilute the sample 1:2 with a standard of known single reagent, sample blanking can be done prior to the incubation phase before
concentration any color develops. Substituting deionized water for sample is done to subtract
C. Substitute saline for the reagent the absorbance of the reagent (reagent blanking). Diluting the sample with a
D. Use a larger volume of the sample standard (standard addition) may be done when the absorbance is below the
minimum detection limit for the assay. Using a larger volume of sample will make
the interference worse.
Which instrument requires a highly regulated DC power C. A spectrophotometer with a photomultiplier tube
supply?
C When AC voltage regulators are used to isolate source lamp power, light
A. A spectrophotometer with a barrier layer cell output fluctuates as the voltage changes. Because this occurs at 60 Hz, it is not
B. A colorimeter with multilayer interference filters detected by eyesight or slow-responding detectors. Photomultiplier tubes are
C. A spectrophotometer with a photomultiplier tube sensitive enough to respond to the AC frequency and require a DC-regulated
D. A densitometer with a photodiode detector power supply.
Which statement regarding reflectometry is true? C. 100% reflectance is set with an opaque film called a white reference
A. The relation between reflectance density and C Reflectometry does not follow Beer's law, but the relationship between
concentration is linear concentration and reflectance can be described by a logistic formula or algorithm
B. Single-point calibration can be used to determine that can be solved for concentration. For example,
concentration K/S = (1 - R) 2/2R, where K = Kubelka-Munk absorptivity constant, S = scattering
C. 100% reflectance is set with an opaque film called a coefficient, R = reflectance density. K/S is proportional to concentration. The
white reference white reference is analogous to the 100%T setting in spectrophotometry and
D. The diode array is the photodetector of choice serves as a reference signal. Dr = log R0/R1, where Dr is the reflectance density,
R0 is the white reference signal, and R1 is the photodetector signal for the test
sample.
MLS Review Harr - Clinical
Chemistry
** multiple choices Questions and
rationale covering all chapters
** Expert-Verified Explanation
** Questions with Verified Answer
** New Edition | Updated
** 100% Guaranteed Pass
** 100% Correct Answers | Graded A+
,Which formula correctly describes the relationship D. All of these options
between absorbance and %T ?
D Absorbance is proportional to the inverse log of transmittance.
A. A = 2 - log %T A = -log T = log 1/T
B. A = log 1/T Multiplying the numerator and denominator
C. A = -log T by 100 gives:
D. All of these options A = log (100/100 X T)
100 X T = %T,
substituting %T for 100 X T gives:
A = log 100/%T
A = log 100 - log %T
A = 2.0 - log %T
For example, if %T = 10.0, then:
A = 2.0 - log 10.0
log 10.0 = 1.0
A = 2.0-1.0 = 1.0
A solution that has a transmittance of 1.0 %T would have B. 2.0
an absorbance of:
B
A. 1.0 A = 2.0 - log %T
B. 2.0 A = 2.0 - log 1.0
C. 1% The log of 1.0 = 0
D. 99% A = 2.0
In absorption spectrophotometry: D. Absorbance is directly proportional to concentration
A. Absorbance is directly proportional to D Beer's law states that A = a × b × c, where a is the absorptivity coefficient (a
transmittance constant), b is the path length, and c is concentration. Absorbance is directly
B. Percent transmittance is directly proportional to proportional to both b and c. Doubling the path length results in incident light
concentration contacting twice the number of molecules in solution. This causes absorbance to
C. Percent transmittance is directly proportional to the double, the same effect as doubling the concentration of molecules.
light path length
D. Absorbance is directly proportional to concentration
Which wavelength would be absorbed strongly by a red- A. 450 nm
colored solution?
A A solution transmits light corresponding in wavelength to its color, and usually
A. 450 nm absorbs light of wavelengths complementary to its color. A red solution transmits
B. 585 nm light of 600-650 nm and strongly absorbs 400-500 nm light.
C. 600 nm
D. 650 nm
A green-colored solution would show highest B. 525 nm
transmittance at:
B Green light consists of wavelengths from 500-550 nm. A green-colored solution
A. 475 nm with a transmittance maximum of 525 nm and a 50-nm bandpass transmits light
B. 525 nm of 525 nm and absorbs light below 475 nm and above 575 nm. A solution that is
C. 585 nm green would be quantitated using a wavelength that it absorbs strongly, such as
D. 620 nm 450 nm.
SITUATION: A technologist is performing an enzyme A. Replace the source lamp
assay at 340 nm using a visible-range
spectrophotometer. After setting the wavelength and A Visible spectrophotometers are usually supplied with a tungsten or quartz
adjusting the readout to zero %T with the light path halogen source lamp. Tungsten lamps produce a continuous range of
blocked, a cuvette with deionized water is inserted. With wavelengths from about 320-2,000 nm. Output increases as wavelength
the light path fully open and the 100%T control at becomes longer peaking at around 1,000 nm, and is poor below 400 nm. As the
maximum, the instrument readout will not rise above lamp envelope darkens with age, the amount of light reaching the photodetector
90%T. What is the most appropriate first course of at 340 nm becomes insufficient to set the blank reading to 100%T. Quartz
action? halogen lamps produce light from 300 nm through the infrared region. Deuterium
or hydrogen lamps produce ultraviolet-rich spectra optimal for ultraviolet (UV)
A. Replace the source lamp work. Mercury vapor lamps produce a discontinuous spectrum that includes a
B. Insert a wider cuvette into the light path high output at around 365 nm that is useful for fluorescent applications. Xenon
C. Measure the voltage across the lamp terminals lamps generate a continuous spectrum of fairly uniform intensity from 300-2,000
D. Replace the instrument fuse nm, making them useful for both visible and UV applications.
Which type of monochromator produces the purest D. A prism and a variable exit slit
monochromatic light in the UV range?
D Diffraction gratings and prisms both produce a continuous range of
A. A diffraction grating and a fixed exit slit wavelengths. A diffraction grating produces a uniform separation of wavelengths.
B. A sharp cutoff filter and a variable exit slit A prism produces much better separation of high-frequency light because
C. Interference filters and a variable exit slit refraction is greater for higher-energy wavelengths. Instruments using a prism
D. A prism and a variable exit slit and a variable exit slit can produce UV light of a very narrow bandpass. The
adjustable slit is required in order to allow sufficient light to reach the detector to
set 100%T.
,Which monochromator specification is required in order D. 5-nm bandpass
to measure the true absorbance of a compound having a
natural absorption bandwidth of 30 nm? D Bandpass refers to the range of wavelengths passing through the sample. The
narrower the bandpass, the greater the photometric resolution. Bandpass can be
A. 50-nm bandpass made smaller by reducing the width of the exit slit. Accurate absorbance
B. 25-nm bandpass measurements require a bandpass less than one-fifth the natural bandpass of
C. 15-nm bandpass the chromophore
D. 5-nm bandpass
Which photodetector is most sensitive to low levels of D. Photomultiplier tube
light?
D The photomultiplier tube uses dynodes of increasing
A. Barrier layer cell voltage to amplify the current produced by the photosensitive cathode. It is
B. Photodiode 10,000 times as sensitive as a barrier layer cell, which has no amplification. A
C. Diode array photomultiplier tube requires a DC-regulated lamp because it responds to light
D. Photomultiplier tube fluctuations caused by the AC cycle.
Which condition is a common cause of stray light? C. Dispersion from second-order spectra
A. Unstable source lamp voltage C Stray light is caused by the presence of any light other than the wavelength of
B. Improper wavelength calibration measurement reaching the detector. It is most often caused by second-order
C. Dispersion from second-order spectra spectra, deteriorated optics, light dispersed by a darkened lamp envelope, and
D. Misaligned source lamp extraneous room light.
A linearity study is performed on a visible D. Stray light
spectrophotometer at 650 nm and the following
absorbance readings are obtained: D Stray light is the most common cause of loss of linearity at high-analyte
concentrations. Light transmitted through the cuvette is lowest when absorption
Concentration of Standard is highest. Therefore, stray light is a greater percentage of the detector response
10.0 mg/dL when sample concentration is high. Stray light is usually most
20.0 mg/dL significant when measurements are made at the extremes of the visible spectrum
30.0 mg/dL because lamp output and detector response are low.
40.0 mg/dL
50.0 mg/dL
Absorbance
0.20
0.41
0.62
0.79
0.92
The study was repeated using freshly prepared
standards and reagents, but results were identical to
those shown. What is the most likely cause of
these results?
A. Wrong wavelength used
B. Insufficient chromophore concentration
C. Matrix interference
D. Stray light
Which type of filter is best for measuring stray light? C. Sharp cutoff
A. Wratten C Sharp cutoff filters transmit almost all incident light until the cutoff wavelength
B. Didymium is reached. At that point, they cease to transmit light. Because they give an "all or
C. Sharp cutoff none effect," only stray light reaches the detector when the selected wavelength
D. Neutral density is beyond the cutoff.
Which of the following materials is best suited for D. Holmium oxide glass
verifying the wavelength calibration of a
spectrophotometer? D Wavelength accuracy is verified by determining the wavelength reading that
gives the highest absorbance (or transmittance) when a substance with a narrow
A. Neutral density filters natural bandpass (sharp absorbance or transmittance peak) is scanned. For
B. Potassium dichromate solutions traceable to the example, didymium glass has a sharp absorbance peak at 585 nm. Therefore, an
National Bureau of Standards reference instrument should give its highest absorbance reading when the wavelength dial
C. Wratten filters is set at 585 nm. Holmium oxide produces a very narrow absorbance peak at
D. Holmium oxide glass 361 nm; likewise, the hydrogen lamp of a UV spectrophotometer produces a
656-nm emission line that can be used to verify wavelength. Neutral density
filters and dichromate solutions are used to verify absorbance accuracy or
linearity. A Wratten filter is a widebandpass filter made by placing a thin layer of
colored gelatin between two glass plates and is unsuitable for
spectrophotometric calibration
, Why do many optical systems in chemistry analyzers B. To minimize error caused by source lamp fluctuation
utilize a reference light path?
B A reference beam is used to produce an electrical signal at the detector to
A. To increase the sensitivity of the measurement which the measurement of light absorption by the sample is compared. This
B. To minimize error caused by source lamp fluctuation safeguards against measurement errors caused power fluctuations that change
C. To obviate the need for wavelength adjustment the source lamp intensity. Although reference beams increase the accuracy of
D. To reduce stray light effects measurements, they do so at the expense of optical sensitivity since some of the
incident light must be used to produce the reference beam
Which component is required in a spectrophotometer in C. Photodiode array
order to produce a spectral absorbance curve?
C There are two ways to perform spectral scanning for compound identification.
A. Multiple monochromators One is to use a stepping motor that continuously turns the monochromator so
B. A reference optical beam that the wavelength aligned with the exit slit changes at a constant rate. A more
C. Photodiode array practical method is to use a diode array detector. This consists of a chip
D. Laser light source embedded with as many as several hundred photodiodes. Each photodiode is
aligned with a narrow part of the spectrum produced by a diffraction grating, and
produces current proportional to the intensity of the band of light striking it
(usually 1-2 nm in range). The diode signals are processed by a computer to
create a spectral absorbance or transmittance curve.
The half-band width of a monochromator is defined by: A. The range of wavelengths passed at 50% maximum transmittance
A. The range of wavelengths passed at 50% maximum A Half-band width is a measure of bandpass made using a solution or filter
transmittance having a narrow natural bandpass (transmittance peak). The wavelength giving
B. One-half the lowest wavelength of optical purity maximum transmittance is set to 100%T (or 0 A). Then, the wavelength dial is
C. The wavelength of peak transmittance adjusted downward, until a readout of 50%T (0.301 A) is obtained. Next, the
D. One-half the wavelength of peak absorbance wavelength is adjusted upward
until 50%T is obtained. The wavelength difference is the half-band width. The
narrower the half-band width, the better the photometric resolution of the
instrument.
The reagent blank corrects for absorbance caused by: A. The color of reagents
A. The color of reagents A When a spectrophotometer is set to 100%T with the reagent blank instead of
B. Sample turbidity water, the absorbance of reagents is automatically subtracted from each
C. Bilirubin and hemolysis unknown reading. The reagent blank does not correct for absorbance caused by
D. All of these options interfering chromogens in the sample such as bilirubin, hemolysis, or turbidity.
A plasma sample is hemolyzed and turbid. What is C. Substitute saline for the reagent
required to perform a sample blank in order to correct the
measurement for the intrinsic absorbance of the sample C A sample blank is used to subtract the intrinsic absorbance of the sample
when performing a spectrophotometric assay? usually caused by hemolysis, icterus, turbidity, or drug interference. On
automated analyzers, this is accomplished by measuring the absorbance after
A. Substitute deionized water for the sample the addition of sample and a first reagent, usually a diluent. For tests using a
B. Dilute the sample 1:2 with a standard of known single reagent, sample blanking can be done prior to the incubation phase before
concentration any color develops. Substituting deionized water for sample is done to subtract
C. Substitute saline for the reagent the absorbance of the reagent (reagent blanking). Diluting the sample with a
D. Use a larger volume of the sample standard (standard addition) may be done when the absorbance is below the
minimum detection limit for the assay. Using a larger volume of sample will make
the interference worse.
Which instrument requires a highly regulated DC power C. A spectrophotometer with a photomultiplier tube
supply?
C When AC voltage regulators are used to isolate source lamp power, light
A. A spectrophotometer with a barrier layer cell output fluctuates as the voltage changes. Because this occurs at 60 Hz, it is not
B. A colorimeter with multilayer interference filters detected by eyesight or slow-responding detectors. Photomultiplier tubes are
C. A spectrophotometer with a photomultiplier tube sensitive enough to respond to the AC frequency and require a DC-regulated
D. A densitometer with a photodiode detector power supply.
Which statement regarding reflectometry is true? C. 100% reflectance is set with an opaque film called a white reference
A. The relation between reflectance density and C Reflectometry does not follow Beer's law, but the relationship between
concentration is linear concentration and reflectance can be described by a logistic formula or algorithm
B. Single-point calibration can be used to determine that can be solved for concentration. For example,
concentration K/S = (1 - R) 2/2R, where K = Kubelka-Munk absorptivity constant, S = scattering
C. 100% reflectance is set with an opaque film called a coefficient, R = reflectance density. K/S is proportional to concentration. The
white reference white reference is analogous to the 100%T setting in spectrophotometry and
D. The diode array is the photodetector of choice serves as a reference signal. Dr = log R0/R1, where Dr is the reflectance density,
R0 is the white reference signal, and R1 is the photodetector signal for the test
sample.
