Chapter 25
The Reflection of Light:
Mirrors
Chapter 25 The reflection of light :Mirrors
25.1 Wave Fronts and Rays
25.2 The Reflection of Light
25.3 The Formations of Images by a Plane Mirror
25.3 Spherical mirrors
25.4 The Formations of Images by Spherical mirrors
25.6 The Mirror Equation and the Magnification
equation
25.1 Wave Fronts and Rays
• For a small spherical object whose surface is pulsating, a sound
wave is emitted that moves spherically outward from the object at
a constant speed
• Wave fronts are surfaces on which all points of a wave are in the
same phase of motion
• Rays are lines that are perpendicular to the wave fronts and point
in the direction of the velocity of the wave
• The concept of a wave fronts and rays can also be used
describe light waves
• Spherical wave front:- the rays are perpendicular to the wave
fronts and diverge
• At large distances from the source, the wave fronts become less
and less curved.
• Plane wave fronts:- the wave fronts are flat surfaces and the
rays are parallel to each other
25.2 The Reflection of Light
• When light reflects from a smooth surface, the reflected light
obeys the law of reflection
a. The incident ray, the reflected ray, and the normal to the
surface all lie in the same plane
b. The angle of reflection 𝜃𝑟 equals the angle of incident 𝜃𝑖
𝜃𝑟 = 𝜃𝑖
• In specular reflection, the reflected rays are parallel to each other.
• In diffuse reflection, the surface reflects light in various directions
25.3 The Formation of Images by a Plane Mirror
The person’s right hand becomes the image’s left hand.
The image has three properties:
1. It is upright (erect).
2. It is the same size as you are.
3. The image is as far behind the mirror are you are in front of it.
• A ray of light from the top of the chess piece reflects from the
mirror.
• To the eye, the ray seems to come from behind the mirror.
• A virtual image is one from which all the rays of light do not
actually come, but only appear to do so.
• The geometry used to show that the image distance is equal to
the object distance.
• The magnitude of object distance 𝑑𝑜 equals the magnitude of
the image distance 𝑑𝑖
• The height of the image equals the height of the object
Conceptual Example 1 Full-Length Versus Half-Length Mirrors
What is the minimum mirror height necessary for her to see her full
image?
Conceptual Example 2 Multiple Reflections
A person is sitting in front of two mirrors that intersect at a right
angle.
The person sees three images of herself. Why are there three, rather
than two, images?
25.4 Spherical Mirrors
• If the inside surface of the spherical mirror is polished, it is a
concave mirror.
• If the outside surface is polished, is it a convex mirror.
• The law of reflection applies, just as it does for a plane mirror.
• The principal axis of the mirror is a straight line drawn
through the center and the midpoint of the mirror.
• A point on the tree lies on the principal axis of the concave
mirror.
• Rays that are close to the principal axis are known as paraxial
rays
• Paraxial rays are no necessarily parallel to the principal axis
• Image is formed where light rays cross the principal axis at a
common point after reflection.
• Light rays near and parallel to the principal axis are reflected
from the concave mirror and converge at the focal point.
• The focal length is the distance between the focal point and the
mirror.
1
𝑓= 𝑅
2
• The focal point of a concave mirror is halfway between the
center of curvature of the mirror C and the mirror at B.
• Rays that are far from the principal axis do not converge to a
single point.
• The fact that a spherical mirror does not bring all parallel rays
to a single point is known as spherical aberration.
Convex (diverging) mirror
• When paraxial light rays that are parallel to the principal axis
strike a convex mirror, the rays appear to originate from the focal
point.
1
𝑓=− 𝑅
2
25.5 The Formation of Images by Spherical Mirrors
using graphical (ray tracing) method
CONCAVE (CONVERGING) MIRRORS
This ray is initially parallel to the principal
axis and passes through the focal point.
This ray initially passes through the focal
point, then emerges parallel to the principal
axis.
This ray travels along a line that passes
through the center.
Image formation and the principle of reversibility
• When an object is located between the focal point and a concave
mirror, an enlarged, upright, and virtual image is produced.
An example is makeup mirror (or shaving mirror)
CONVEX (DIVERGING) MIRRORS
• Ray 1 is initially parallel to the principal axis and appears to
originate from the focal point.
• Ray 2 heads towards the focal point, emerging parallel to the
principal axis.
• Ray 3 travels toward the center of curvature and reflects back on
itself.
• Convex mirrors always produce virtual images that are
diminished in size and upright, relative to the object.
25.6 The Mirror Equation and Magnification
1
1
1
=
+
𝑓 𝑑 𝑜 𝑑𝑖
ℎ𝑖
𝑑𝑖
𝑚=
=−
ℎ𝑜
𝑑𝑜
Summary of Sign Conventions for Spherical Mirrors
1
1
1
=
+
𝑓 𝑑𝑜 𝑑 𝑖
ℎ𝑖
𝑑𝑖
𝑚=
=−
ℎ𝑜
𝑑𝑜
NOTE THAT:
• 𝑓 is positive for a concave mirror and negative for a convex
mirror
• 𝑑𝑜 is positive if the object is in front of the mirror and negative
if it is behind the mirror
• 𝑑𝑖 is positive if the image is in front of the mirror (real image)
and negative if it is behind the mirror (virtual image)
• 𝑚 is positive for an upright image and negative for an inverted
image
Example 5 A Virtual Image Formed by a Convex Mirror
A convex mirror is used to reflect light from an object placed 66 cm
in front of the mirror. The focal length of the mirror is -46 cm.
Find the location of the image and the magnification.
Solution
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
𝑓 = −46𝑐𝑚
𝑑𝑜 = +66𝑐𝑚
1
1
1
=
−
𝑑𝑖 −46 𝑐𝑚 +66 𝑐𝑚
1 1 1
= −
𝑑𝑖 𝑓 𝑑𝑜
𝑑𝑖 = −27 𝑐𝑚
−𝑑𝑖
𝑀=
𝑑𝑜
− −27 𝑐𝑚
𝑀=
= 0.41
66 𝑐𝑚
An object is located 14 cm in front of a convex mirror, the image
being 7 cm behind the mirror. A second object, twice as tall as the
first one, is placed in front of the mirror, but at a different location.
The image of this second object has the same height as the other
image.
How far in front of the mirror is the second object located?
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
1
1
1
=
+
𝑓 14𝑐𝑚 −7𝑐𝑚
𝑓 = −14
The image of this second object has the same height as the other image
ℎ𝑖2 = ℎ𝑖1
ℎ𝑖
𝑑𝑖
𝑚=
=−
ℎ𝑜
𝑑𝑜
−𝑑𝑖2 ℎ𝑜2 −𝑑𝑖1 ℎ𝑜1
=
𝑑𝑜2
𝑑𝑜1
∴ ℎ𝑖 =
∴ 𝑑𝑖2 = 𝑑𝑜2
A second object, twice as tall as the first one
𝑑𝑖1
𝑑𝑜1
−𝑑𝑖
ℎ𝑜
𝑑𝑜
ℎ𝑜1
ℎ𝑜2
ℎ𝑜2 = 2ℎ𝑜1
Example 7
An optometrist is using a keratometer to measure the radius of
curvature of the cornea of the eye. When an object is placed 9.0 cm in
front of the cornea, the magnification of the corneal surface is
measured to be 0.046. (The corneal surface acts like a convex mirror)
Determine the radius of the cornea.
Solution
1
1
1
=
+
𝑓 𝑑 𝑜 𝑑𝑖
𝑓=−
𝑅
2
1
2
=−
𝑓
𝑅
𝑑𝑖
𝑚=−
= 0.046
𝑑𝑜
∴ 𝑑𝑖 = −0.046𝑑𝑜
2
1
1
− =
+
𝑅 𝑑𝑜 𝑑𝑖
2 1
1
− = −
𝑅 9
0.046 9
2
− = −2.30
𝑅
2
∴𝑅=
= 0.87 𝑐𝑚
2.30
An object is located 14 cm in front of a convex mirror, the image
being 7 cm behind the mirror. A second object, twice as tall as the
first one, is placed in front of the mirror, but at a different location.
The image of this second object has the same height as the other
image.
How far in front of the mirror is the second object located?
1
1
1
=
+
𝑓 𝑑𝑜 𝑑𝑖
1
1
1
=
+
𝑓 14𝑐𝑚 −7𝑐𝑚
𝑓 = −14
The image of this second object has the same height as the other image
ℎ𝑖2 = ℎ𝑖1
ℎ𝑖
𝑑𝑖
𝑚=
=−
ℎ𝑜
𝑑𝑜
−𝑑𝑖2 ℎ𝑜2 −𝑑𝑖1 ℎ𝑜1
=
𝑑𝑜2
𝑑𝑜1
∴ ℎ𝑖 =
∴ 𝑑𝑖2 = 𝑑𝑜2
A second object, twice as tall as the first one
𝑑𝑖1
𝑑𝑜1
−𝑑𝑖
ℎ𝑜
𝑑𝑜
ℎ𝑜1
ℎ𝑜2
ℎ𝑜2 = 2ℎ𝑜1
𝑑𝑖1 = −7𝑐𝑚
𝑑𝑜1 = 14𝑐𝑚
∴ 𝑑𝑖2 = 𝑑𝑜2
𝑑𝑖2 = 𝑑𝑜2
𝑑𝑖1
𝑑𝑜1
−7𝑐𝑚
14𝑐𝑚
1
1
1
=
+
𝑓 𝑑𝑖2 𝑑𝑜2
ℎ𝑜2 = 2ℎ𝑜1
ℎ𝑜1
ℎ𝑜2
𝑑𝑖2 = 𝑑𝑜2
1
2
𝑑𝑖2 = −0.250𝑑𝑜2
1
1
1
=
+
−14𝑐𝑚 −0.25𝑑𝑜2 𝑑𝑜2
𝑑𝑜2 = 42𝑐𝑚
𝑓 = −14𝑐𝑚
𝑑𝑖1
𝑑𝑜1
ℎ𝑜1
2ℎ𝑜1
THE END