When light interacts with objects or a medium, such as glass or water, it displays certain
properties: it can either be reflected, absorbed or
transmitted.
Reflection (ESBN2)
When you smile into a mirror, you see your own face smiling back at you. This is caused by
the reflection of light rays on the mirror. Reflection occurs when an incoming light ray
bounces off a surface. In the following figure, a still lake reflects the landscape
surrounding it.
To describe the reflection of light, we will use the following terminology. The incoming
light ray is called the incident ray. The light ray moving away from
the surface is the reflected ray. The most important characteristic of
these rays is their angles in relation to the reflecting surface. These angles are
measured with respect to the normal of the surface. The normal is an
imaginary line perpendicular to the surface. The angle of incidence,
\(\theta_i\) is measured between the incident ray and the surface
normal. The angle of reflection, \(\theta_r\) is measured between the
reflected ray and the surface normal. This is shown in Figure 5.3.
When a ray of light is reflected, the reflected ray lies in the same plane as the incident
ray and the normal. This plane is called the plane of incidence and is
shown in Figure 5.4.
Law of reflection
The angle of incidence is equal to the angle of reflection:
\[\theta_{i} = \theta_{r}\]
and the incident ray, reflected ray, and the normal, all lie in the same
plane.
The simplest example of the law of reflection is if the angle of incidence is
\(\text{0}\)\(\text{°}\). In this case, the angle of reflection is also
\(\text{0}\)\(\text{°}\). You see this when you look straight into a mirror.
Applying what we know from the law of reflection, if a light ray strikes a surface at
\(\text{60}\)\(\text{°}\) to the normal to the surface, then the angle that the
reflected ray makes with the normal must also be \(\text{60}\)\(\text{°}\) as shown
in Figure 5.6.
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Real world applications of reflection
A parabolic reflector is a mirror or dish (e.g. a satellite dish) which has a
parabolic shape. Some examples of very useful parabolic reflectors are car
headlamps, spotlights, telescopes and satellite dishes. In the case of car
headlights or spotlights, the outgoing light produced by the bulb is reflected
by a parabolic mirror behind the bulb so that it leaves as a collimated beam
(i.e. all the reflected rays are parallel). The reverse situation is true for a
telescope where the incoming light from distant objects arrives as parallel rays
and is focused by the parabolic mirror to a point, called the focus, where an
image can be made. The surface of this sort of reflector has to be shaped very
carefully so that the rays all arrive at the same focal point.
Rays and Reflection
Textbook Exercise 5.1
Are light rays real? Explain.
Light rays are not real.
In physics we use the idea of a light ray to indicate the
direction in which light travels. In geometrical optics,
we represent light rays with straight arrows to show how
light propagates. Light rays are not an exact
description of the light; rather they show the direction
in which the light wavefronts are travelling.
The diagram shows a curved surface. Draw normals to the
surface at the marked points.
Which of the labels, A-H, in the diagram, correspond to the
following:
normal
E
angle of incidence
C
angle of reflection
D
incident ray
B
reflected ray
A
State the Law of Reflection. Draw a diagram, label the
appropriate angles and write a mathematical expression
for the Law of Reflection.
The law of reflection states that the angle of incidence is
equal to the angle of reflection and the incident ray,
reflected ray, and the normal, all lie in the same
plane.
We can write this mathematically as: \(θ_{i} =
θ_{r}\).
Draw a ray diagram to show the relationship between the angle
of incidence and the angle of reflection.
The diagram shows an incident ray \(I\). Which of the other 5
rays (A, B, C, D, E) best represents the reflected ray
of \(I\)?
Ray B looks like it is reflected at the same angle as the
incident ray.
A ray of light strikes a surface at
\(\text{15}\)\(\text{°}\) to the normal to the
surface. Draw a ray diagram showing the incident ray,
reflected ray and surface normal. Calculate the angles
of incidence and reflection and fill them in on your
diagram.
We are told that the ray of light strikes the surface at
\(\text{15}\)\(\text{°}\) to the normal to the
surface, so the angle of incidence is
\(\text{15}\)\(\text{°}\). Since the angle of
incidence is equal to the angle of reflection, the angle
of reflection is also \(\text{15}\)\(\text{°}\).
Putting this onto a ray diagram (not to scale!):
A ray of light leaves a surface at
\(\text{45}\)\(\text{°}\) to the normal to the
surface. Draw a ray diagram showing the incident ray,
reflected ray and surface normal. Calculate the angles
of incidence and reflection and fill them in on your
diagram.
We are told that the ray of light leaves the surface at
\(\text{45}\)\(\text{°}\) to the normal to the
surface, so the angle of reflection is
\(\text{45}\)\(\text{°}\). Since the angle of
reflection is equal to the angle of incidence, the angle
of incidence is also \(\text{15}\)\(\text{°}\).
Putting this onto a ray diagram (not to scale!):
A ray of light strikes a surface at
\(\text{25}\)\(\text{°}\) to the surface. Draw a
ray diagram showing the incident ray, reflected ray and
surface normal. Calculate the angles of incidence and
reflection and fill them in on your diagram.
We are told that the ray of light strikes the surface at
\(\text{25}\)\(\text{°}\) to the surface. To find
the angle of incidence we note that the normal is drawn
at \(\text{90}\)\(\text{°}\) to the surface. So the
angle of incidence is \(\text{90}\text{°} -
\text{25}\text{°} = \text{65}\text{°}\). Since
the angle of incidence is equal to the angle of
reflection, the angle of reflection is also
\(\text{65}\)\(\text{°}\).
Putting this onto a ray diagram (not to scale!):
A ray of light leaves a surface at
\(\text{65}\)\(\text{°}\) to the surface. Draw a
ray diagram showing the incident ray, reflected ray and
surface normal. Calculate the angles of incidence and
reflection and fill them in on your diagram.
We are told that the ray of light leaves the surface at
\(\text{65}\)\(\text{°}\) to the surface. To find
the angle of reflection we note that the normal is drawn
at \(\text{90}\)\(\text{°}\) to the surface. So the
angle of reflection is \(\text{90}\text{°} -
\text{65}\text{°} = \text{25}\text{°}\). Since
the angle of reflection is equal to the angle of
incidence, the angle of incidence is also
\(\text{25}\)\(\text{°}\).
Putting this onto a ray diagram (not to scale!):
A beam of light (for example from a torch) is generally not
visible at night, as it travels through air. Try this
for yourself. However, if you shine the torch through
dust, the beam is visible. Explain why this happens.
To see the beam from a torch the light rays emitted by the
torch need to be reflected off something for our eye to
see it. At night the light rays are not being reflected
off anything and so we do not see the beam of the torch.
If we shine the torch through dust, then we can see the
beam as the light rays reflect off the dust particles.
If a torch beam is shone across a classroom, only students in
the direct line of the beam would be able to see that
the torch is shining. However, if the beam strikes a
wall, the entire class will be able to see the spot made
by the beam on the wall. Explain why this happens.
To see the beam from a torch the light rays emitted by the
torch need to be reflected off something for our eye to
see it. So only the students in the direct path of the
light will see that the torch is shining. However as
soon as the torch beam is reflected off the wall, the
learners are all able to see the spot where the beam
reflects.
A scientist looking into a flat mirror hung perpendicular to
the floor cannot see her feet but she can see the hem of
her lab coat. Draw a ray diagram to help explain the
answers to the following questions:
Will she be able to see her feet if she backs away
from the mirror?
She will not be able to see her feet. Your eye sights
along a line to see your feet. This line must
intersect the mirror.
The solid lines show the path of light from her
labcoat hem to her eyes, while the dotted line
shows the path that light would have to take to
reach her eyes from her feet. As she moves
further away the only thing that changes is the
angles of incidence and reflection.
What if she moves towards the mirror?
She still will not be able to see her feet. Your eye
sights along a line to see your feet. This line
must intersect the mirror.
The solid lines show the path of light from her
labcoat hem to her eyes, while the dotted line
shows the path that light would have to take to
reach her eyes from her feet. As she moves
closer the only thing that changes is the angles
of incidence and reflection.
Absorption
In addition to being reflected, light can also be absorbed. Recall
from grade 10 that visible light covers a range of wavelengths in the
electromagnetic spectrum. The colours we see with our eyes correspond to light
waves with different wavelengths or frequencies.
Imagine that you shine a torch or other source of white light onto a white piece of
paper. The light is reflected off the paper and into our eyes and we see the
colour of the paper as white. Now if we were to shine the same light onto a red
apple we will notice that the colour of the apple appears red. This means that
the surface of the apple is only reflecting red light into our eyes. All the
other wavelengths in the incident white light are absorbed by the apple's skin.
If you touch the apple, it will feel warm because it is absorbing the energy
from all the light it is absorbing. Because of absorption and reflection, we can
perceive colours of different objects. White objects reflect all or most of the
wavelengths of light falling on them, coloured objects reflect particular
wavelengths of light and absorb the rest. Black objects absorb all the light
falling on them. This is why wearing a white t-shirt outside in the sun is
cooler than wearing a black t-shirt, since the white t-shirt reflects most
of the light falling on it, while the black t-shirt will absorb it and heat up.
Transmission
A further property of light is that it can be transmitted through
objects or a medium. Objects through which light can be transmitted are called
transparent and objects which block out light or that light cannot pass
through are called opaque.
For example, glass windows allow visible light to pass through them which is why we
can see through windows. The light rays from things outside the window can pass
through or be transmitted through the glass and into our eyes. Brick walls on
the other hand are opaque to visible light. We cannot see through brick walls
because the light cannot be transmitted through the wall into our eyes. The
transmission of light through an object depends on the wavelength of the light.
For example, short wavelength visible light cannot be transmitted through a
brick wall whereas long wavelength radio waves can easily pass through walls and
be received by a radio or cell phone. In other words, a brick wall is
transparent to radio waves!