Brief Overview of Mirrors and Lenses

Brief Overview of Mirrors and Lenses
Paul Avery
PHY 3400
Oct. 7, 2000
I Notation................................................................................................................................. 1
II Images and optics .................................................................................................................. 2
III Plane mirror....................................................................................................................... 3
IV Focal points of mirrors and lenses..................................................................................... 4
IV.1 Spherical mirror focal point .......................................................................................... 4
IV.2 Lens focal point............................................................................................................. 5
IV.3 Lens maker’s equation................................................................................................... 5
V Finding images ...................................................................................................................... 6
V.1 Lens equation ................................................................................................................ 6
V.2 Sign convention for distances ....................................................................................... 7
V.3 Real and virtual images ................................................................................................. 7
V.4 Magnification ................................................................................................................ 8
V.5 Principal ray diagrams................................................................................................... 9
V.5.a Mirror ....................................................................................................................9
V.5.b Lens ..................................................................................................................... 10
I
Notation
I use the variables p or q to refer to a distance from a mirror or lens.
p = object distance
q = image distance
f = focal length
R = radius of curvature
n = index of refraction
m = magnification = image size / object size. m can be positive or negative
m θ = angular magnification = angular size of image / angular size of object
h = height of the object (positive or negative)
h′ = height of the image (positive or negative)
1

II
Images and optics
It is a remarkable fact that when light rays emerging from an object are reflected or refracted they
can appear to come from a different location, called the image. It turns out that only specially
shaped mirrors and lenses can form sharp images. We shall deal in this course with basic shapes
such as plane mirrors, spherical mirrors and thin lenses.
All geometric optics can be derived from the laws of reflection and refraction. Consider
reflection first. When light hits a surface, the incident and reflected ray make equal angles from
the normal to the surface, where the incident ray, reflected ray and normal form a plane, as shown
in the figure below. Note that this law holds even when the surface is not flat. In that case, one
uses the normal at the point of incidence.
Reflection from a surface.
θ
θ
Refraction is the bending of light when it goes from one medium into another. Refraction obeys
Snell’s law, n1sinθ1 = n2sinθ2, where n1
is the index of refraction of the first medium, n 2 is the
index of refraction of the second medium, θ1
is the angle of the ray from the normal in the first
medium and θ 2 is the angle of the ray from the normal in the second medium. The two rays and
the normal form a single plane, just as for reflection. The laws of refraction determine the
properties of lenses. Note that when n2 > n1, the ray in medium 2 bends towards the normal
θ 1
n 1
θ 2
n 2
Refraction at a surface.
2

III
Plane mirror
We are familiar with the images from plane mirrors. If we stand a distance p from the mirror,
then our image is upright, identical in size and an equal distance p behind the mirror. These facts
can be deduced from simple geometry. By careful construction, using the fact that the angle of
incidence is the same as the angle of reflection, you can see that light rays emitted from the tip of
the object and bouncing off the mirror can be extended to the other side so they appear to be
coming from the same position on the other side of the mirror. A person standing a distance away
would “see” the image on the other side of the mirror.
Exercise: Try to use one of the rays and simple geometry to “prove” that the image is located as
shown below.
Equal angle
reflection
1
Observer sees rays
coming from here
Observer
2
3
Object
Image
Mirror
3

IV
Focal points of mirrors and lenses
The focal point of a lens or mirror is one of its most important characteristics. Rays of light
moving parallel to the optic axis (the dotted line perpendicular to the mirror or lens in the figures
below) are bent in such a way that they pass through the focal point. In the case of a convex
mirror (see figure on the right), the rays are reflected in such a way that they appear to emerge
from the focal point.
IV.1 Spherical mirror focal point
R > 0
R < 0
C
f
f
C
Parallel rays pass reflect from mirror,
pass through focal point.
Parallel rays pass reflect from mirror,
appear to pass through focal point.
A spherical mirror has a single focal point, with focal length given by f = 1 2
R, where R is the
radius of curvature of the mirror. Assume rays of light are incident from the left. Then R is
positive when the mirror is concave to the left (above left) and negative when concave to the
right (above right). The focal point is always on the concave side, on the same side as the center
of the mirror. Light rays incident along the optic axis either reflect and pass through the focal
point (above left) or reflect and appear to pass through the focal point (above right).
Example 1: One makes a spherical mirror with a radius of curvature of 50 cm. The focal length is
f = 50 /2 = 25 cm. The focal point is located on the inside of the curve.
Example 2: A plane mirror has an infinite radius of curvature. Why? Because of this fact the
focal length is infinity.
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IV.2 Lens focal point
f
f
A lens always has two focal points, each located a distance f from the lens center, where f is the
focal length. Incident rays moving parallel to the optic axis are refracted in such a way that they
pass through the focal point. A typical lens is shown above.
IV.3 Lens maker’s equation
If you are given a lens made out of material of a known index of refraction and ground to a
certain shape, then it is possible to calculate the focal length. The formula below is known as the
“lens-maker’s equation”:
1 ⎛ 1 1 ⎞
= ( n − 1)
⎜ + ⎟
f ⎝r1 r2
⎠
where n is the index of refraction of the lens material and r 1 and r 2 are the radii of curvature of
the front and back face of the lens, respectively. The radius is positive when concave to the
center and negative otherwise. Thus in the figure above r 1 and r 2 are both positive. Note that
turning a lens around 180 degrees has no effect on its focal length. Why?
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Example: A lens is made from a glass having index of refraction 1.5. The front end is flat and the
back end is convex outward with radius of curvature 40 cm (see figure below). Find the focal
length.
The front end is flat, so its radius of curvature is infinite. The back end has a positive radius of
curvature because it is concave to the center. Thus, r 1 =∞and r 2 = 40 cm. The focal length is
then
1 ⎛ 1 1 ⎞ 1
= ( 1.5 − 1)
⎜ + ⎟=
f ⎝∞
40 ⎠ 80
or f = 80 cm.
V
Finding images
V.1 Lens equation
The lens equation is the basic formula relating object distance, image distance and focal length.
The equation applies equally to mirrors or lenses. All distances are measured from the center of
the mirror or lens
1 1 1
+ =
p q f
This equation is accurate in the paraxial approximation, that is, when all angles relative to the
optical axis are small.
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Example: Imagine a mirror with a focal length of 20 cm and an object at 30 cm from the mirror.
Then the image location is found from
1 1 1 1 1 1
= − = − =
q f p 20 30 60
or q = 60 cm from the mirror. As we will see below, the fact that q is positive means that the
image is in the direction of the outgoing rays, on the same side as the object.
V.2 Sign convention for distances
It’s best to be consistent about signs. Orient the diagram so that the object is to the left of the
lens/mirror, as in the figures below. Then the object distance and image distance are measured as
follows.
Object: p > 0 when object on same side as incoming rays.
Image: q > 0 when image on same side as outgoing rays.
Example: Consider a mirror with radius of curvature 100 cm with its concave side facing an
object 60 cm away. Find the position of the image.
The focal length is f = +100/2 = 50 cm. The image distance is found from the lens equation
1 1 1
+ =
60 q 50
or q = +300 cm. The image is on the same side of the mirror as the object, but much further away
from the mirror.
V.3 Real and virtual images
An image is real when it is located where real light rays intersect, i.e., one can project it on a
sheet of paper. This corresponds to q > 0. An image is virtual when the image is located where
light rays cannot propagate. A virtual image has q < 0. The images below are both real.
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Object
p
p
Real image
C
f
f
f
Real Image
q
q
Example: In the previous example, the object distance was p = 60 cm and the image distance was
q = 300 cm. Both the object and image are therefore real.
V.4 Magnification
The magnification of a lens or mirror is defined by m= h′ / h, where h ′ is the size of the image
and h is the size of the object. Each can be positive or negative. For a lens or mirror the
magnification is
m =− q/
p
Thus for real images the image is inverted while for virtual images it is upright.
Example: Refer to the previous example again. The object distance from the mirror (f = 50 cm)
was p = 60 cm and the image distance was found to be q = 300 cm. This gives a magnification of
m = –q / p or m = –5. The image is thus inverted and 5 times as large as the original object.
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V.5 Principal ray diagrams
An image can be constructed from an object graphically by drawing three rays from the object to
the mirror or lens. Principal rays are a terrific way of checking the location of the image
calculated from the lens equation. In particular, you can tell if the image is real or virtual and
approximate position, size and orientation of the image. The three rays for a mirror and lens are
shown below
V.5.a Mirror
1. Ray from object moves parallel to optic axis, strikes mirror and passes through focus.
2. Ray from object passes though focus, strikes mirror, passes back parallel to axis.
3. Ray passes from object through center, strikes mirror, bounces back on itself.
These rays are shown below for a concave (left) and convex (right) mirror. Note that for convex
mirrors the focus and center are on the opposite side.
1
1
2
3
2
1
Virtual rays
2
C
f
f
C
3
Real image
3
Virtual image
9

V.5.b Lens
1. Ray from object moves parallel to optic axis, through lens, passes through focus.
2. Ray from object passes though focus, through lens, moves parallel to axis.
3. Ray passes from object through center in straight line.
The rays are show below for a converging lens (f positive) and a diverging lens (f negative).
1
2
f
3
Real image
f
1
Virtual image
2
f
f
3
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Example: An 2 cm tall object is 80 cm from a mirror of 100 cm radius of curvature concave
towards the object. Locate the image using the principal 3 rays.
The focal length is f = +100 / 2 = 50 cm.
3
1
C
f
2
Measuring from the diagram gives a distance of q = 130 cm and a real image. The magnification
is about –1.6 (inverted image). The correct answer is found from the lens equation
1 1 1
+ =
80 q 50
or q = +133.3 cm. The magnification is 133.3/ 80 1.67
m =− =− . Thus the true size of the image
is 1.67 × 2 = 3.34 cm and inverted.
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