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OCT-C study materials
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OCT-C practice questions

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Question 1Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

In OCT, axial resolution is determined primarily by which property of the light source?

  1. The spectral resolution of the spectrometer
  2. The source center wavelength and bandwidth Correct answer.
  3. The responsivity of the photodetector
  4. The numerical aperture of the sample objective

Explanation

Axial resolution follows from the coherence length of the source, approximately 0.44·λ0²/Δλ for a Gaussian spectrum, so it scales with center wavelength and bandwidth. Focusing optics and detector properties do not enter this relationship.

Question 2Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Transverse (lateral) resolution in OCT is governed mainly by:

  1. The focused spot size set by the sample-arm optics Correct answer.
  2. The bandwidth of the light source
  3. The number of pixels in the spectrometer line camera
  4. The optical path length of the reference arm

Explanation

Lateral resolution is a confocal property, roughly λ/(2·NA), and is therefore fixed by the focusing optics in the sample arm. Source bandwidth drives axial resolution instead.

Question 3Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

A single depth profile acquired at one transverse position is called:

  1. A B-scan
  2. A C-scan
  3. An en face image
  4. An A-scan Correct answer.

Explanation

An A-scan is the one-dimensional reflectivity profile along the beam axis. B-scans combine adjacent A-scans into a cross-section, and C-scans (en face images) are formed from a depth slab of a volumetric data set.

Question 4Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Which condition must be met for an interference signal to be detected in an OCT interferometer?

  1. The sample and reference arms are matched to within one optical cycle
  2. The detector bandwidth must exceed the optical carrier frequency
  3. The sample must exhibit birefringence
  4. The optical path difference between the arms is within the source coherence length Correct answer.

Explanation

Interference is observable only while the path mismatch remains shorter than the coherence length, which is what provides optical sectioning. Matching the arms to within a single optical cycle would be impractical and is not required.

Question 5Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Which light source is characteristic of swept-source OCT?

  1. A wavelength-swept laser with narrow instantaneous linewidth Correct answer.
  2. A mode-locked femtosecond laser producing a fixed frequency comb
  3. A continuous-wave narrowband gas laser
  4. A broadband superluminescent diode with a fixed output spectrum

Explanation

Swept-source OCT uses a laser whose emission wavelength is tuned rapidly in time; the instantaneous linewidth is narrow, but the full sweep spans a broad optical bandwidth that determines axial resolution.

Question 6Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

In spectrometer-based spectral-domain OCT, the interference spectrum is recorded by:

  1. A lock-in amplifier with an optical chopper
  2. A single-element photodiode
  3. A photon-counting avalanche photodiode
  4. A line-scan camera inside a spectrometer Correct answer.

Explanation

Spectral-domain OCT needs the spectrum resolved in wavelength, which is done by dispersing light onto a line-scan camera. A single photodiode records only total intensity and cannot resolve spectral fringes.

Question 7Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

How is an A-scan generated in a time-domain OCT system?

  1. Scanning the beam transversely across the sample with galvanometers
  2. Varying the detector gain synchronously with depth
  3. Translating the reference arm and recording signal versus time Correct answer.
  4. Fourier transforming the optical spectrum measured by a spectrometer

Explanation

Time-domain OCT moves the reference mirror so that the coherence gate sweeps through the sample depth; the detected interference amplitude as a function of time is the A-scan. Spectral encoding is the alternative used in Fourier-domain systems.

Question 8Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Compared with time-domain OCT at equal optical power, Fourier-domain detection provides which main advantage?

  1. A signal-to-noise ratio gain of roughly 20 dB at equal power Correct answer.
  2. Immunity to sample motion
  3. Deeper penetration into highly scattering tissue
  4. Higher axial resolution

Explanation

Fourier-domain OCT records all depths simultaneously, and the Fourier transform concentrates the noise, yielding a sensitivity advantage of roughly 20–30 dB. Axial resolution and penetration depth are set by the source, not by the detection scheme.

Question 9Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

A source has a center wavelength of 840 nm and a bandwidth of 45 nm, with an approximately Gaussian spectrum. Which value best approximates the axial resolution in air?

  1. About 1.5 µm
  2. About 0.8 µm
  3. About 7 µm Correct answer.
  4. About 19 µm

Explanation

For a Gaussian spectrum, axial resolution is approximately 0.44·λ0²/Δλ. Substituting 0.84 µm and 0.045 µm gives 0.44 × 0.7056 / 0.045 ≈ 6.9 µm.

Question 10Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

In spectral-domain OCT, if the spectral sampling interval in wavenumber is halved while the total wavenumber range is unchanged, the maximum unaliased imaging depth:

  1. Increases by a factor of about 1.4
  2. Is unchanged
  3. Is halved
  4. Is doubled Correct answer.

Explanation

Maximum unaliased depth is inversely proportional to the wavenumber spacing between adjacent samples. Halving that spacing therefore doubles the depth range, at the cost of acquiring twice as many samples for the same bandwidth.

Question 11Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

In spectral-domain OCT, the Fourier transform of a real-valued spectral interferogram is Hermitian, which produces:

  1. Polarization fading along the fiber-optic path
  2. A progressive loss of sensitivity with increasing depth
  3. A complex-conjugate mirror artifact that halves usable depth Correct answer.
  4. Random axial motion artifacts from sample pulsation

Explanation

A real-valued spectrum transforms into a Hermitian signal, so structure at positive depth is duplicated at negative depth. This mirror image occupies the opposite half of the depth range unless removed by phase-shifting or complex reconstruction.

Question 12Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

A dispersion imbalance between the sample and reference arms primarily causes:

  1. Broadening of the axial point spread function Correct answer.
  2. A reduction of transverse resolution
  3. An increase in the maximum imaging depth
  4. A shift of the source center wavelength

Explanation

Unmatched material dispersion adds quadratic phase across the spectrum, which broadens the coherence peak and degrades axial resolution. The effect is normally compensated numerically by applying a phase correction.

Question 13Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Why is resampling (k-linearization) of the acquired spectrum necessary before Fourier transformation in spectral-domain OCT?

  1. The source emits unpolarized light
  2. Shot noise must be removed before detection
  3. Spectrometers sample uniformly in wavelength, not in wavenumber Correct answer.
  4. The number of A-lines per B-scan must be increased

Explanation

The Fourier transform requires samples evenly spaced in wavenumber. A diffraction grating spectrometer samples evenly in wavelength, so the data must be interpolated onto a uniform wavenumber grid to avoid axial broadening and depth-dependent artifacts.

Question 14Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

The sensitivity roll-off with depth observed in spectrometer-based OCT is caused mainly by:

  1. The finite spectral resolution of the spectrometer Correct answer.
  2. Coupling losses in the reference fiber path
  3. Defocus of the sample beam at large depths
  4. Absorption of light within the sample

Explanation

Finite spectral resolution, from pixel width and optical blur, low-pass filters the interference fringe and attenuates the high fringe frequencies that encode large depths. Balanced detection and higher-resolution spectrometers reduce this roll-off.

Question 15Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Increasing the numerical aperture of the sample-arm objective will:

  1. Improve axial resolution
  2. Reduce specular reflections at the sample surface
  3. Increase the maximum imaging depth
  4. Improve transverse resolution but shorten the depth of focus Correct answer.

Explanation

A larger numerical aperture reduces the diffraction-limited spot size, improving lateral resolution, but the confocal parameter scales roughly as λ/NA² and therefore shrinks. This trade-off limits the useful imaging depth for high-NA systems.

Question 16Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

For imaging deep within highly scattering tissue, which center wavelength is generally preferred?

  1. 1310 nm Correct answer.
  2. 400 nm
  3. 840 nm
  4. 488 nm

Explanation

Tissue scattering decreases approximately as a power of wavelength, so longer wavelengths penetrate more deeply. Wavelengths near 1050–1310 nm are therefore used for deep targets such as the choroid.

Question 17Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Dual-balanced detection in swept-source OCT is used mainly to:

  1. Suppress common-mode source intensity noise and DC terms Correct answer.
  2. Improve the axial resolution of the system
  3. Compensate for dispersion mismatch between the arms
  4. Increase the sweep repetition rate of the laser

Explanation

The two detectors receive interference fringes of opposite sign while source intensity fluctuations are common to both, so subtraction cancels the intensity noise and the non-interfering DC terms. This brings sensitivity close to the shot-noise limit.

Question 18Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

In a Fourier-domain OCT A-scan, the strong signal centered at zero depth corresponds to:

  1. DC (reference-only and sample-only) and autocorrelation terms Correct answer.
  2. The complex-conjugate mirror image of true structures
  3. Multiple scattering from deep within the sample
  4. Shot noise arising in the detector circuit

Explanation

The detected spectrum contains the source envelope plus self-interference of each arm, and these transform to zero delay. Balanced detection or background subtraction removes them and frees the top of the depth range for imaging.

Question 19Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

A raster scan protocol in OCT refers to:

  1. A circular scan centered on the optic disc
  2. Parallel B-scans displaced by a fixed step to form a volume Correct answer.
  3. A radial spoke pattern through the fovea
  4. Repeated A-scans at a single location, averaged to reduce noise

Explanation

Raster scanning steps the beam in one transverse direction between successive B-scans, generating a regularly sampled three-dimensional data set. Circular and radial patterns are alternative protocols used for specific clinical views.

Question 20Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Two sources share a center wavelength; source A has twice the bandwidth of source B. Assuming Gaussian spectra, the coherence length of source A is:

  1. Half that of source B Correct answer.
  2. Twice that of source B
  3. Unchanged relative to source B
  4. Four times that of source B

Explanation

Coherence length is proportional to λ0²/Δλ, so doubling the bandwidth halves it. Since axial resolution is directly proportional to coherence length, source A also provides twice the axial resolution of source B.

Question 21Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

In a swept-source OCT system without balanced detection, the noise floor at typical sample powers is usually dominated by:

  1. Relative intensity noise (excess noise) of the source Correct answer.
  2. Quantization noise of the digitizer
  3. Shot noise from the local oscillator
  4. Johnson noise of the transimpedance amplifier

Explanation

Unbalanced detection leaves source intensity fluctuations that scale with reference-arm power and exceed shot noise at typical operating levels. Balanced detection suppresses this term so the system can approach the shot-noise limit.

Question 22Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

During rapid transverse beam scanning in OCT, fringe washout occurs because:

  1. The coherence length shrinks as the beam moves across the sample
  2. The digitizer sampling rate exceeds the fringe frequency
  3. Phase drift during the detection window reduces fringe contrast Correct answer.
  4. The effective source bandwidth narrows at high scan speeds

Explanation

Interference contrast requires a stable phase over the detection interval. If transverse motion causes the phase to change appreciably within the camera exposure or sweep time, the fringe averages toward zero and signal is lost.

Question 23Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Zero-padding the spectral interferogram before Fourier transformation:

  1. Improves axial resolution beyond the source limit
  2. Interpolates the A-scan without adding new information Correct answer.
  3. Extends the maximum unaliased imaging depth
  4. Removes the complex-conjugate mirror artifact

Explanation

Zero-padding lengthens the transform and yields a finer sample spacing along depth, which is equivalent to spectral interpolation. The axial point spread function width remains set by the measured optical bandwidth.

Question 24Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Why are axial and transverse resolution largely independent in OCT?

  1. Axial resolution comes from source coherence; transverse from focusing Correct answer.
  2. Both are determined by the numerical aperture of the objective
  3. Both depend only on the bandwidth of the detector electronics
  4. Axial resolution is set by the speed of the scanning mirrors

Explanation

OCT decouples the two resolutions because axial sectioning arises from low-coherence interferometry, governed by the source spectrum, while transverse resolution is confocal and set by the sample-arm optics. This permits fine axial resolution with moderate numerical aperture and long working distance.

Question 25Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

For a fixed number of spectrometer pixels, widening the source bandwidth to improve axial resolution will, all else equal:

  1. Reduce the maximum imaging depth Correct answer.
  2. Leave the maximum imaging depth unchanged
  3. Increase the maximum imaging depth
  4. Improve the sensitivity roll-off with depth

Explanation

With a fixed pixel count, a wider bandwidth means each pixel spans a larger wavenumber interval, and maximum depth is inversely proportional to that interval. There is therefore a trade-off between axial resolution and depth range in spectrometer-based systems.

Question 26Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Most clinical OCT systems derive their interference signal from which optical arrangement?

  1. A Young double-slit interferometer illuminated through a pair of pinholes
  2. A Michelson interferometer with sample and reference arms Correct answer.
  3. A Fabry–Perot cavity formed by two partially reflecting mirrors
  4. A Sagnac interferometer in which two beams counter-propagate around a loop

Explanation

OCT measures interference between light backscattered from the tissue and light that has traveled a known reference path. This requires a beam splitter or fiber coupler that divides the source light, a reference reflector, and a recombination point feeding the detector. The Michelson geometry satisfies these requirements and is the basis of both time-domain and Fourier-domain instruments.

Question 27Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

A peripapillary scan for retinal nerve fiber layer thickness around the optic disc is typically acquired with which scan pattern?

  1. A single stationary A-scan held at one point
  2. A raster scan covering a rectangular area
  3. A radial scan of straight lines through the fovea
  4. A circular scan centered on the optic disc Correct answer.

Explanation

A circular pattern traces a closed ring at a fixed radius around the disc, so the resulting unwrapped B-scan samples the nerve fiber layer along the entire peripapillary circumference. Thickness can then be read directly at each clock hour of the ring. Raster and radial patterns cross the region but do not follow the circumference.

Question 28Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

A reflective layer lies 500 µm beneath the tissue surface in tissue with a group refractive index of 1.4. The system converts round-trip delay into a one-way displayed depth. At what displayed depth does the layer appear?

  1. 357 µm
  2. 1000 µm
  3. 500 µm
  4. 700 µm Correct answer.

Explanation

The interferometer measures a round-trip optical delay of 2nd, where n is the group index and d the physical depth. Halving the delay to report a one-way depth gives n·d = 1.4 × 500 µm = 700 µm. Structures therefore appear deeper than their geometric depth by the factor n.

Question 29Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Why is an optical circulator often used in a fiber-based swept-source OCT system?

  1. It removes the need for a reference arm by generating a synthetic reference reflection internally
  2. It blocks reference-arm light so that only light backscattered from the sample reaches the detector
  3. It amplifies the light returning from the sample before detection to raise the signal-to-noise ratio
  4. It directs light returning from the sample to the detector instead of back to the source Correct answer.

Explanation

In a simple two-by-two coupler, sample light returning toward the source is split, with roughly half lost on the path back to the laser. A circulator passes light from source to sample port and routes the returning light to the detector, recovering about 3 dB and protecting the source from back-reflections. It is a routing element, not an amplifier or filter.

Question 30Fundamentals of OCT: physics, light sources, interferometry, scan types and instrumentation

Why do fiber-based OCT systems commonly include polarization controllers in the sample and reference arms?

  1. To convert a broadband source into a rapidly tunable narrowband source
  2. To compensate automatically for dispersion imbalance between the two arms
  3. To broaden the source spectrum and improve axial resolution
  4. To match the polarization states of the sample and reference beams Correct answer.

Explanation

Interference is strongest when the two recombining fields share a polarization component, and this visibility scales with the degree of alignment between their states. Fiber birefringence, bending and sample birefringence all alter the polarization state, so adjustable paddles are used to restore a high-contrast fringe. Dispersion balancing is a separate concern handled numerically or with hardware.

Question 31Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Which retinal layer lies immediately adjacent to the vitreous cavity?

  1. External limiting membrane
  2. Internal limiting membrane Correct answer.
  3. Nerve fiber layer
  4. Bruch membrane

Explanation

The internal limiting membrane is the innermost boundary of the retina, a basement membrane elaborated by Müller cells, and it is the first retinal structure bordering the vitreous.

Question 32Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

The nuclei of the photoreceptors are located in which retinal layer?

  1. Outer plexiform layer
  2. Ganglion cell layer
  3. Outer nuclear layer Correct answer.
  4. Inner nuclear layer

Explanation

Photoreceptor cell bodies are stacked in the outer nuclear layer, with their inner and outer segments extending outward toward the RPE and their synapses lying in the outer plexiform layer.

Question 33Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

The hyperreflective outer retinal band that represents the photoreceptor inner/outer segment junction is termed the:

  1. Ellipsoid zone Correct answer.
  2. Interdigitation zone
  3. Retinal pigment epithelium
  4. External limiting membrane

Explanation

The ellipsoid zone corresponds to the junction between photoreceptor inner and outer segments and lies between the external limiting membrane and the interdigitation zone.

Question 34Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Which choroidal layer lies directly external to Bruch membrane?

  1. Choriocapillaris Correct answer.
  2. Sattler layer
  3. Suprachoroidal space
  4. Haller layer

Explanation

The choriocapillaris is the innermost choroidal layer, a dense capillary bed sitting immediately outside Bruch membrane that supplies the outer retina.

Question 35Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Which statement best describes the foveola on OCT?

  1. It contains both rods and cones with a full inner retina
  2. It has the thickest ganglion cell layer in the retina
  3. It contains cone photoreceptors but lacks inner retinal layers Correct answer.
  4. It overlies the optic disc margin

Explanation

At the foveola the inner retinal layers are displaced laterally, leaving only cone photoreceptors, their nuclei and Müller cell processes, which produces the foveal depression.

Question 36Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

In the macula, the Henle nerve fiber layer is located within which retinal layer?

  1. Outer plexiform layer Correct answer.
  2. Nerve fiber layer
  3. Inner plexiform layer
  4. Inner nuclear layer

Explanation

Henle nerve fiber layer consists of the obliquely running axons of foveal cones and lies within the outer plexiform layer, which is why the outer plexiform layer is thick in the macula.

Question 37Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

According to the ISNT rule for peripapillary retinal nerve fiber layer thickness, which quadrant is normally the thickest?

  1. Inferior Correct answer.
  2. Superior
  3. Temporal
  4. Nasal

Explanation

The ISNT rule describes the normal pattern of peripapillary nerve fiber layer thickness as Inferior greater than Superior greater than Nasal greater than Temporal.

Question 38Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

On enhanced-depth OCT of the normal optic nerve head, the lamina cribrosa appears as:

  1. A uniformly hyporeflective band without internal structure
  2. Hyperreflective bands with hyporeflective pores Correct answer.
  3. A thin hyperreflective line parallel to the internal limiting membrane
  4. A dome-shaped hyperreflective mass anterior to the disc

Explanation

The lamina cribrosa is a sieve-like scleral structure whose connective tissue beams are hyperreflective and whose neural pores, through which ganglion cell axons pass, are hyporeflective.

Question 39Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

The external limiting membrane is formed by:

  1. Tight junctions between capillary endothelial cells
  2. Basement membrane of the retinal pigment epithelium
  3. Junctions between Müller cells and photoreceptors Correct answer.
  4. Desmosomes between bipolar and ganglion cells

Explanation

The external limiting membrane is not a true membrane but a row of junctional complexes linking Müller cell processes to photoreceptor inner segments, appearing as a hyperreflective line just external to the outer nuclear layer.

Question 40Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Which OCT finding indicates that a posterior vitreous detachment has occurred?

  1. A faint hyperreflective line lifted off the internal limiting membrane Correct answer.
  2. A hyperreflective band within the outer choroid
  3. Thickening of the peripapillary nerve fiber layer
  4. Loss of the foveal depression with preserved layers

Explanation

When the posterior vitreous cortex separates from the internal limiting membrane it becomes visible as a faint hyperreflective line in the vitreous cavity, clearly detached from the retinal surface.

Question 41Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Which cells have their cell bodies located in the inner nuclear layer?

  1. Photoreceptors
  2. Retinal ganglion cells
  3. Astrocytes of the nerve fiber layer
  4. Bipolar and horizontal cells Correct answer.

Explanation

The inner nuclear layer contains the cell bodies of bipolar, horizontal and amacrine cells together with Müller cell nuclei, while photoreceptors and ganglion cells have their bodies in other layers.

Question 42Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

The interdigitation zone on OCT is located between the:

  1. Outer nuclear layer and the ellipsoid zone
  2. Inner and outer segments of the photoreceptors
  3. Apices of the retinal pigment epithelium and the outer segment tips Correct answer.
  4. Internal limiting membrane and the nerve fiber layer

Explanation

The interdigitation zone represents the interlocking of apical RPE microvilli with photoreceptor outer segment tips and therefore lies between the ellipsoid zone and the RPE band.

Question 43Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Which statement about normal choroidal thickness on enhanced-depth OCT is correct?

  1. It is thinnest subfoveally and increases with age
  2. It is uniform across the macula at about 100 µm
  3. It is greatest subfoveally and decreases with age Correct answer.
  4. It increases markedly with increasing axial length

Explanation

In normal eyes the choroid is thickest beneath the fovea and thins with advancing age and with increasing axial length, so it is not uniform across the macula.

Question 44Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

On OCT, Bruch membrane is best described as the:

  1. Outermost hyperreflective band of the RPE complex Correct answer.
  2. Middle layer of the choroid vascular network
  3. Innermost hyperreflective line of the retina
  4. Hyporeflective space between RPE and choroid

Explanation

Bruch membrane forms the outermost reflective band of the RPE-Bruch complex, lying between the retinal pigment epithelium and the choriocapillaris.

Question 45Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Which retinal layer appears hyporeflective on a normal OCT B-scan?

  1. Inner plexiform layer
  2. Nerve fiber layer
  3. Outer nuclear layer Correct answer.
  4. Retinal pigment epithelium

Explanation

Nuclear layers contain cell bodies and are hyporeflective, whereas layers rich in axons, dendrites and synapses, such as the nerve fiber layer and plexiform layers, and the RPE appear hyperreflective.

Question 46Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

The optic cup on OCT is defined as:

  1. The central depression within the optic nerve head Correct answer.
  2. The termination of Bruch membrane at the disc margin
  3. The peripapillary scleral canal
  4. The hyperreflective tissue rim surrounding the disc

Explanation

The optic cup is the central depression of the optic nerve head, and the cup-to-disc ratio is derived from its horizontal or vertical extent relative to the disc.

Question 47Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

The macular ganglion cell complex measured on OCT comprises the:

  1. Inner nuclear layer, outer plexiform layer and outer nuclear layer
  2. Nerve fiber layer, ganglion cell layer and inner plexiform layer Correct answer.
  3. Photoreceptor layers and the retinal pigment epithelium
  4. Ganglion cell layer, inner nuclear layer and outer plexiform layer

Explanation

The ganglion cell complex consists of the three innermost layers of the macula: the nerve fiber layer, the ganglion cell layer and the inner plexiform layer.

Question 48Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

Beta-zone peripapillary atrophy on OCT is characterized by:

  1. A hyperreflective membrane over the optic disc
  2. Thickening of the peripapillary nerve fiber layer
  3. Cystic spaces within the lamina cribrosa
  4. Absence of RPE and photoreceptors temporal to the disc Correct answer.

Explanation

Beta-zone peripapillary atrophy corresponds to complete loss of the RPE and photoreceptor layers adjacent to the disc, allowing the sclera and choroid to be seen more directly; alpha zone shows RPE irregularity only.

Question 49Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

At the foveola, photoreceptor outer segments are:

  1. Absent, leaving bare retinal pigment epithelium
  2. Replaced by Müller cell processes at the pit
  3. Shorter than elsewhere to limit light scatter
  4. Elongated, forming the central bouquet of cones Correct answer.

Explanation

Foveolar cone outer segments are longer than those elsewhere and are packed into the so-called central bouquet, which enhances photon capture at the visual axis.

Question 50Normal anatomy on OCT: retinal layers, vitreoretinal interface, choroid, and optic nerve head

On the standard ETDRS macular thickness grid, the inner ring is typically:

  1. Equal in thickness to the outer ring
  2. Thicker than the outer ring Correct answer.
  3. Thinner than the outer ring
  4. Thinner than the central subfield

Explanation

Outside the fovea, the inner ETDRS ring contains thick retina because the ganglion cell and nerve fiber layers are still well developed, while the more peripheral outer ring is thinner.