Chapter 30 The Law of Reflection
“It follows that each little region of a luminous body,
such as the Sun, a candle, or a burning coal, generates
its own waves of which that region is the center. Thus
in the flame of a candle, having distinguished the
points A, B, C, concentric circles described about each
of these points represent the waves which come from
them. And one must imagine the same about every
point of the surface and of the part within the flame.”
Christian Huygens
30.1 Light as an Electromagnetic Wave
In chapter 29 we saw that light is an electromagnetic wave; a transverse wave that
travels through empty space at the speed of 3.00 × 108 m/s or 186,000 miles/s. The
relation between the wavelength λ, frequency ν, and speed c of the light is given by
the fundamental equation of wave propagation as
λν = c
(30.1)
Notice that we are now using the Greek lower case letter ν (nu) to represent
frequency rather than the letter f, we used previously in the study of wave motion.
This usage is customary in physics, especially when dealing with electromagnetic
radiation in modern physics.
The wavelength of visible light varies from about 3.8 × 10−7 m for violet light
to the longer wavelength of red light at 7.2 × 10−7 m. Since the wavelength of light
is so small, we commonly use the nanometer1 (nm), where
1 nm = 10−9 m
In these units visible light varies from about 380.0 nm to about 720.0 nm.
A great deal of research went into optics, the study of light, before its
electromagnetic character was known. Hence, an elementary treatment of optics
can be followed without recourse to electromagnetic theory. A light wave is
represented in figure 30.1. The electric and magnetic vectors are not shown and the
magnetic portion of the wave is completely missing.
If a monochromatic point source of light (one of a single wavelength) is turned
on at a particular instant, then a spherical wave emanates from the source. A
two-dimensional view of the wave is shown in figure 30.2. A wave front is a line
connecting points all having the same phase of vibration and some are shown in
figure 30.2. Far away from the source of light the circular fronts look more like plane
A nonstandard unit for the wavelength of light that is still in common usage is the angstrom,
ÿ . One angstrom is equal to 10−10 m. In this unit the wavelength of light varies from
abbreviated A
ÿ.
about 3800 to 7200 A
1
30-1
Chapter 30 The Law of Reflection
Figure 30.1 A light wave.
Figure 30.2 The wave front.
fronts. The waves are then called plane waves. These plane waves can then be shown
as in figure 30.3. A line drawn perpendicular to the wave front is called a ray of light
Figure 30.3 Plane waves.
30-2
Chapter 30 The Law of Reflection
and represents the direction of propagation of the light wave. Note that the ray of
light travels in a straight line. The wavelength of light is so small that in a
relatively short distance away from the point source of light, the waves appear
plane. Even if the source of light is not a point, when we are sufficiently far away
from the source, the waves are effectively plane. In all the subsequent discussions
we will assume that all the light waves are plane waves.
This analysis of light into waves and rays allows us two different descriptions
of light. When only the light rays are dealt with in the analysis of an optical system,
the description is called geometrical optics. When the analysis of an optical system
is done in terms of waves, the description is called wave optics or physical optics.
Still another description of light is possible by treating light as little bundles of
electromagnetic energy, called photons. Such a description is called quantum optics
and will be discussed later in chapter 35. Except for the analysis of the laws of
reflection and refraction by wave motion in sections 30.2, 31.1, and 31.2, chapters
30 and 31 discuss geometrical optics. Chapter 32 deals with wave optics.
At a large distance from a point source of light, the light waves are plane
waves propagating as in figure 30.3. If these plane waves were to strike an obstacle
of some kind, the wave front becomes distorted. In order to find the new location
and shape of the distorted wave front after a passage of time, use is made of a
principle, called Huygens’ principle, named after Christian Huygens (1629-1695).
Huygens’ principle states that each point on a wave front may be considered as a
source of secondary spherical wavelets. These secondary wavelets propagate in the
forward direction at the same speed as the initial wave. The new position of the wave
front at a later time is found by drawing the tangent to all of these secondary
wavelets at the later time.
As an example, consider the wave front in figure 30.4(a), labeled AC. The
(a)
(b)
Figure 30.4 Huygens’ construction.
30-3
Chapter 30 The Law of Reflection
points marked on the wave front, AC, act as a source of secondary waves. The
secondary wavelets travel a distance R, given by
R = v∆t
where v is the speed of both the initial wave AC and the wavelet and ∆t is the time
interval considered. The point B in figure 30.4 represents the source of one such
wavelet. The tangent to all these secondary wavelets is shown as the new wave
front in the figure. A similar case is shown in figure 30.4(b) for plane waves. An
important application of Huygens’ principle is found in the law of reflection.
30.2 The Law of Reflection
Consider a plane wave advancing toward a smooth surface such as a glass or a
mirror, as shown in figure 30.5(a). A light ray AO, which is perpendicular to the
Figure 30.5 The law of reflection.
wave fronts, is shown making an angle of incidence i with the normal N. The side B1
of the advancing wave front B1B2 has just made contact with the surface. While B2 is
moving the distance R = v∆t, a secondary wavelet emanates from B1 and also travels
the distance R = v∆t and is shown in the figure. At the end of the time interval ∆t,
B2 will have just made contact with the surface, at the point B’2. The new reflected
wave front is found from Huygens’ principle by drawing the new wave front such
that it touches the point B’2 and is tangent to the secondary wavelet from B1. The
point of tangency is called B’1. The new wave front is shown in figure 30.5(a) as the
orange line B’1B’2. The wave is said to have been reflected. The new reflected ray OC
is drawn perpendicular to the wave front B’1B’2. The rest of the reflected wave fronts
30-4
Chapter 30 The Law of Reflection
are drawn parallel to B’1B’2 and perpendicular to ray OC and are shown in the
figure. The reflected ray OC makes an angle r called the angle of reflection, with the
normal N. Because the incident ray AO is perpendicular to the wave front B1B2, the
angle that the wave front B1B2 makes with the surface is the same as the angle of
incidence i that incident ray AO makes with the normal N. This can be seen by
looking at triangle B1EO. For the moment, call angle EB1O, the angle θ. The sum of
all the angles in triangle B1EO must equal 1800. That is,
θ + α + 900 = 1800
but
or
Therefore,
and
α + i = 900
α = 900 − i
θ + (900 − i) + 900 = 1800
θ=i
as stated. Similarly, since reflected ray OC is perpendicular to reflected wave front
B’1B’2, the reflected wave front B’1B’2 makes an angle r with the reflecting surface.
We see in the figure that in triangle B1B2B’2,
sin i = v∆t
B1B’2
(30.2)
While from triangle B1B’1B’2, we see that
sin r = v∆t
B1B’2
(30.3)
Since the right-hand side of equation 30.2 is equal to the right-hand side of equation
30.3, we have
sin i = sin r
or
i=r
(30.4)
Equation 30.4 is a statement of the first law of reflection and it says that the angle
of incidence i is equal to the angle of reflection r. The second law of reflection was
implied in the derivation and it says that the incident ray, the normal, and the
reflected ray all lie in the same plane.
We have derived the law of reflection by Huygens’ principle and the wave
nature of light. Having thus established the law, we can simplify the procedure for
its use by drawing the ray diagram as in figure 30.5(b). An incident ray, making an
30-5
Chapter 30 The Law of Reflection
angle i with the normal, is reflected such that the reflected ray makes an angle r,
which is equal to the angle i. In the rest of this chapter, we will apply the law of
reflection by using ray diagrams only.
30.3 The Plane Mirror
An object O is placed a distance p (called the object distance) in front of a plane
mirror RT, as shown in figure 30.6. As you know, if you are the object O, you will
see your image in the mirror. How is this image formed and how far behind the
mirror is the image located?
Figure 30.6 A plane mirror.
To determine how an image is formed, we need consider only two of the many
light rays that emanate from the point O. Consider first the ray OA. It makes an
angle of incidence i1 with the normal to the mirror. By the law of reflection, this ray
is reflected as AC such that the angle r1 is equal to the angle i1. Anyone standing in
front of the mirror at C sees this reflected ray, but to this observer it appears to
have come from behind the mirror as indicated by the dashed line.
The second incident ray OB makes an angle of incidence i2 with the mirror
and is reflected as ray BD at an angle r2 such that angle r2 is equal to angle i2. To an
observer in front of the mirror at D this reflected ray also appears to come from
behind the mirror, as indicated by the dashed line. The two reflected rays appear to
have come from the point I behind the mirror where the two dashed lines intersect.
The point I is called the image of the object O. The distance from the image to the
mirror is called the image distance and is designated by the letter q. The image is
called a virtual image because the light rays do not actually come from that point,
they only appear to come from that point. If you were to walk behind the mirror,
you would not find any real image there. A real image can be projected onto a screen
while a virtual image cannot. In general, a real image is one on which all the rays
are converging, while a virtual image is one from which all the rays are diverging.
30-6
Chapter 30 The Law of Reflection
Thus, rays AC and BD are diverging away from the virtual image I and will never
cross on the left-hand side of the mirror.
We should also note that there is a third ray that can be used. This is the ray
that is perpendicular to the mirror. Since its angle of incidence is zero, its reflected
angle is also zero, and the ray is reflected back along the normal. This ray appears
to come from directly behind the mirror and is also indicated by a dashed line that
comes from the image point I.
In the triangle OSA the angle AOS is also equal to i1 because
alternate-interior angles of two parallel lines are equal. Using similar reasoning,
angle AIS of triangle SIA is equal to angle r1. From triangle OSA we have
tan i1 = opposite = SA
adjacent
p
(30.5)
while from triangle SIA
tan r1 = SA
q
(30.6)
But the law of reflection states that the angle r1 is equal to the angle i1. Equating
30.5 to 30.6 leads to
q=p
(30.7)
Equation 30.7 says that the image is as far behind the mirror as the object is in front
of it.
In general, to describe an optical image three words are necessary: its nature
(real or virtual), its orientation (erect, inverted, perverted), and its size (enlarged,
true, or reduced). Recall that when you look into a mirror your image is reversed.
That is, if you hold up your right hand in front of the mirror, your image appears as
though the left hand was held up. This inversion of left-right symmetry is called
perversion. Thus, a plane mirror produces a virtual, perverted, true image.
Example 30.1
An image in a plane mirror. If an object 15.0 cm high is placed 20.0 cm in front of a
plane mirror, where is the image located and how high is the image?
Solution
The example is illustrated in figure 30.7. A ray from the top and bottom of the
object is drawn normal to the mirror. These rays are reflected upon themselves, as
shown. The top and bottom of the object appear to come from the dashed lines
behind the mirror. Another ray OB is reflected from the mirror as ray BC. If this
ray is produced backward it intersects the dashed line from the top of the object at I,
producing the image.
30-7
Chapter 30 The Law of Reflection
Figure 30.7 Finding the height of an image for a plane mirror.
From the triangles we have
tan i = ho
p
(30.8)
where ho is the height of the object and p is the object distance. Also from the figure,
tan r = hi
q
(30.9)
where hi is the height of the image and q is the image distance. The image distance
for the plane mirror, found from equation 30.7, is
q = p = 20.0 cm
And since i = r, by the law of reflection, equation 30.9 is equal to equation 30.8, and
hence
hi = ho
(30.10)
Equation 30.10 says that for a plane mirror, the height of the image hi is equal to the
height of the object ho. For this example, ho = 15.0 cm, therefore, hi is also equal to
15.0 cm. When the height of the image is the same height as the object, the image is
said to be true.
Example 30.2
Minimum size of a mirror. What is the minimum height of a plane mirror such that
a woman of height ho can see herself full length?
30-8
Chapter 30 The Law of Reflection
Solution
The minimum height is most easily found with the help of figure 30.8. The woman
is represented by the arrow. Let us assume that the distance from her eyes to the
Figure 30.8 Minimum height of a plane mirror.
top of her head is small enough to be neglected. The normal ray AB from the top of
her head, when produced backward, locates the top of the head of the image. The
ray CD, emanating from her feet, strikes the mirror at point D and is reflected back
to her eyes at point A. Ray CDA appears to have come from point F, along the ray
FDA. A vertical line from F that intersects the produced ray AB at G is the entire
image of the woman. The minimum length of the mirror for the woman to see her
entire height is just the length BD, which we must now determine.
The tangent of the angle of incidence i is
tan i = EC
ED
whereas the tangent of the angle of reflection r is
tan r = AE
ED
30-9
Chapter 30 The Law of Reflection
Because the angle i is equal to the angle r by the law of reflection, these two
tangents are equal, and hence,
EC = AE
ED ED
or
EC = AE
(30.11)
But, as we can see from the figure
AE + EC = ho
the height of the object. Using equation 30.11, this becomes
AE + AE = ho
2AE = ho
AE = ho
2
But, as we can see from the figure, AE = BD, the length of the mirror. Therefore, the
minimum height of the mirror is
BD = ho
2
That is, the woman needs a mirror that is at least half her size, in order to see the
entire length of her body in the mirror.
Determining the image of an object with a plane mirror is, as we have seen,
rather simple. But what if the reflecting surface is not a plane? How do we find the
image then? We will now consider a spherical reflecting surface and find the image
of an object for that spherical surface. But remember there are many other
reflecting surfaces that could be considered such as parabolas, cylinders, and so on.
30.4 The Concave Spherical Mirror
A spherical mirror is a reflecting surface, whose radius of curvature is the radius
of the sphere from which the mirror is formed, figure 30.9. Here C is the center of
curvature of the mirror and R is its radius of curvature. The line going through the
center of the mirror, the vertex in figure 30.10, is called the principal axis, or optical
axis, of the mirror. Light rays that are parallel and close to the principal axis of the
concave mirror converge to a point called the principal focus F of the mirror. We
should note here that if a light source were placed at the principal focus, light rays
from the principal focus would retrace the same path, coming out parallel to the
principal axis after reflection. This is a principle of optics sometimes referred to as
30-10
Chapter 30 The Law of Reflection
Figure 30.9 A concave spherical mirror.
Figure 30.10 Parallel light converges to the principal focus of a concave
spherical mirror.
the principle of reversibility. It can be stated formally as: if a ray traces a certain
path through an optical system in one direction, then a ray sent backward through
the system along the same path, traverses the original path and comes out along the
line that the original ray entered. Let us now determine the location of the principal
focus of a concave spherical mirror.
Focal Length of a Concave Spherical Mirror
Consider a single ray AB parallel to the principal axis, as shown in figure 30.11. The
normal N to the reflecting surface at point B is actually an extension of the radius of
curvature since the radius is always perpendicular to its arc. The incoming ray AB,
therefore, makes an angle of incidence i with the normal. By the law of reflection,
the angle of reflection r is equal to this angle of incidence i and the reflected ray is
shown as BD. The point where this reflected ray crosses the principal axis is called
the principal focus and is designated by the letter F. Because AB is parallel to the
principal axis, the angle θ is equal to the angle i. However, the angle i is equal to
30-11
Chapter 30 The Law of Reflection
Figure 30.11 Center of curvature and principal focus of a concave spherical mirror.
the angle r by the law of reflection. Therefore, the angle θ is equal to the angle r,
and triangle CBF is an isosceles triangle, with side CF equal to side BF. If the
incident ray is fairly close to the principal axis, then the length BF is approximately
equal to the length VF. That is,
CF = BF = VF
and, therefore,
CF = VF
(30.12)
From the figure we see that
CF + VF = R
(30.13)
where R is the radius of curvature of the mirror. Combining equations 30.13 and
30.12, gives
VF = R
2
The distance VF from the vertex of the mirror to the principal focus F is called the
focal length f of the mirror. Using this notation,
f= R
2
(30.14)
That is, the focal length of a concave spherical mirror is equal to one-half the
radius of curvature of the mirror. (Remember that, in this derivation we assumed
that the incident ray was relatively close to the principal axis. If the width of the
mirror is comparable to its radius of curvature, all incident parallel rays are not
close to the principal axis, and all these rays, after reflection, do not pass through
the principal focus. This defect is called spherical aberration. See problem 36 at the
end of this chapter.)
30-12
Chapter 30 The Law of Reflection
Example 30.3
The focal length of a spherical mirror. To what radius should you grind a spherical
concave mirror in order to get a focal length of 10.0 cm?
Solution
The radius of the mirror, found from equation 30.14, is
R = 2f = 2(10.0 cm) = 20.0 cm
To go to this Interactive Example click on this sentence.
Location of an Image by Ray Diagram
Let us find the image produced by an object placed in front of a concave spherical
mirror, as shown in figure 30.12. The object OP, in brown, is placed a distance p in
Figure 30.12 Ray diagram for a concave spherical mirror.
front of the mirror, called the object distance. The center of curvature is located at
C, and the principal focus F is located at a distance R/2 from the mirror. The image
is found by drawing the following three standard rays.
1. Ray (1), in red, is drawn parallel to the principal axis. On striking the mirror,
it is reflected through the principal focus F.
2. Ray (2), in green, is drawn through the principal focus F. On striking the
mirror, it is reflected parallel to the principal axis because all rays that emanate
from the principal focus come out parallel to the principal axis after reflection.
3. Ray (3), in blue, is drawn through the center of curvature C. It is reflected upon
itself, since it lies along the normal.
30-13
Chapter 30 The Law of Reflection
The intersection of these three rays defines the top of the image, whereas the
principal axis defines the bottom of the image. The image QI is shown in yellow in
the figure. The distance of this image from the vertex of the mirror is called the
image distance and is designated by the letter q. The image is real, since the rays of
light actually converge at this point. If we were to place a screen at this point, we
would see a sharp image of the original object located at this point. Note that the
image is inverted.
The Mirror Equation
We can determine the location of the image analytically by making use of the
geometry of the ray diagram. Figure 30.13(a) shows a portion of ray (2) of figure
30.12 as it goes through the principal focus and hits the mirror at location D’. Since
Figure 30.13 The geometry for the mirror equation.
only a small portion of the mirror, which is close to the principal axis, is ever used
(in order to avoid spherical aberration), the arc VD’ of the mirror can be represented
by the straight line VD. The length of the line VD is the same as the height of the
image hi. There are two triangles, OPF and FVD. The tangents of the angle θ1 of
these triangles are given by
tan θ1 = ho
(30.15)
p−f
and
tan θ1 = hi
(30.16)
f
Since the left-hand sides of equations 30.15 and 30.16 are equal, they can be
equated, to give
h o = hi
p−f
f
Rearranging terms, we can write
ho = p − f
(30.17)
hi
f
We will return to this equation shortly.
30-14
Chapter 30 The Law of Reflection
Now consider the ray from the object that goes through the vertex V of the
mirror, figure 30.13(b). This ray is reflected and intersects the image at the point I.
Because V lies along the principal axis, which is normal to the mirror at V, the
angle of incidence OVP is equal to the angle of reflection IVQ by the law of the
reflection. Let us call these angles θ2. From figure 30.13(b) we can see that
tan θ2 = ho
p
and
tan θ2 = hi
q
Since the left-hand side of these two equations are equal they can be equated, to
give
ho = hi
p
q
Rearranging terms,
ho = p
(30.18)
hi
q
Since the left-hand side of equation 30.17 is equal to the left-hand side of equation
30.18, they can be equated to give
p= p−f = p−1
q
f
f
Dividing each term by p, gives
1 = 1 − 1
(30.19)
q
f
p
Rearranging terms, the equation becomes
1 = 1 + 1
f
p q
(30.20)
Equation 30.20 is called the mirror equation, and it shows the relation between the
focal length f of the mirror, the object distance p, and the image distance q.
Example 30.4
Finding the image for a concave spherical mirror. An object 5.00 cm high is placed
30.0 cm in front of a concave spherical mirror of 10.0-cm focal length. Where is the
image located?
30-15
Chapter 30 The Law of Reflection
Solution
The location of the image, found from equation 30.19, is
1 = 1 − 1
q
f
p
=
1
−
1
10.0 cm 30.0 cm
−1
= 0.100 cm − 0.0333 cm−1 = 0.0667 cm−1
and
q = 15.0 cm
To go to this Interactive Example click on this sentence.
Magnification
Another important relationship in an optical system can be obtained by defining the
linear magnification M of the mirror as the ratio of the size of the image hi to the
size of the object ho. That is,
M = hi
ho
Thus, the magnification tells how much larger the image is than the object. Using
equation 30.18, we can rewrite this as
M = hi = − q
ho
p
(30.21)
When we know the magnification M, we find the height of the image hi from
equation 30.21 as
hi = Mho
(30.22)
The minus sign in equation 30.21 is one of convention. Notice from example
30.4 and the ray diagram of figure 30.12, that a positive value of q gives an inverted
image. The minus sign in equation 30.21 thus signifies that when M is negative, the
image is inverted. We will see a little later that it is possible to have a negative
value of q. That case gives us a positive value for M, and thus signifies that the
image is erect or right side up.
30-16
Chapter 30 The Law of Reflection
Example 30.5
The magnification of a spherical mirror. Determine the magnification and size of the
image in example 30.4.
Solution
The magnification of the system, found from equation 30.21, is
M = − q = −15.0 cm = −0.500
p
30.0 cm
The height of the image, found from equation 30.22, is
hi = Mho
= (−0.500)(5.00 cm)
= −2.50 cm
The minus sign indicates that the image is inverted.
To go to this Interactive Example click on this sentence.
To summarize the possible cases of magnification:
if M > 1 the image is enlarged
if M = 1 the image is true
if M < 1 the image is reduced
if M > 0 the image is erect
if M < 0 the image is inverted.
Some Special Cases of the Concave Spherical Mirror
A great deal of insight into the concave mirror can be obtained by first placing the
object at infinity (or at least a very far distance away from the mirror). Now we
move the object in toward the mirror and observe what happens to the image.
Case 1: The Object Is Located at Infinity
When the object is at a very large distance from the mirror, the object distance p is
effectively infinite. The location of the image, found from equation 30.19, is
1 = 1 − 1
q
f
p
30-17
Chapter 30 The Law of Reflection
= 1 − 1
f
∞
But 1/∞ is equal to zero. Therefore,
1 = 1
q
f
and
q=f
(30.23)
Equation 30.23 says that when the object is located at infinity the image is located
at the principal focus. The ray diagram for this case was shown in figure 30.10. The
magnification is M = q/p = q/∞ = 0 for this case, showing that all the rays converge
to a point and the height of the image is hi = 0.
Case 2: The Object Is between Infinity and the Center of Curvature
This is a general case, the ray diagram of which was shown in figure 30.12. The
image location is found from equation 30.19 when the appropriate values of f and p
are introduced. Note from figure 30.12 that the image, which was at F when the
object was at infinity, has now moved to the left of F away from the mirror. The
image is inverted and real and the magnification (M = q/p) is less than 1 because q
is less than p. The image is therefore smaller than the object, but still greater than
its value of zero, when the image was at the focal point F.
Case 3: The Object Is Located at the Center of Curvature of the Mirror
When the object is placed at the center of curvature of the mirror, the object
distance p is equal to the radius of curvature R, which is equal to 2f. The image is
found from
1 = 1 − 1
q
f
p
= 1 − 1= 1− 1 = 1
f
R f 2f 2f
and
q = 2f = R
(30.24)
Equation 30.24 says that when the object is located at the center of curvature of the
mirror, the image is also located there. Figure 30.14 shows the ray diagram for this
case. Because q = p for this case, the magnification becomes
M = − q = − p = −1
p
p
The minus sign for the value of M indicates that the image is inverted, as seen in
figure 30.14. The absolute value of the magnification is equal to 1, hence, the height
of the image hi is the same size as the height of the object ho. Again note that as the
30-18
Chapter 30 The Law of Reflection
object moves closer to the mirror, the image moves farther away from the mirror.
The image has gotten bigger until it is now the same size as the object.
Figure 30.14 Ray diagram for an object placed at the center of curvature of the
mirror.
Case 4: The Object Is Located between the Center of Curvature and the
Principal Focus of the Mirror
The image for this case is found by the ray diagram in figure 30.15. Note from the
diagram that the image distance q is now greater than the object distance p.
Therefore, the magnification, M =− q/p, is negative, which says that the image is
inverted, and since the absolute value of the magnification is now greater than 1, the
image is now enlarged. Also note that as the object moves closer to the mirror, the
image keeps moving farther away from the mirror. We can find numerical values for
q from equation 30.19 when the values of f and p are specified. The magnification is
found from equation 30.21 and the height of the image from equation 30.22.
Figure 30.15 An object is placed between the center of curvature C and the focal
point F.
30-19
Chapter 30 The Law of Reflection
Case 5: The Object Is Located at the Principal Focus
When the object is located at the principal focus, p = f, and the image, found from
equation 30.19, is
1 = 1 − 1 = 1 − 1 =0
q
f
p
f
f
The only way for 1/q to equal zero is for q to become infinite. Therefore,
q=∞
That is, when the object moves into the principal focus, the image moves out to
infinity.
Case 6: The Object Lies within the Principal Focus
To see what happens when the object is moved within the principal focus, let us
draw a new ray diagram, as shown in figure 30.16.
Figure 30.16 Ray diagram for an object placed within the principal focus.
1. Ray (1) is drawn parallel to the principal axis as before. It is reflected from the
mirror and goes through the principal focus F.
2. There are an infinite number of rays emanating from the tip of the arrow. Ray
(2) is that ray that is emitted from the tip of the arrow but is aligned with an
imaginary ray that comes from the principal focus to the tip of the arrow. Ray (2)
thus acts as though it is a ray that comes from the principal focus and hence, it
is reflected from the mirror parallel to the principal axis.
3. Ray (3) is drawn as though it came from the center of curvature C. It is
reflected upon itself, as shown.
Notice that these three rays do not intersect anywhere on the left-hand side
of the mirror, and hence they cannot form a real image anywhere on that side of the
mirror. However, if we dash these rays backward, through the mirror, we will see
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Chapter 30 The Law of Reflection
that they do intersect on the right-hand side of the mirror. When we look into the
mirror, it will appear as though these three rays are emanating from that point on
the other side of the mirror. This point, therefore, locates the image; but we now call
it a virtual image, since light rays only appear to come from that point. The image is
not real; it cannot be projected on any screen. Note that the image is now erect and
enlarged. This example points out very clearly the difference between real and virtual
images. For a real image, rays converge to that point; for a virtual image, rays
diverge away from that point. If the mirror equation were solved for the image
distance q, it would come out as a negative quantity.
Example 30.6
An object is placed within the principal focus. An object, 5.00 cm high, is placed 7.00
cm in front of a concave spherical mirror of 10.00-cm focal length. Find the location
of its image and its final size.
Solution
The image, found from equation 30.19, is
and
1 = 1 − 1 =
1
−
1
q
f
p 10.00 cm 7.00 cm
−1
= 0.1000 cm − 0.1428 cm−1 = −0.0428 cm−1
q = −23.3 cm
Notice that for the object placed within the principal focus, the image distance q is
negative, as stated previously. A negative value of q indicates that the image is
virtual. The magnification is
M = − q = − (−23.3 cm) = +3.33
p
7 cm
Note in this case that since q is negative, the magnification M becomes positive and
hence, the image is erect. The final height of the image is
hi = Mho = (3.33)(5.00 cm)
= 16.7 cm
To go to this Interactive Example click on this sentence.
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Chapter 30 The Law of Reflection
A practical example of a concave spherical mirror used in this way is found in
a man’s shaving mirror or a woman’s make-up mirror. The focal length is designed
such that when you place your head in front of the mirror, it is located within the
principal focus of the mirror. You therefore see your face erect and enlarged.
30.5 The Convex Spherical Mirror
A convex spherical mirror is a reflecting surface formed from a portion of the outside
of a sphere, as shown in figure 30.17. The Christmas tree ball decoration is a good
example of such a mirror. An incident ray, parallel to the principal axis of a convex
spherical mirror, is shown in figure 30.18. The incident ray makes an angle i with
Figure 30.17 A convex spherical mirror.
Figure 30.18 A parallel ray of light diverges away from the principal axis for a
convex spherical mirror.
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Chapter 30 The Law of Reflection
the normal to the surface. The normal is the extension of the radius of curvature
from the center of curvature C. By the law of reflection, the reflected ray makes an
angle r with the normal and the angle r is equal to angle i. The reflected ray is thus
seen to diverge away from the principal axis. Hence, for a convex spherical mirror,
all rays parallel to the principal axis are reflected such that they diverge away from
the principal axis. (Recall that for a concave spherical mirror, parallel rays
converged to the principal axis.) If the reflected ray is dashed backward, it
intersects the principal axis at the point F. All rays parallel to the principal axis
intersect at this same point, which is called the principal focus of the convex
spherical mirror. Thus, rays that are parallel to the principal axis appear to diverge
from the principal focus F. The distance from V, the vertex of the mirror, to F is
called the focal length f of the mirror. The radius of curvature of the convex
spherical mirror is considered to be negative because the center of curvature is on
the other side of the mirror. (Recall that for the concave mirror, the center of
curvature was on the same side as the reflecting surface and the radius of curvature
was positive.) Since the radius of curvature is negative for the convex spherical
mirror, the focal length f, which is equal to one-half the radius of curvature, is also
negative. With this convention, the mirror equation, equation 30.20, is applicable to
either type of mirror.
Location of an Image by a Ray Diagram
The location of the image for a convex spherical mirror is found by drawing the
three standard rays in the ray diagram shown in figure 30.19.
Figure 30.19 Ray diagram for a convex spherical mirror.
1. The first ray (1), in red, is drawn parallel to the principal axis and is reflected
from the mirror as if it came from the principal focus F.
2. The second ray (2), in green, is drawn straight toward the principal focus F. It
is reflected parallel to the principal axis before it can get to the principal focus.
This is another example of the principle of reversibility; all parallel rays are
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Chapter 30 The Law of Reflection
diverged away from the principal focus, while all rays converging to F are
reflected parallel to the principal axis.
3. The third ray (3), in blue, is drawn through the center of curvature C of the
mirror. Any line through C lies along the radius of curvature of the mirror and is
therefore normal to the reflecting surface. Any ray along the normal is reflected
upon itself.
As we can see from figure 30.19, the three standard rays diverge away from
the principal axis and hence never intersect to the left of the mirror to form a real
image. However, if each of these lines are dashed backward behind the mirror, they
intersect there to form a virtual image. Also note that the image is erect and
reduced in size. If the location of the image were determined by the mirror equation,
the image distance q would be negative.
Example 30.7
Finding the image for a convex spherical mirror. An object 5.00 cm high is placed
30.0 cm in front of a convex spherical mirror of −10.0-cm focal length. Where is the
image located?
Solution
The location of the image, found from equation 30.19, is
1 = 1 − 1
q
f
p
=
1
−
1
−10.0 cm 30.0 cm
= −0.100 cm−1 − 0.0333 cm−1 = −0.133 cm−1
q = −7.52 cm
The negative value of q indicates that the image is virtual. Thus, we can use the
mirror equation for a convex spherical mirror but the focal length and the image are
negative.
To go to this Interactive Example click on this sentence.
Example 30.8
The magnification of a convex spherical mirror. Find the magnification and the size
of the image in example 30.7.
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Chapter 30 The Law of Reflection
Solution
The magnification is
M = − q = −(−7.52 cm) = +0.251
p
30.0 cm
Note that M is positive indicating an erect image. The size of the image is
hi = Mho = (0.251)(5.00 cm)
= 1.26 cm
To go to this Interactive Example click on this sentence.
If we consider the different possible cases of moving the object from infinity
toward the mirror, as in the case of the concave mirror, we find for a convex
spherical mirror that for a real object, the image is always virtual, erect, and reduced
in size.
Table 30.1 is a summary of the sign conventions for spherical mirrors. In
general, images are real if rays converge to them, whereas images are virtual if rays
Table 30.1
Sign Conventions for Spherical Mirrors
Quantity
Positive
Negative
Convex mirror
R and f
Concave mirror
R and f
Real image
q
Virtual image
q
Erect image
M
Inverted image
M
Real object
p
Virtual object
p
diverge from them. Objects, on the other hand, are real if rays diverge from them,
whereas objects are virtual if rays are converging toward them. It is possible that a
real image can be a virtual object for a second optical surface. The focal length of all
spherical mirrors is equal to one-half the radius of curvature of the mirror.
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Chapter 30 The Law of Reflection
The Language of Physics
Optics
The study of light (p. ).
Wave front
A line connecting points all having the same phase of vibration. Wave fronts from a
point source of light are spherical. Far away from the source, the waves appear
plane (p. ).
Geometrical optics
The analysis of an optical system in terms of light rays that travel in straight lines
(p. ).
Wave optics or physical optics
The analysis of an optical system in terms of the wave nature of light (p. ).
Quantum optics
The study of light in terms of little bundles of electromagnetic energy, called
photons (p. ).
Huygens’ principle
Each point on a wave front may be considered as a source of secondary spherical
wavelets. These secondary wavelets propagate at the same speed as the initial
wave. The new position of the wave front at a later time is found by drawing the
tangent to all of these secondary wavelets at the later time (p. ).
The law of reflection
The angle of incidence is equal to the angle of reflection. The incident ray, the
normal, and the reflected ray all lie in the same plane (p. ).
Real image
An image formed by rays converging to a point. A real image can be projected onto a
screen (p. ).
Virtual image
An image formed by rays diverging from a point. A virtual image cannot be
projected onto a screen (p. ).
Optical image
A reproduction of an object by an optical system. To describe an image three words
are necessary: its nature (real or virtual), its orientation (erect, inverted, or
perverted), and its size (enlarged, true, or reduced) (p. ).
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Chapter 30 The Law of Reflection
Spherical mirror
A reflecting surface whose radius of curvature is the radius of the sphere from
which the mirror is formed. A plane mirror is a special case of a spherical mirror
with a radius of curvature that is infinite (p. ).
Principle of reversibility
If a ray traces a certain path through an optical system in one direction, then a ray
sent backward through the system along the same path, traverses the original path
and exits along the line that the original ray entered (p. ).
Focal length
The point where all rays parallel and close to the principal axis converge. The focal
length of a concave spherical mirror is equal to one-half of its radius of curvature
(p. ).
Magnification
The ratio of the size of an image to the size of its object (p. ).
Summary of Important Equations
λν = c
The wavelength, frequency, and speed of light
Law of reflection
i=r
Focal length of a spherical mirror
Mirror equation
Magnification
Height of image
(30.1)
(30.4)
f = R/2
(30.14)
1 = 1 + 1
f
p q
(30.20)
M = hi = − q
ho
p
(30.21)
hi = Mho
(30.22)
Questions for Chapter 30
1. If you hold this textbook in front of a plane mirror the letters will all be
reversed from left to right. Why are they not also reversed from top to bottom?
2. Explain the difference between a real image and a virtual image.
*3. How can Huygens’ principle be justified?
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Chapter 30 The Law of Reflection
4. What effect does changing the object distance have on the size of an image
in a plane mirror?
5. It is easy to see an image formed by the reflection of light from a smooth
surface such as a mirror. What type of reflection would occur if the surface were
rough?
6. In the barbershop, mirrors are placed on the wall in front and behind the
customer. How are the mirrors arranged so that the customer can see the back of
his head?
7. Explain the process of formation of an image for a concave spherical mirror
as the object starts at infinity and moves toward the mirror. What happens to the
magnification?
8. You would like to place a mirror in the corner of your store so that you can
see every thing going on in the store. What type of mirror should you use?
*9. Using Huygens’ principle, describe what happens to a wave front of light
when it hits the edge of a surface.
*10. Two plane mirrors are placed at 900 to each other. Explain how, when
you look directly toward the vertex of the two mirrors, you see yourself as others see
you. That is, left is left and right is right. Draw a diagram to help in the analysis.
11. How can you make a toy periscope with two plane mirrors?
12. The Hubble Space Telescope, launched by NASA in May 1990 to take
pictures of the universe never seen before, developed a serious flaw. The telescope
was not able to focus properly. The cause of the problem was spherical aberration of
either the primary or secondary mirror. What is spherical aberration and how does
it affect the telescope?
The 2.4-m main mirror from the Hubble Space Telescope.
Problems for Chapter 30
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Chapter 30 The Law of Reflection
30.1 Light as an Electromagnetic Wave
1. What is the frequency of (a) violet light of 380.0-nm wavelength and (b) red
light of 720.0-nm wavelength?
2. What is the wavelength of the electromagnetic radiation of the following
frequencies: (a) 100 kHz, (b) 10.0 MHz, (c) 4.00 × 1014 Hz, and (d) 6.00 × 1015 Hz.
3. How many light waves of 450-nm wavelength can you fit into a distance of
1.00 cm?
Diagram for problem 3.
4. How long does it take light to reach earth from (a) the sun and (b) the
moon?
30.3 The Plane Mirror
5. An object 10.0 cm high is placed 20.0 cm in front of a plane mirror. Where
is the image located and how big is it?
6. What is the minimum height of a mirror such that a student, 1.73 m tall,
can see his entire body?
7. Repeat problem 6, but take into account that the eyes of the student are
10.0 cm below the top of his head.
8. A student stands in front of a plane mirror that is equal to half of her
height, but when looking straight ahead, her eyes look directly into the center of the
mirror. How far back from the mirror must she move in order to see her entire body
in the mirror?
Diagram for problem 8.
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Chapter 30 The Law of Reflection
30.4 The Concave Spherical Mirror
9. What is the focal length of a spherical mirror if the radius of curvature is
25.0 cm?
10. An object 10.0 cm high is placed 50.0 cm in front of a concave spherical
mirror of 15.0-cm focal length. Find the image by (a) a ray diagram and (b) the
mirror equation. Is the image real or virtual? Is the image erect or inverted? What
is the size of the image?
*11. Find the image with a concave spherical mirror of 10.0-cm focal length if
the object is located at (a) 60.0 cm, (b) 40.0 cm, (c) 20.0 cm, (d) 10.0 cm, and (e) 5.00
cm. Draw a ray diagram for each case.
12. Find the magnification for each case in problem 11.
13. If the object is 5.00 cm high, find the height of the images in each case in
problem 12.
*14. A concave spherical mirror has a focal length of 20.0 cm. Find the image
distance, magnification, and height of the image when the object is located at (a)
100 cm, (b) 80.0 cm, (c) 40.0 cm, (d) 20.0 cm, and (e) 10.0 cm. Draw a ray diagram
for each case.
15. An object is placed 15.0 cm in front of a concave spherical mirror mounted
on an optical bench in the laboratory. A screen is moved along the optical bench
until the object and image are located at the same point. Find the focal length of the
mirror.
16. Find the focal length of a concave spherical mirror that has a
magnification of 2.00 when an object is placed 20.0 cm in front of it.
17. An object 10.0 cm high is placed 10.0 cm in front of a concave spherical
mirror of 15.0-cm focal length. Find the image by a ray diagram and the mirror
equation. How high is the image?
18. Where should an object 5.00 cm high be placed in front of a 25.0-cm
concave spherical mirror in order for its image to be erect and 10.0 cm high?
19. For a concave spherical mirror of 20.0-cm focal length, find two locations
of an object such that the height of the image is four times the height of the object.
20. An object is placed 40.0 cm in front of a concave spherical mirror and its
image is found 25.0 cm in front of the mirror. What is the focal length of the mirror?
21. A concave spherical mirror has a focal length of 15.0 cm. Where should an
object be placed such that the height of the image is a quarter of the height of the
object?
22. An object is placed 10.0 cm in front of a concave spherical mirror of
15.0-cm focal length. Find the location of the image and its magnification.
23. Find the radius of curvature of a shaving mirror such that when the
object is placed 15.0 cm in front of the mirror, the image has a magnification of 2.
24. A concave spherical mirror has a focal length of 15.0 cm. Where should an
object be placed to give a magnification of (a) 2.00 and (b) −2.00?
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Chapter 30 The Law of Reflection
30.5 The Convex Spherical Mirror
25. A reflecting Christmas tree ball has a diameter of 8.00 cm. What is the
focal length of such an ornament?
26. An object 10.0 cm high is placed 30.0 cm in front of a convex spherical
mirror of −10.0-cm focal length. Find the image by a ray diagram and the mirror
equation. Find the height of the image.
*27. Find the image with a convex spherical mirror of 10.0-cm focal length if
the object is located at (a) 60.0 cm, (b) 40.0 cm, (c) 20.0 cm, (d) 10.0 cm, and (e) 5.00
cm. Draw a ray diagram for each case.
28. An object is 12.0 cm in front of a convex spherical mirror, and the image is
formed 24.0 cm behind the mirror. Find the focal length of the mirror.
29. Where should an object be placed in front of a convex spherical mirror of
15.5-cm focal length in order to get a virtual image with a magnification of one-half?
30. The distance between a real object and a virtual image formed by a
convex spherical mirror is 50.0 cm. If the focal length of the mirror is f = −25.0 cm,
find the two possible positions for the mirror.
Additional Problems
*31. A plane mirror is rotated through an angle θ. Show that the reflected ray
will always be rotated through an angle of 2θ.
32. Show that a plane mirror is a special case of a concave spherical mirror
whose radius of curvature is infinite. What does the mirror equation reduce to?
*33. Two mirrors make an angle of 900 with each other. Show that if a ray of
light is incident on the first mirror at an angle of incidence i, the reflected ray from
the second mirror makes an angle of reflection of 900 − i.
Diagram for problem 33.
Diagram for problem 34.
*34. Two mirrors make an angle of θ with each other. Show that if a ray of
light is incident on the first mirror at an angle of incidence i, the reflected ray from
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Chapter 30 The Law of Reflection
the second mirror makes an angle of reflection of θ − i.
*35. Show that when an object is placed in front of a concave spherical mirror
of focal length f and experiences a magnification M, the image is located at the
image distance given by
q = f(1 − M)
36. An object is placed 25.0 cm in front of a concave spherical mirror. The
image is found to be a quarter of the size of the object. Find the focal length of the
mirror.
37. An object is magnified by a factor of 2 when it is placed 15.0 cm in front of
a concave spherical mirror. Find the radius of curvature of the mirror.
38. A dentist uses a small concave spherical mirror to see a cavity in a tooth.
If the image is to be magnified by a factor of 3 when the tooth is 2.50 cm in front of
the mirror, what should be the focal length of such a mirror?
39. An optical system is designed so that an object for a convex spherical
mirror of focal length f = −15.0 cm is 20.0 cm behind the mirror (a virtual object).
Find the image distance and the magnification, and determine whether the image is
real or virtual, erect or inverted.
40. Repeat problem 39 with an object located only 5.00 cm behind the mirror.
*41. Use a compass to draw a concave spherical mirror 10.0 cm in radius.
Draw light rays parallel to the principal axis at every 1.00 cm above and below the
principal axis. Using a protractor, carefully measure the angles of incidence and
reflection for each of these rays, and see where they cross the principal axis. What
does this tell you about the underlying assumption in the mirror equation? How
does this relate to spherical aberration of the mirror?
*42. Draw a graph of the image distance q as a function of the object distance
p for a concave spherical mirror of focal length 10.0 cm. Show the regions that
represent the concave mirror and the convex mirror. Show the regions where the
images are real and where they are virtual.
*43. Draw a graph of the magnification M of a concave spherical mirror as a
function of the object distance p. Repeat for a convex spherical mirror.
Interactive Tutorials
44. Light as a wave. A tower on earth transmits a laser beam of frequency f =
5.00 × 1014 Hz to a spaceship at a distance x = 7.40 × 1011 m. Calculate (a) the
wavelength λ of the laser beam and (b) the time t for the beam to reach the
spaceship.
45. Spherical mirror with known radius of curvature. An object of height ho =
3.00 cm is placed at the object distance p = 10.0 cm of a spherical mirror of radius of
curvature R = 8.00 cm. Find (a) the focal length f of the mirror, (b) the image
distance q, (c) the magnification M, and (d) the height hi of the resulting image.
46. Spherical mirror with known focal length. An object of height ho = 8.50 cm
is placed at the object distance p = 35.0 cm of a spherical mirror of focal length f =
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Chapter 30 The Law of Reflection
15.0 cm. Find (a) the radius of curvature R of the mirror, (b) the image distance q,
(c) the magnification M, and (d) the height hi of the resulting image.
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clicking on this sentence.
30-33