Objectives
After working through this chapter, you should be able to:
Explain what vergence and refraction are
Explain how sign convention works, and be able to determine whether a distance should be positive or negative
Understand how refractive index and curvature of a surface can affect the emergent vergence
Be able to calculate use vergence equations to determine where an image will form after refraction
Introduction
In chapter 1 we began to learn about light – what it is, how it interacts with materials and how light allows us to see colour and objects.
The current chapter will seek to explore these elements a little further by going over some key terms and introducing the principles of how light changes direction when it meets certain materials (vergence and refraction).
Key terms relating to light
Let’s start with some discussion of ‘light’. As discussed in chapter 1 , light possesses wave-particle duality , meaning it exhibits characteristics of both waves and particles. If we take a point light source, we can imagine that light will be emitted in all directions, and we can liken this to ripples on a pond after throwing a pebble into the water. However, with light, these ‘ripples’ are referred to as wavefronts . Much like our water ripples, these wavefronts remain consistently spaced (unless interfered with – see chapter 10 ), and as the distance from the light source increases, the wavefronts will start to get relatively ‘flatter’. In terms of describing the path of the light, this can be depicted as a wave (a wiggly line) or a ray (an arrow), as shown in Fig. 2.1 .
This book uses wavefronts, waves and rays interchangeably in diagrams to illustrate key points, but please remember that they are all connected.
Vergence
As discussed in chapter 1 , when light travels unimpeded in a homogenous medium, it will travel in a straight line (rectilinear propagation of light); however, this is a slight over-simplification. As you can see from Fig. 2.1 , any individual light ray will be travelling in a straight line, but if we draw a number of light rays to show how the light is truly leaving the light source, we can see that the light would be travelling in straight lines in all directions ( Fig. 2.2 A). So, if we select a few rays close together, known as a pencil , the individual rays travel in their respective straight lines, but this means that they are actually travelling further away from each other ( Fig. 2.2 B).
The difference in direction of each ray within a pencil represents the vergence , which can be classified as one of three types ( Fig. 2.3 ):
- •
Divergence (diverging rays) – individual light rays travel away from one another
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Parallel (parallel rays) – individual rays remain in line with one another (this requires a light source at infinity)
- •
Convergence (converging rays) – individual light rays travel towards one another
It is important to note that the vergence of the rays is associated with the wavefronts and how they are sampled ( Fig. 2.4 ) – all wavefronts are curved (like ripples from throwing a stone in a pond), but the closer they’re sampled to the source, the more curved they will appear to be. In Fig. 2.4 you can see that the closer the observer is to the light source, the more curved the wavefronts will be when they reach the observer’s pupil. Ultimately, then, if the sampled wavefronts appear to be curved, then divergence or convergence is indicated, whereas wavefronts that appear to be flat indicate parallel vergence. This also shows that the further away the wavefronts are from the source, the flatter they appear to become, so parallel vergence is most often associated with sources that are a reasonable distance away, referred to in optics as infinity . Therefore, for example, light from the sun, which is pretty far away, would be considered to have parallel vergence when it finally reaches the Earth. However, if the object is closer to the observer than optical infinity (<6 metres), then wavefronts will appear to be diverging, as shown in Figs 2.3 and 2.4 (and this also nicely highlights how distance is related to vergence, which we will discuss in the next section).
Now, just to make things slightly more complicated, in optics, as well as thinking of light as being emitted from a light source (e.g. a lamp), we can also consider the light to be coming from an object , as we know that in order for us to be able to see an object, we need light to be reflecting off it. Helpfully though, just like with a light source, the light reflected from objects has parallel vergence when the object is at infinity and diverging rays when it’s in proximity, so the principle remains the same.
Vergence calculations
Up to this point, we have only considered vergence in descriptive terms, but it’s also important to be able to calculate values of vergence. Now, we already know that vergence is associated with the distance of wavefront from the light source/object, as a greater distance will lead to flatter wavefronts. So you’ll be pleased to know that the calculation for object vergence (the vergence of light reflected from an object) mostly revolves around the distance of the object from the point of measurement.
Mathematically, vergence is given the symbol L, is measured in dioptres (D) , and is calculated using Equation 2.1 . However, in order to calculate vergence we need to know two things: the refractive index of the primary medium (in which the object exists), indicated by n and, as mentioned above, the distance of the object from the point of measurement (usually a surface of some kind, e.g. a lens), represented by l.
L=nl
objectvergence=refractive index of mediumobject distance from surface
Refractive index is a term that relates to the density of the material that the light is moving through, so the value of the refractive index indicates how fast (or slow) light will travel through that particular material, relative to how fast it would travel in a vacuum. The refractive index of air is 1.00, meaning that light will travel (approximately) as fast in air as it would in a vacuum, whereas the refractive index of crown glass is 1.523, meaning light would travel 1.523 times slower in crown glass compared to a vacuum. Importantly, this means that refractive index of a material will never be below 1.00. Some of the most common examples of materials and their respective refractive indices are listed for your information in Table 2.1 .
| Material | Refractive Index ( n ) |
|---|---|
| Air | 1.00 |
| Water | 1.333 |
| Plastic (CR39) | 1.498 |
| Crown glass | 1.523 |
Now, when we say that the letter n in Equation 2.1 indicates the refractive index of the primary medium, we mean the refractive index of the medium in which the object exists. So, for example, in Fig. 2.5 the refractive index changes when the mug is placed in water or encased in glass.
Finally, as a top tip, if an optics question doesn’t state otherwise, it is safe to assume that the object exists in air (so, if you are not provided with a material or a refractive index, assume a value of n = 1.00).
When considering object distance (l) we need to remember that, in optics, distances are always measured from the surface/lens to the object and are always measured in metres . This is because dioptres are SI (International System) Units and so the distance also needs to be the correct SI unit in order for the numbers to come out correctly. Distance measures will also need to be assigned either a positive or negative value depending on the location of the object from the surface. In typical optics convention, we assume that light will always travel from left to right across the page, as shown in Fig. 2.6 , and this is called linear sign convention . The reason we do this is because it helps to make diagrams easy to read, and it helps to make the mathematical equations work.
Now, assuming a direction of light where light begins on the left and travels to the right, and understanding that distances are always measured from the surface/lens to the object, if a distance is measured with the direction of light (e.g. from left to right) then the distance is assigned a positive value or ‘sign’ (e.g. +0.16 m). However, if the distance is measured in the opposite direction of that of light, then the distance is assigned a negative value or ‘sign’ (e.g. −0.08 m). Be very careful not to forget the minus sign when recording these measurements in your calculations, as it’s one of the most likely areas for errors to occur! In the example image ( Fig. 2.7 ) the object is left of the surface, which means that we measure from the surface on the right to the object on the left, a direction that is opposite to the direction of light ( Fig 2.6 ). Thus the distance (l) would be recorded as negative.
Assuming a refractive index of 1.00 then, Equation 2.1 tells us that any object distance which is negative should produce a negative vergence value, and any positive distance should produce a positive vergence value. Importantly, negative values for vergence (e.g. −2.00 D) describe diverging rays, and positive values (e.g. +3.00 D) describe converging rays.
If an object is placed 10 cm in front of a surface, what is the object’s vergence at the point where light from the object meets the surface?
Step 1: Determine what we need to calculate
object vergence, L
Step 2: Define variables
l = −0.10 m (the question stated the object was 10 cm ‘in front of’, meaning it’s to the left of, the surface. Therefore, it will have a negative distance and we need to convert to metres)
n = 1.00 (nothing is specifically mentioned so we assume the object is in air)
Step 3: Determine necessary equation
L = n/l ( Equation 2.1 )
Step 4: Calculate
L = n/l
L = 1.00/ −0.10
L = −10.00D
(don’t forget the ± sign and the units!)
Practice questions 2.1:
2.1.1 If an object is placed 12 cm in front of a surface, what is the object’s vergence at the point where light from the object meets the surface?
2.1.2 If an object is placed 7 cm in front of a surface, what is the object’s vergence at the point where light from the object meets the surface?
However, it is important to note that as we always assume light is travelling from left to right, and that light will always originate from a light source or an object, this means that our sources and objects will always be pictured on the left in any diagrams (which will be important to remember in future chapters). This means that the source or object should always have a negative (or infinite) distance and will therefore always have diverging or parallel light rays – so how do we produce convergence? The answer is that when a light ray travels through a surface (e.g. a lens or a glass block or water), it’s path may be altered by either a change in refractive index or a change in power (see ‘Spherical Curved Surfaces’) The view of the object after the rays change direction is called the image , and an image can have converging, diverging or parallel rays. The distance of the image from the surface is denoted as l ′ (pronounced little el prime or little el dash).
In Fig. 2.8 , the image is on the right of the surface, which gives it a positive distance and indicates the light rays are converging to produce it.
Now, although light usually travels in straight lines, it can change direction if it travels into a different medium. This change in direction is called refraction and is one of the core underlying principles of visual optics.
Refraction
If a light ray changes direction, this is called refraction . This can occur if there is a change in velocity of the light as it travels from one material (primary medium, e.g. air) to another (secondary medium, e.g. glass) where the refractive indices of the materials are different. A real-life example of this is when you try to look at your arms or legs under water when you’re in a swimming pool – water has a different refractive index to the air and so the image of your limbs looks distorted. This is refraction in action! You can try it at home by placing a pencil in a glass of water (or by watching the demo video on the associated Elsevier web content) – does it look distorted in any way? Please also see chapter 22 to complete your own demonstration of how refractive index can influence an image, and see Box 2.1 to investigate why we can see ice in water.
As a general rule, when light travels within a single medium it travels in a straight line, providing the refractive index doesn’t change (see section on ‘ The Weird and Wonderful World of Refraction Through a Single Material ’ for more discussion on this). Similarly, if light travelled from one medium to another of an identical refractive index, no refraction (bending of the light) would occur.
Now, in the case of ice cubes in a glass of water, ice is literally made of water, and yet it refracts light differently – this suggests it probably has a different refractive index somehow (which indeed it does). This is because refractive index is determined by both density and temperature. Ice is less dense than water (we know this because ice floats in water – a clear indication of lower density) and usually a lot colder, which means its refractive index is lower, sitting at approximately 1.309 (relative to water’s 1.333). This means light can technically travel slightly faster through ice than it can through water – which will affect the refraction that we see. Next time you have an ice cube in a drink, see if you can link it back to what you’ve learned here!
In Fig. 2.9 , light rays are seen approaching a glass block. Another way of phrasing this is to say the light rays are incident upon the surface of the glass block, and so these are called the incident light rays . You will hopefully notice that the incident light rays are approaching the glass block at a particular angle relative to the normal , which is a hypothetical line drawn perpendicularly to the surface itself ( Box 2.2 ). This particular angle between the incident ray and the normal is known as the angle of incidence (i). Now, because the glass block (n′ 1.523) has a different refractive index to air (n 1.00), the light rays will change direction (refract) as they enter it and thus alter the angle of the light ray relative to the normal. This new angle is called the angle of refraction (i ′) , but please be careful here, as students often confuse the angle of refraction with the angle of deviation, so it is important to remember that both the angle of incidence and the angle of refraction are always measured relative to the normal. In this example, the light rays have refracted (or bent) towards the normal, meaning that the angle of refraction (i′) ( Box 2.3 ) is smaller than the angle of incidence. In contrast to this, the angle of deviation describes the angular difference between the emergent (refracted) light ray and the original path of the incident light ray, but don’t worry too much about that right now.
The ‘normal’ is a hypothetical line that we imagine exists relative to the surface of refractive or reflective material. Crucially the normal always exists at a 90° angle to the surface and intersects the point where the light ray is meeting the surface. In this textbook it is always represented as an orange dashed line.
It is important to remember how to draw the normal so that we can remember how to measure our angles of incidence (i) and refraction (i′) correctly. Remember that this is true even if the surface is curved. See Fig. B2.1 for a demonstration.
In optics, the apostrophe is a prime (or dash), meaning that if we were talking about the angle of refraction, for example, represented as (i′), we would pronounce this as ‘i prime’ (though I like to call it ‘ little i prime’ to save myself the confusion when we sometimes use capital letters). As a handy tip for you, try to remember that the apostrophe usually means that the variable is representing something after a refractive or reflective process has taken place, and so it is related to the secondary (after refraction) variables.
As an example of this, the refractive index of air might be written as n = 1.00, but if the light moves from air (the primary medium) to a glass block (the secondary medium) we would write the refractive index of the glass block as n′ = 1.523.
Calculating angles
In order to mathematically calculate the angles of incidence or refraction, we can use something called Snell’s law ( Equation 2.2 ) to work out one from the other if we know the refractive indices of the materials (see Table 2.1 ). Also, just as a word of caution, in this text I have defined the angles of incidence and refraction as i and i′ respectively, but you may encounter other resources that refer to them as θ and θ′ instead – they represent the same thing and simply depend upon whichever convention the author prefers to use!
n(sini)=n'(sin i’)
primary refractive index(sin×angle of incidence)=secondary refractive index(sin×angle of refraction)
Snell’s law helpfully shows us that the refractive indices of the primary (without the primes) and secondary (with the primes) mediums will affect the size of the angles. Essentially, without getting into too much detail, if the light ray moves from a low refractive index to a high one, the light ray will bend towards the normal (producing a smaller angle of refraction), whereas if the light ray moves from a high refractive index to a low one, the light ray will bend away from the normal (producing a larger angle of refraction). This is shown in Fig. 2.10 .
If a light ray is incident on a glass block (refractive index 1.523) at an angle of 28°, what is the angle of refraction?
Step 1: Determine what we need to calculate
angle of refraction, i’
Step 2: Define variables
i = 28° (angle of incidence)
n = 1.00 (nothing is specifically mentioned so we assume the primary medium is air)
n′ = 1.523 (refractive index of secondary medium – glass block)
Step 3: Determine necessary equation
n (sin i) = n′ (sin i′) ( Equation 2.2 )
Step 4: Calculate
n (sin i) = n′ (sin i′)
1.00 (sin 28) = 1.523 (sin i′)
0.4695.. = 1.523 (sin i′) (solve the left side)
0.4695../ 1.523 = (sin i′) (rearrange n′)
0.3083.. = (sin i′) (solve the left side)
sin-1 (0.3083..) = i′ (rearrange sin – this makes it an inverse sin)
i′ = 17.95°
(don’t forget the units!)
Also remember to double check your answer. If n′ is larger than n (as in this case) then i′ should be smaller than i.
Practice questions:
2.2.1 If a light ray is incident on a glass block (refractive index 1.523) at an angle of 35°, what is the angle of refraction?
2.2.2 If a light ray is incident on a piece of plastic (refractive index 1.498) at an angle of 25°, what is the angle of refraction?
2.2.3 If a light ray refracts out of a glass block (refractive index 1.523) at an angle of 20°, what was the angle of incidence?
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