# Features of Light

Optics Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
12.1Features of Light

In this Physics tutorial, you will learn:

• What are rays? What are beams? Where do they differ from each other?
• How does light propagate?
• Why light is an EM wave?
• What are the main components of an EM wave?
• Which are the colours of visible spectrum of light? How are they determined?
• How many types of materials are there concerning their ability to allow light pass through them?
• What is shadow? Which are the conditions for the shadow to produce?
• What are the sub-categories of shadow? What are their features?

## Introduction to Features of Light

What shape do the light beams have? Do they contradict what you already know about the shape of EM waves whose part is visible light? Explain your opinion.

Do you think all colours of light have the same features (wavelength, frequency, speed, etc.)? Why?

Are we able to see all possible light rays emitted by the Sun? Why?

Do you like to expose yourself in direct sunlight during a very hot day? What do you do in this case?

The above questions will be discussed in this tutorial, in which a number of light features will be discussed. This is done to emphasize the visible light as the part of EM spectrum that is closer to our needs.

You are not completely unfamiliar with light features because many properties of light are similar or the same as those of waves in general and EM waves in particular, we covered this in detail in our Physics tutorials on Waves, if you have not read through these, we suggest you do so if you are not familiar with the following properties. More specifically, you should already know that:

1. Light is an electromagnetic wave that travels at very high speed (in vacuum c = 300 000 km/s = 3 × 108 m/s). Most of its propagation occurs in empty space. Therefore, light does not need any material medium to propagate.
2. The speed of light is constant when it travels through the same medium. The speed of light in vacuum is considered as the highest speed in the universe.
3. We (humans) are able to see only a short segment of the entire spectrum of EM radiation produced. More precisely, we can see about 1/73 of the spectrum of EM radiation. Thus, we can see EM waves of frequencies ranging from about 4.2 × 1014 Hz to 7.5 × 1014 Hz (the corresponding range of wavelengths is from 4 × 10-7 m to 7 × 10-7 m (400 nm to 700 nm).
4. The equation of light waves is the same as for all the other EM waves, i.e.
c = λ × f
where λ is the wavelength (in metres) and f is the frequency (in Hertz).
5. Light waves possess all features a wave has, like amplitude, period, frequency, wavelength, speed, energy, power, phase, etc. The approach used to calculate these quantities is similar to those used in other kinds of waves.
6. Light is a transverse wave, i.e. its amplitude is perpendicular to the wavelength as shown in the figure below. 7. The form of light waves is sinusoidal as all the other transverse waves. This helps us find calculate many light features using the sinusoidal functions.

## Rays and Beams. Propagation of Light

We often use terms like "ray" and "beam" to represent light and its propagation is space. For example, we say "light rays enter in my room through the window" or "I am producing a light beam when turning on my torch". However, ray and beam they are not synonyms; rather, they represent different concepts. First, it is important to understand that ray and beam are artificial constructs. Ray is usually associated with geometric optics and helps us understand the light propagation, and so, it helps to visualize how light propagates through an optical system. In this consideration, the ray is one-dimensional and has a zero cross-sectional area (zero thickness), and therefore it is not physical. Beams have a finite cross-section and they are typically associated with lasers, for example. Hence, if we use the language of geometry we can compare rays with straight lines and beams with long cylinders of light. In this sense, you can think of the beam as being "made up of many rays" - so beams can be thought of as "bundles of rays". Both rays and beams contain a very large number of light waves travelling at a very narrow space. As a result, waves overlap and we are not able to discern the waveform anymore. (Another reason why we are not able to see the waveform of light is the very high frequency (the oscillation rate) of light. As mentioned above, light waves oscillate at frequencies of range 1014 Hz. This means light waves oscillate at 1014 times per second. Even if we had powerful microscopes, we would not be able to "catch" such fast oscillations.)

Thus, we can say that light rays and beams are a proof that light propagates in straight lines, despite their sinusoidal form.

## Why Light is an Electromagnetic Wave? What does EM Wave Really Mean?

Light, as a kind of EM wave carries energy, not matter. As such, it is produced by a source which when it is very hot becomes incandescent (emits light). Light is part of EM spectrum, i.e. unlike mechanical waves, it possesses both an electric and magnetic component (E and H) which are perpendicular to each other. These components are represented by the respective fields (electric and magnetic field). Thus, all EM waves contain two sinusoidal equations: one for each field. They are:

E = Emax × sin(k × x - ω × t)

for electric field and

H = Hmax × sin(k × x - ω × t)

for magnetic field.

The electric field component is much greater than the magnetic one. The ratio between them gives the speed of light c.

c = E/H

The value 3 × 108 m/s for the speed of light in vacuum is a rounded value. Accurate measurements give the value c = 299 792 458 m/s.

The figure below shows the two components of EM wave where the magnetic component is much greater than it really is for demonstration purpose. This is the reason why light is an EM wave, i.e. because it is a combination of an electric and magnetic sinusoidal component.

## Visible Light Spectrum. Colours

As explained earlier, humans can see EM waves of wavelength ranging from 400 nm to 700 nm. Shorter the wavelength, more powerful the wave. Our eyes perceive the strength of light waves through colours. Thus, the most powerful light waves we can see are of violet (purple) colour. This is the reason why UV radiation is the first category of EM waves that are more powerful than our ability to see ("ultra" means "beyond", i.e. ultraviolet rays means rays beyond the violet part of visible light spectrum).

Similarly, the weakest light rays we can see are of red colour as they are near the infrared part of EM spectrum (infra means "under" or "below", therefore infrared radiation means radiation that is below (less powerful than) the red part of visible light.

The order of colours from the least powerful to the most powerful (from the less energetic to the most energetic) is:

Red → Orange → Yellow → Green → Cyan → Blue → Violet

Look at the figure below. The following values give the range of wavelengths for each colour:

Red → 700 nm - 635 nm

Orange → 635 nm - 590 nm

Yellow → 590 nm - 560 nm

Green → 560 nm - 520 nm

Cyan → 520 nm - 490 nm

Blue → 490 nm - 450 nm

Violet → 450 nm - 400 nm

Obviously, there are no fixed borders between colours (the transition or switch between colours is gradual). Therefore, we must refer to values at middle of each range to be sure we have the colour we are looking for. This means we have some typical values for the wavelength of each colours, which represent them in exercises. For example, we can take 540 nm as typical wavelength for green light, 470 nm for blue light and so on.

Remark! The above values are valid for wavelengths of light waves in vacuum as when light passes through another medium it slows down and therefore the wavelengths decrease (frequency does not change).

### Example 1

A light wave has a frequency of 5.2 × 1014 Hz. What colour has this light wave?

### Solution 1

From the equation of light waves

c = λ × f

and given that c = 3 × 108 m/s, we obtain for the required wavelength

λ = c/f
= 3 × 108/5.2 × 1014
= 0.577 × 10-6 m
= 577 × 10-9 m
= 577 nm

From the above list, it is obvious this light wave belongs to yellow part of visible light spectrum (590 nm - 560 nm).

## Media of Light Propagation

Light cannot pass through all media. Some materials allow light pass through them. They are known as transparent materials or media. Vacuum, air, glass, water etc. are examples of transparent materials.

Any light beam penetrates regularly through transparent materials. As a result, we see everything through them. The opposite of transparent is opaque, i.e. in such materials light cannot pass through. Metals, concrete, wood, etc. are examples of opaque materials. There is also a third category of materials in which light passes only partially across. Furthermore, the beam is not regular anymore after passing through them. These materials are known as translucent. As you see from the figure, the light is dimmer after passing through a translucent material. This means not all light waves are able to pass through such materials because some of the light waves falling on a translucent material are absorbed by the material itself (light is transmitted only partially through them). Furthermore, they will diffuse after leaving the translucent medium.

Examples of translucent materials include thin plastic materials such as plastic bags, dirty water, clouds, blurry glass, etc.

Since light propagates in straight lines, it primarily illuminates objects that are in direct exposure to the light source. (We will see in the next section than objects are also illuminated indirectly through diffuse reflection, which helps us see objects that are not directly exposed to the light source). As a result, when we expose an opaque object to any light source, only the frontal side of the object, which is directly exposed to the light source, will be fully illuminated. Since opaque objects do not allow light rays to pass through them, the space behind such objects will be less illuminated. As a result, a dark region called shadow will exist in that part, as shown in the figure. There are two types of shadows: umbra and penumbra. Umbra represents a complete shadow; it is formed when the light source is very small or when it is far away from the opaque objects (such as the Sun). In this case, an umbra is formed behind the object. Furthermore, the umbra borders are sharp, like in the above figure.

When the light source is relatively large and it cannot be considered as a point light source, or when there are two or more point sources emitting light, there will be regions behind an opaque object, which are partially illuminated. As a result, a dim shadow (half-shadow or penumbra) is formed on the ground in these regions as shown in the figure. An example of penumbra is when you see a football match in the evening. The field is illuminated by multiple light sources. As a result, you see a number of dim shadows (penumbra) formed around the players because their opaque body blocks the light from one of sources but that part is illuminated by the other sources that are in different places around the field. ### Example 2

What kind of region (umbra, penumbra or illuminated) will form at the points A, B, C, D and E shown in the figure? ### Solution 2

Points A and B are illuminated only by the source 1. Therefore, a penumbra is formed at these points.

Points D and E are illuminated only by the source 2. Therefore, a penumbra is formed at these points.

Point C is not illuminated by any source as the opaque barrier blocks the light emitted by them. Therefore, an umbra is formed at this point.

Look at the figure: ## Whats next?

Enjoy the "Features of Light" physics tutorial? People who liked the "Features of Light" tutorial found the following resources useful:

1. Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Optics Revision Notes: Features of Light. Print the notes so you can revise the key points covered in the physics tutorial for Features of Light
3. Optics Practice Questions: Features of Light. Test and improve your knowledge of Features of Light with example questins and answers
4. Check your calculations for Optics questions with our excellent Optics calculators which contain full equations and calculations clearly displayed line by line. See the Optics Calculators by iCalculator™ below.
5. Continuing learning optics - read our next physics tutorial: Reflection of Light

## Physics Calculators

You may also find the following Physics calculators useful.