Below are some representative answers to the ‘Test Your Knowledge’ questions that appear at the end of every chapter. They’re designed to help you focus your learning, so make sure you have a go at answering them before you take a peek at the answers!
Chapter 1
TYK.1.1 How do we measure ‘wavelength’?
From two equal points on the wave, for example, from peak to peak or trough to trough.
TYK.1.2 Does ultraviolet light have higher or lower energy than visible light?
Ultraviolet light has higher energy than visible light (because it has a shorter wavelength and higher frequency).
TYK.1.3 Does ultraviolet light have a longer or shorter wavelength than visible light?
Ultraviolet light has a smaller wavelength than visible light.
TYK.1.4 Think about a regular table. Do you think light approaching the table is absorbed, reflected, transmitted or a combination of a few of these?
Light needs to be reflected for us to be able to see it, but it will be absorbing some of the light as well. This is particularly true for black or brown tables.
TYK.1.5 What colour would be produced if we combined green and red wavelengths?
Yellow.
TYK.1.6 What is a shadow?
A shadow is an absence of (at least some) light produced when an obstacle blocks the path of the light source.
Chapter 2
TYK.2.1 What is a collection of light rays called?
A pencil.
TYK.2.2 What does parallel vergence tell us about the origin of the light rays?
The rays have come from far away (infinity) because the wavefronts are flat relative to one another (zero vergence).
TYK.2.3 How do we decide whether an object distance will be negative or positive?
Object distance will always be negative because light starts at the object, and our convention states that light travels from left (where the object is) to right. We also always measure from the surface, so object distances will always be measured against the direction of light and therefore always be negative.
TYK.2.4 If a light ray moved from a medium with a refractive index of 1.00 into a medium with a refractive index of 1.523, would it bend towards or away from the normal?
It would bend towards the normal.
TYK.2.5 If you work out that following refraction an image will have a vergence of +4.00D, are the rays converging or diverging?
Converging (positive vergence suggests they are converging).
TYK.2.6 Would a concave spherical surface possess a negative or positive power? Why?
A concave spherical surface would possess a negative power because the centre of curvature will be on the left, meaning the radius of curvature (r) will be negative.
Chapter 3
TYK.3.1 What is the definition of a thin lens?
A lens in which the thickness is small enough (relative to the radius of curvature of each surface) that it’s assumed that the refractive index of the lens material has a negligible effect on the power, so it can be ignored.
TYK.3.2 What would a magnification of −0.6 tell us about the nature of the image?
It’s negative, so that tells us it’s inverted (real), and it’s between 0 and −1, so that tells us it’s minified.
TYK.3.3 How would you calculate equivalent power of two thin lenses in contact with one another?
Add the two powers together (F e = F 1 + F 2 ).
TYK.3.4 What is a principal plane?
A principal plane is the location where an equivalently powered lens would need to be placed within a multiple lens system in order to coincide with the secondary (or primary) focal point of the system.
TYK.3.5 What is the difference between back vertex focal length and secondary equivalent focal length?
The back vertex focal length (f e ′) is the distance between the back lens and the secondary focal point (F′), whereas the secondary equivalent focal length is the distance between the secondary principal plane (H′P′) and the secondary focal point (F′).
TYK.3.6 What determines whether we use step-along vergence or Newton’s formulae to find image distance with a multiple lens system?
The variable we’ve been given for object distance determines which formulae to use. If we’ve been given the object distance relative to the first lens (l) then we use step along, but if we’ve been given the object distance relative to the primary focal point (x) then we use Newton’s formulae.
Chapter 4
TYK.4.1 What is the definition of a thick lens (relative to a thin lens)?
A thick lens takes the refractive index of the lens into account when solving the equations, whereas a thin lens does not.
TYK.4.2 What is the difference between the sag and the edge thickness?
The sag is the height of the segment of the sphere that makes a lens surface, whereas the edge thickness is the literal thickness at the edges of the lens.
TYK.4.3 In an equiconcave lens, does the back surface have a positive or negative radius of curvature?
In an equiconcave lens, the back surface would have a positive radius of curvature.
TYK.4.4 What is a virtual object?
A virtual object is formed when an image acts as a new object for another surface or lens.
TYK.4.5 What causes the resolution of a Fresnel lens to be reduced relative to a ‘regular’ lens?
Image quality will be slightly reduced due to diffraction occurring at the ridges between each curved ring that makes up the Fresnel lens.
Chapter 5
TYK.5.1 What is the presumed power of the reduced eye?
+60.00D
TYK.5.2 What is the difference between spherical and cylindrical refractive error?
Spherical refractive errors are the same in all orientations, whereas cylindrical refractive errors have a different error along a specific axis.
TYK.5.3 If the far-point of a patient’s eye is −25.00 cm, what is their refractive error?
Their refractive error is −4.00DS (L = 1 / −0.25m).
TYK.5.4 Which meridian in a cylindrical lens is written as the ‘axis’?
The axis meridian (the axis with no power).
Chapter 6
TYK.6.1 What are the two laws of reflection?
- (1)
The incident light ray and the reflected light ray lie in one plane.
- (2)
The angle of incidence (i) is equal to the angle of reflection (i′).
TYK.6.2 If an object is 50 cm in front of a plane mirror, where will the image form?
50 cm within the mirror (image distance is equal to object distance in a plane mirror).
TYK.6.3 Describe the nature of an image formed in a plane mirror.
The image is the same size as the object, virtual (upright), reversed and laterally inversed.
TYK.6.4 Explain why the formula for calculating the power (F) of a spherical mirror has a ‘minus sign’ in it.
Unlike with lenses, mirrors reflect light back in the direction it came, meaning that after reflection, without including the minus sign, all variables would be assigned the opposite (incorrect) sign.
TYK.6.5 If a spherical mirror has a radius of curvature of +20 cm, what is the focal length (f)?
+10 cm (f = r/2)
Chapter 7
TYK.7.1 Using your knowledge of optical systems, can you explain why a P-ray will pass through F′?
Because the incident light is parallel to the optical axis (and therefore suggesting it has zero vergence), and light with zero vergence focuses at the secondary focal point (F′).
TYK.7.2 Explain how you could use a ray diagram to decide whether an image distance would be positive or negative.
If the ray diagram shows that the image forms on the right of the lens or mirror, then the distance will be positive, whereas if the ray diagram shows that the image forms on the left of the lens or mirror, then the distance will be negative.
TYK.7.3 In the diagram below, which image is drawn correctly? Explain your answer.
‘A’ is drawn correctly because it’s formed at the intersection of the backwards-projected, refracted rays. The ‘B’ intersection is incorrect because it is intersecting a refracted ray with an incident ray.
TYK.7.4 Draw a ray diagram to show where an image would form if an object was placed at 2F in front of a positively powered lens.
TYK.7.5 Draw a ray diagram to show where an image would form if an object was placed between F′ and the lens in front of a negatively powered lens.
TYK.7.6 Draw a ray diagram to show where an image would form if an object was placed left of C in front of a positively powered mirror.
TYK.7.7 Draw a ray diagram to show where an image would form if an object was placed at any location in front of a negatively powered mirror.
TYK.7.8 To get an inverted, minified image, what kind of lens would be needed (positive or negative) and where would the object need to be placed?
An inverted image can only be produced by a positively powered lens (negative lenses always produce upright images), and minified, inverted images can only be produced when the object is between 2F and infinity.
Chapter 8
TYK.8.1 What wavelength refracts the most – red or blue?
Blue because it’s a shorter wavelength (blue bends best).
TYK.8.2 Explain how raindrops produce a rainbow.
When white light from the sun encounters a raindrop, each constituent wavelength is refracted slightly different amounts. The light begins to disperse, but when it reaches the other side of the raindrop, some of the light is reflected back again, and so it is then refracted a second time as it leaves the raindrop. It is the combination of the two refraction opportunities that disperse the light to produce the rainbow.
TYK.8.3 Would dispersion occur if we shone a red laser through a prism? Explain your answer.
No – dispersion describes the splitting of light into constituent wavelengths. A laser light only possesses one single wavelength, so although the light will refract and change direction, it will not disperse.
TYK.8.4 What is chromatic aberration?
An ‘aberration’ is a reduction in quality of an image. The ‘chromatic’ description suggests it is related to colour, and so ‘chromatic aberration’ describes imperfections in an image related to the splitting of colours.
TYK8.5 If a person is slightly under-minussed in their glasses prescription, will green or red wavelengths be more likely to focus on the retina?
If a person is under-minussed, their eye is focusing light with too much power. This means that it is likely that all the incoming wavelengths of light will be focused earlier in the eye than expected, and therefore the red wavelengths (which usually focus behind the eye) will be more likely to focus on the retina in this case.
Chapter 9
TYK.9.1 Will light deviate towards the base or the apex of the prism?
Light deviates towards the base (the image deviates towards the apex).
TYK.9.2 What is the ‘critical angle’?
A ‘critical angle’ of incidence describes conditions in which light will leave the surface along the surface itself (indicating an angle of refraction of 90°). It can only occur when moving from a material with a high refractive index to a material with a lower refractive index.
TYK.9.3 If a prism was submerged in water, would it deviate light differently than if it was in air? Explain your answer.
Yes – when light leaves the prism, the amount it is refracted is partly dictated by the refractive index difference between the prism and the surrounding medium. If we alter the refractive index of the surrounding medium, it will change how the light leaves the prism.
TYK.9.4 Would increasing the apical angle increase the power of the prism or decrease it?
Increase.
Chapter 10
TYK.10.1 What is the ‘phase’ of a wave?
The phase of a wave is defined as the location of a point on the wave within a cycle (measured in degrees).
TYK.10.2 If two identical waves are 180° ‘out of phase’, what will happen?
Destructive interference will occur, with a net amplitude of zero. This results in no light being produced.
TYK.10.3 What does Huygen’s principle predict about light when it gets blocked by an obstacle?
Huygen’s principle predicts that the wavelets on the primary wavefronts will permit the light to ‘bend’ slightly around the obstacle to produce light in the geometric shadow of the obstacle – this is diffraction.
TYK.10.4 If we increased the number of slits in a diffraction experiment from one slit to five slits, what do you think would happen to the diffraction pattern?
The diffraction pattern would have more interference patterns present (as it would be like having five light sources) and so the bright spots within the maxima would get smaller.
TYK.10.5 Why do soap bubbles look multicoloured sometimes?
Soap bubbles are made of a thin film, which means that as incident white light reaches the ‘film’ of the bubble, some of the light reflects at the front surface of the film, but some passes through to the back surface. Then at the back surface, some light transmits through, but some is reflected, which means that even though we started with one light source, there will be (in this case), two waves that reflect towards us (the observer). This means that we’ll have two waves reaching us, one of which has now travelled a slightly greater distance than the first. These light rays will produce constructive interference if in phase, and destructive interference if out of phase for each of the wavelengths that make up the white light.
Chapter 11
TYK.11.1 What does a focimeter do?
A focimeter determines the spherical power, cylindrical power (and corresponding axis), prismatic power and the optical centre of a lens.
TYK.11.2 What is the graticule?
A graticule is the network of lines in the eyepiece of the focimeter (shown in black in the diagrams) that act as a measuring scale. In this case, the lines represent meridians and degree of prismatic effects.
TYK.11.3 Why is it important to focus the eyepiece before attempting to use a focimeter?
Each observer may have a small amount of uncorrected refractive error, which might make the target appear blurry, even though it is actually in focus. This could lead to errors.
TYK.11.4 If the target falls below the centre of the graticule, would this indicate base-down or base-up prism?
This would indicate base-down prism.
Chapter 12
TYK.12.1 What does ‘photometry’ mean?
Measurement of light.
TYK.12.2 Describe a ‘solid angle’.
A solid angle is a 3-dimensional angle which describes the field of view from a particular point or apex (likened to the angle at the top of a circular cone).
TYK.12.3 What does ‘luminous flux’ mean and what is it measured in?
Luminous flux defines the measure of power (or perceived power) emitted by a light source and is measured in lumens (lm).
TYK.12.4 Which of the following statements uses photometry terms correctly, and why:
‘ The cup is poorly illuminated’ or ‘ The cup is poorly luminated ’?
‘The cup is poorly illuminated’ – because illumination describes light falling onto the cup, whereas luminance is the perceived brightness of light coming off the cup.
TYK.12.5 What does a high number of Kelvins (6000 K) suggest about a light source?
It will appear to look ‘cold’ (towards the blue end of the colour spectrum).
Chapter 13
TYK.13.1 What is an ‘optical instrument’?
An optical instrument is any device or equipment, which can alter an image for enhancement or viewing purposes.
TYK.13.2 Could the human eye be classed as an optical instrument? Explain your answer.
Yes – because it refracts light to focus it in a specific place (the back of the eye) and can adjust the power (accommodation of the lens) in circumstances where needed.
TYK.13.3 If we wanted to focus our camera on something very far away, would we choose a 24 mm or 300 mm lens? Explain your answer.
We would choose the 300 mm (longer focal length) lens because it will produce a higher magnification and allow us to see distant objects more clearly.
TYK.13.4 Would you expect a Galilean telescope to have a positive or negative magnification? Explain your answer.
Positive magnification. Galilean telescopes produce an upright image, so the magnification should always be positive ( see chapter 3 for revision on this ).
Chapter 14
TYK.14.1 Define ‘unpolarised light’.
Unpolarised light is light that is oscillating/vibrating in all orientations (or a large range of orientations), which means the electric field changes randomly over time.
TYK.14.2 Explain the difference between circular and elliptical polarisation.
Both types of polarisation require two perpendicular linearly polarised waves, but for circular, they need to have the same amplitude and be 90° out of phase, whereas elliptical polarisation can be produced with different amplitudes or different degrees-of-phase difference.
TYK.14.3 What type of light would be emitted through two identically oriented polarising filters?
Polarised light which is polarised according to the orientation of the filters. If both filters are at the same angle, then they will let through the same light.
TYK.14.4 Is light reflected off a lake more likely to be polarised in the horizontal or vertical plane?
Horizontal – when light is polarised by reflection, it is polarised parallel to the surface from which it is reflected. Lakes are usually horizontal!
TYK.14.5 Explain why the sun appears red at sunset.
As the unpolarised sunlight passes through the molecules in the atmosphere, the wavelengths are scattered. This scattering affects shorter wavelengths first, so when the sunlight travels a large distance through the atmosphere (as it does at sunrise/sunset) then more of the wavelengths will be scattered out of the light and sent into the atmosphere. This means that only the longer wavelength light is left, which will make the sun appear red.
Chapter 15
TYK.15.1 Why is it not possible to see the back of a patient’s eye without the help of a special device?
The inside of the eye is very dark (meaning you’d need a light to illuminate the inside of the eye to see it), and the pupil is very small, which restricts the field of view.
TYK.15.2 Is a slit-lamp biomicroscope a form of direct or indirect ophthalmoscopy?
Indirect.
TYK.15.3 Why is it advantageous to the clinician to ask the patient to move their gaze when performing ophthalmoscopy?
The field of view when performing ophthalmoscopy is relatively small (even with indirect), so clinicians can ask the patient to change the position of their eyes to allow them to see different parts of the back of the eye.
TYK.15.4 Explain why the anterior angle of the eye is not visible without the help of a special lens.
Because of the refractive index difference between the front surface of the cornea and the air, light from the anterior angle exceeds the critical angle, meaning that it experiences total internal reflection (all the light reflects back into the eye, making it invisible to the external viewer).
TYK.15.5 On a three-mirror goniolens, which mirror would allow the clinician to view the anterior angle?
The D-shape (59°) lens.
TYK.15.6 Explain what an against movement would look like during retinoscopy.
If an against movement is present, the reflex (light from inside the eye) would move in the opposite direction to that of the streak. For example, if the streak was moving from right to left, the reflex would be seen to be moving from left to right.
TYK.15.7 If a clinician performed retinoscopy on a patient at a working distance of 50 cm, what spherical power would they need to account for in the final refractive error?
They would need to account for −2.00DS when determining their final refractive error (L = n/l = 1/0.5 = 2).
TYK.15.8 What equation does applanation tonometry rely on for calculating intraocular pressure (IOP)?
Pressure = force / area
Chapter 16
TYK.16.1 What is an aberration?
An imperfection in the quality of an image – could be from distortion, blurring or both.
TYK.16.2 Which part of a lens induces the greatest amount of aberration?
The periphery (edges) – as distance increases away from the optical centre, aberrations become larger.
TYK.16.3 In terms of Zernike polynomials, what ‘order’ of aberration is defocus ?
2 (because it’s <SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='Z20′>𝑍02Z20
Z 2 0
).
TYK.16.4 Explain how a Scheiner disc can identify the presence of a refractive error.
The Scheiner disc is an occluder with two small, adjacent pinholes. The occluder is placed in front of the patient’s eye and a distant (parallel vergence) light is shone at the occlude. The pinholes should then only let through two small spots of light, which, if no refractive error is present, should focus on the back of the eye to form an image of a single spot. However, if the patient’s eye is inducing aberrations, then the light will focus incorrectly and will appear as two separate spots of light.
TYK.16.5 Explain how adaptive optics systems can take high-resolution images.
Adaptive optics systems utilise a wavefront sensor and a rapidly deformable mirror to constantly monitor aberrations (sensor) and compensate for them (deformable mirror).
Chapter 17
TYK.17.1 What does OCT stand for?
Optical coherence tomography.
TYK.17.2 Explain how far the mirror in a Michelson interferometer would need to move to produce a path difference of a whole wavelength.
The mirror would need to move a distance equivalent to half a wavelength in order to produce a full wavelength of path difference, because moving the mirror half a wavelength will add (or remove) half from the incident ray and also the reflected ray (totalling a whole wavelength).
TYK.17.3 Explain how a TD-OCT system works.
TD-OCT utilises a low-coherence, near-infrared light source, which is split into two beams (a reference beam and a measurement beam) by a beam splitter. One beam (measurement) will enter the patient’s eye, whilst the other beam (reference) will reflect off a moveable mirror. The measurement beam is reflected from the back of the eye with different delay times which are dependent on the optical properties of the tissue and the distance away from the light source. The software within the system can then interpret the interference fringe profiles as they pass through the detector in order to determine the depth of the tissue and produce the nice black-and-white image.
TYK.17.4 Name one difference between a TD-OCT and an FD-OCT system.
Moveable mirror (TD-OCT) versus diffraction grating (FD-OCT).
Low-coherence near-infrared light course (TD-OCT) versus broadband light source (FD-OCT).
TYK.17.5 Name one clinical application of OCT.
Checking the health of the layers of the retina; measuring distances (thickness of cornea, axial length, thickness of retinal nerve fibre layer (RNFL)); measuring the shape of the cornea; checking the fit of a contact lens.
Chapter 18
TYK.18.1 Explain how a pinhole produces an upside-down image.
Light rays have to travel in straight lines, but a pinhole is very small; therefore, light coming off the tip of an object will travel in a straight line through the pinhole and end up near the floor, whilst light rays from the base of the object travel in a straight line through the pinhole and end up near the ceiling.
TYK.18.2 Explain why moving the screen away from the pinhole produces a larger image.
If the object remains still, then the angle of the rays on the other side of the pinhole should remain constant. So, if the screen is closer to the pinhole, it will see more of the scene than if it is farther away.
Chapter 19
TYK.19.1 Explain why the torch looks more yellow as we add more milk.
Adding more milk causes more scattering of the light within the glass. This makes the shorter (blue) wavelengths of light scatter into the milk, causing the torchlight to look more yellow (towards the longer wavelengths) and making the milk look more blue.
TYK.19.2 Explain why using a red light source wouldn’t work.
A red light source wouldn’t work because it only contains red (long wavelengths). Longer wavelengths are more difficult to scatter (so would require so much milk the light wouldn’t be visible through the murkiness) but also lack the full range of wavelengths (as with white light), so there are no blue wavelengths to scatter and make the milk blue.
Chapter 20
TYK.20.1 Explain why the white light produces a rainbow when it passes through the setup described in this chapter.
When white light passes through a prism (a material with two or more refracting surfaces), the individual wavelengths within the white light refract to greater or lesser amounts depending on their wavelength (e.g. red wavelength refracts the least, but blue refracts the most). When the mirror is placed in the tray of water, it turns the tray of water into a prism and splits the wavelengths through dispersion.
TYK.20.2 Explain what you think would happen if you changed the angle of the light approaching our homemade prisms.
As we change the angle of incidence (angle of approaching light), the angle of deviation in the prism changes as well, which means that the dispersed light will move along the paper.
Chapter 21
TYK.21.1 Explain why the distance between the hot spots is equal to half the wavelength of a microwave.
As the amplitude increases upwards or downwards away from the midpoint, the energy also increases, meaning that both the peak and the trough of the microwave profile will produce the highest amount of intensity (heat). Therefore, the microwave will produce two hotspots – one at the peak and one at the trough. The distance between these points is equal to half a wavelength.
TYK.21.2 Explain why we needed to take the turntable out of the microwave for it to work.
The turntable allows the interference produced by the microwave to be evenly spread around the food as it rotates through the hot spots. For this experiment to work we need to make sure the hot spots don’t move, so the turntable needed to be removed.
Chapter 22
TYK.22.1 Explain how the human eye can focus distant light onto the back of the eye.
The human eye can focus light because it has a convex front surface (the cornea) with a small radius of curvature, and the cornea has a different refractive index to the air (~1.376).
TYK22.2 Why did we use water for the refractive index difference?
Water is clear which means the image will still be visible through the ‘lenses’, and it has a refractive index very similar to that of the human eye (water, 1.333; cornea, 1.376).
TYK.22.3 Explain why the image is sometimes a little distorted near the edges of the ‘cornea’ in our demonstration (see Fig. 22.7 for an example of this distortion).
These are examples of wavefront aberrations that become increasingly pertinent as light travels through the peripheral edges of a lens. In a way, this makes our homemade ‘cornea’ even more realistic, as a real cornea also induces more aberrations as light moves away from the optical centre.
TYK.22.4 If we had filled our lenses with glycerine (a viscous, clear liquid of refractive index 1.47), do you think the results would have been the same? Discuss your thoughts.
The experiment would have been a lot stickier (!), but the results would largely have been the same, with the flat lens not having any impact on the image, and the ‘cornea’ producing a magnified image. However, our homemade cornea would have had more power with the glycerine (relative to the water). This is shown in Equation 2.4 :
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