Chapter 32
OPTICAL IMAGES
32.1 Mirrors
The point P’ is called the image or the
virtual image of P (light does not
emanate from it)
The left-right reversal in the mirror is also
called the depth inversion (the front and
the back are also inverted)
1
The mirror transforms a right-handed coordinate
system
i, j , k , i × j = k
into a left-handed system with
i × j = −k
The image of an arrow. To locate the image
of any point P, one can follow two rays:
(1) perpendicular to the mirror (is reflected
onto itself)
(2) ray at some angle
with the normal
. an equal angle
to the mirror (is reflected at
θ
θ
).
The intersection of these two rays gives the
image P’. The image is at the same distance
behind the mirror as the origin in front, is
upright, and is the same size as the origin.
2
A set of mirrors provides multiple images. P1’ is
the image of P in mirror 1, P2’ is the image of P in
mirror 2, and P1,2’ is the image of P with reflections
first in mirror 1 and then in mirror 2.
A ray reflected from two perpendicular
mirrors is reflected back along a parallel
path irrespective of an angle of incidence
3
Spherical mirrors
The rays from a point source P on the axis
of concave mirror converge in the (real)
image P’ (the light actually emanate from
this point!) and then continue as if it is a real
point source
Spherical aberration. Only the rays that
are close to the axis (paraxial rays)
converge in the image point; the faraway
rays blur the image. To sharpen the
image one must block non-central parts of
the mirror.
4
Finding location of the image. The
ray through the center of the sphere
is reflected onto itself. The ray
striking at point A at angle θ with
with the normal is reflected at angle θ
with the normal. The geometry dictates:
β = α + θ , γ = α + 2θ ⇒ 2β = α + γ ,
α ≈ l / s, β ≈ l / r , γ ≈ l / s ' ⇒ 1s + s1' = 2r
This equation relates the distance to the image to the distance to the
object and the radius of the sphere
Solving this equation for s’ yields
s' =
sr
2 s −r
≡
r
2− r / s
5
Distance of the image from the axis.
Two right triangles in the Figure are
similar. Thus
y '/ y = − s '/ s ⇒ y ' = − y ( s '/ s )
(the negative sign means that P and P’ are on the opposite sides of the axis)
When the object is far away from the mirror, s >> r, 1/s << 2/r and the image is
at the point
s' ≈ r /2
which is called the focal point F of the mirror ( y ' = − y ( s '/ s ) → 0 ).
Using the distance to the focal point f = r/2 instead of r, one gets the mirror
equation
1/ s + 1/ s ' = 1/ f
6
Parallel rays striking a concave mirror
are reflected to a point on the focal
plane. In reverse, if a source is on the
focal plane the reflected rays are parallel
to each other. This illustrates the
principle of reversibility of reflection
(and refraction)
Example 1. An object is 24 cm from concave mirror of radius 16 cm, 2 cm
above the axis. Find the position of the image and the focal length of the mirror.
The focal length of the mirror f = r/2 = 8 cm. Using the mirror equation, one gets:
1/ s ' = 1/ f − 1/ s = (1/ 8 − 1/ 24) cm −1 = 0.0833 cm −1
s ' = 12 cm;
y ' = − y ( s '/ s ) = −2 ⋅ (12 / 24 ) cm = −1 cm
7
Ray diagrams for mirrors
Three principal rays:
(1) Ray parallel to the axis after
reflection goes through the focal and
image point
(2) Ray going through the focal
point, after reflection goes parallel to
the axis through the image point
(3) Ray going through the center C is
normal to the mirror and, after
reflection, reverses itself and goes
through the image point.
The image point lies at the intersection of these three rays. Any two of
them can be used in construction for finding the image point.
8
Typical configurations
Central point is between the object and the
mirror: The image is between the center
and the focal point, is reversed, and is
smaller than the object
Object is between the central point and the
mirror: The image is behind the mirror, is
direct (not reversed), and is bigger than the
object (here s’ is negative)
9
Sign conventions for the mirror equation
(1) s is positive if the object is on the incident-light side of the mirror
(2) s’ is positive if the object is on the reflected-light side of the mirror
(3) r (or f) is positive if the mirror is concave (center of curvature is on the
reflected-light side of the mirror)
The lateral magnification m is the ratio of the image size to the object size:
m = y '/ y ≡ − s '/ s
(m is negative if the image is inverted).
10
Convex mirrors
are described by the same mirror equation with the proper sign conventions
Ray diagram for a convex mirror. In onstruction,
one should use the same (three) principal rays
Example 2. An object 1.5 cm high is 24 cm away from a convex mirror of
radius 12 cm. Find the image.
The focal distance is –12/2 cm = - 6 cm. The mirror equation yields:
1/ s ' = 1/ f − 1/ s = ( −1/ 6 − 1/ 24 ) cm −1 = −0.2083 cm −1 ,
s ' = −4.8 cm;
y ' = − y ( s '/ s ) = −1.5 ⋅ ( −4.8 / 24 ) cm = 0.3 cm,
m = − s '/ s = − ( −4.8 / 24 ) = 0.2
The image is behind the mirror (virtual image; s’ < 0), is upright (y’, m > 0), and
is smaller than the object (|m| < 1).
11
32.2 LENSES
Images formed by refraction
The ray construction for cylinder
with spherical surface made of
material with refraction index n2.
Light comes from the medium with
refraction index n1. The angles are
related by Snell’s law. For paraxial
refraction, the angles are small.
The distance to the image is
n1
n2
s
s'
n2 − n1
r
In refraction, the real images are formed in back of the surface (refracted-light
side), the virtual images – in front of the surface.
+
=
Rules
(1) s is positive for objects on the incident side
(2) s’ is positive for images on refracted-light side
(3) r is positive if the center is on the refracted-light side
12
Example 3. Derive an expression for magnification m = y’/y by a spherical
refracting surface.
Snell’s law in a small angle approximation has a
form
n1 sin θ1 = n2 sin θ 2 ⇒ n1θ1 = n2θ 2
From geometry,
θ1 ≈ tan θ1 = sy ; θ 2 ≈ tan θ 2 = − sy''
Combining,
m=
y'
y
=−
s 'θ 2
sθ1
=−
s ' n1
sn2
Example 4. An object is 3 m below the flat surface of water. Where is the image?
In equation
n1
s
+ ns2' =
n2 − n1
r
n1 = 1.33 (water), n2 = 1 (air), s = 3 m, r is infinite
(flat surface). Thus, n1 ≡ 1 = − 1.33 , s ' = − s /1.33 = −2.5 m
s'
s'
s
The negative sign means that the image is virtual – on the opposite side from
the refracted-light side (under water). The magnification is
m = − nn12ss' =
1.33⋅ 2.5
3
=1
which is not surprising (flat surface!)
13
Example 5. The object is in the air, 20 cm from a spherical container with water
(r = 12 cm). Where is the image and what is the magnification?
The main equation is again
n1
s
+
n2
s'
=
n2 − n1
r
n1 = 1.0 (air), n2 = 1.33 (water), s = 0.2 m, r = 0.12 m. Then
1.33
s'
−1
−1
−1.0
= ( 1.330.12
− 1.0
m
=
−
2.25
m
,
0.2 )
s ' = − 1.33
2.25 m = −0.59 m
meaning that the image is in the air, in front of the sphere.
The magnification is,
m=
y'
y
=−
s ' n1
sn2
=
0.59
0.2 i1.33
= 2.22
14
Thin Lenses
A lens has two curved surfaces. The refraction at each surface should be
considered separately.
Example 6. Consider a thin lens of
index of refraction n in the air. The
curvatures are r1 and r2, the object is
at distance s from the (first surface
of the) lens. Find the image.
The distance s1’ of the image due to the refraction at the first surface is
n
n −1
1
s
r1
s1'
The first surface image becomes the object for the second surface. This light is
refracted at the second surface with the “object” being, in this case, at the
negative distance s1’ from it (“virtual object” to the left of the second surface):
+ =
n
− s1'
+ s1' = 1r−2n
Adding these equations eliminates s1’,
15
1
s
+ s1' =
n −1
r1
+ 1r−2n ≡ 1f ,
1
f
= ( n − 1)  r11 − r12 
This is the thin-lens equation.
The equation for the focal length f in terms of the radii of the surfaces is called
the lens-maker’s equation. The focal length of a thin lens gives the image
distance when the object distance is infinite.
For a lens in this figure, r1 is positive and r2 is
negative, so f is positive (a double convex or
converging lens or positive lens).
In general, the image distance s’ is positive when the
image is on the refracted-light side of the lens
16
This figure shows a double concave or
diverging lens (a negative lens) the plane
wave diverges after refraction. The focal
distance is negative. Any lens that is thinner in
the middle is negative.
Example 7. Find the focal length of this
double-convex lens.
1
f
= ( n − 1)  r11 − r12  = (1.5 − 1) 101 − −115  cm1 = 121cm
17
Each lens has two focal points: convergence
(or divergence) points for light coming from
the right or left. Both points are at the same
distance, but on the opposite sides of the lens
defined as the first and second focal
points depending on direction of the incident
light. For a positive lens, the first focal point is
on the incident-light side.
If parallel rays are incident at a small
angle with the axis, the light is focused at
a point in the focal plane a distance f
from the lens.
The reciprocal of the focal length is called the power of lens, P = 1/f. When f is
in m, the power is measured in reciprocal meters, m-1, which are called
18
diopters [D]
Example 8. Find the focal length
and the power of this lens.
Here both are r1 and r2 are positive:
P=
1
f
1
1
= ( n − 1)  r11 − r12  = (1.5 − 1)  0.1
− 0.13

= 1.15 D;
f = 1/ P = 0.87 m = 87 cm
In the lab, it easier to measure the focal length f than measure r1 and r2 and
then calculate f.
19
Ray diagram
In construction, it is easier to follow three principal rays which converge to the
image point:
(1) The central ray (undeflected)
(2) The focal ray – emerges parallel to the axis
(3) The parallel ray – the emerging ray is directed toward (or away from) the
second focal point.
The ray diagram for a negative lens is here:
20
The lateral magnification of the lens is
m = y '/ y = − s '/ s
Negative magnification means that the image is inverted.
Example 9. An object h = 0.8 cm high is placed 6 cm from a double convex
lens with f = 8 cm. Find the image graphically and algebraically, find
magnification, and determine whether the image is real or virtual.
The object is between the
focus and the lens. The
diagram looks like this.
1
6 cm
1
1
1
+ s1' = 8 cm
, s1' = 8 cm
− 6 cm
= −0.042 cm−1 ,
s ' = −24 cm; m = − s '/ s = 4
The height of the image h’ = mh = 4 x 0.8 cm = 3.2 cm
21
Combination of lenses
Example 10. The object is 4 cm from a double convex lens with f = 12 cm. The
second lens with f = 6 cm is placed 12 cm to the right of the first. Find the final
image.
By accident (or by design?) the
second lens is placed into the
focus of the first Thus, the
parallel ray of the first lens
is the central ray for the
second. The focal ray of the
first lens is always the parallel
ray of the second. The
construction is simple.
The object for the second lens is the (virtual) image of the first located at
1
4 cm
+ s1' = 121cm ,
1
s'
1
1
= 121cm − 4cm
= − 6cm
, s ' = −6 cm; m = −s '/ s = 1.5
This image / object is s2 = 6 cm + 12 cm = 18 cm away from the second lens,
1
18 cm
+ s1' =
2
1
6 cm
,
1
s2 '
=
1
6 cm
− 181cm =
1
9 cm
, s2 ' = 9 cm; m2 = − s2 '/ s2 = −0.5
22
Example 11. Two lenses of focal length 15 cm each are 22.5 cm apart. Find
the image of an object 22.5 cm from one of the lenses.
The first construction yields the distance
to the “first image”:
1
22.5 cm
1
s'
+ s1' = 151cm ,
= 151cm − 22.51 cm =
1
45 cm
,
s ' = 45 cm;
m = − s '/ s = −2
The rays for the second construction are
continuation of the first. The final image
is at
1
( −45+ 22.5) cm
+ s12 ' = 151cm ,
1
s2 '
= 151cm + 22.51 cm =
1
9 cm
,
s2 ' = 9 cm; m2 = − s2 '/ s2 = 0.4; m = m1 ⋅ m2 = −0.8
23
The final image is inverted and real (see the diagram).
Compound lenses
When two thin lenses are compounded (brought close together), the effective
focus length is
1/ f eff = 1/ f1 + 1/ f 2 , Peff = P1 + P2
Proof. For the first lens,
1
s
+ s1' = 1f1 .
For the second one, the object distance is the negative of the first image
distance,
1
−s'
+ s12 =
1
f2
.
Adding these two equations, we get
( 1s + s1' ) + ( −1s ' + s1 ) = ( 1s + s1 ) = 1f + f1
2
2
1
2
=
1
f eff
.
24
Aberrations
Blurring that results from
rays not focused at a
single focus point is called
aberration. The aberration
is usually caused by large
values of angles between the rays and the axis. The circle of least confusion
is at point C where the diameter of the image of a point object is minimum. This
type of aberration in a thin lens is called spherical aberration.
25
32.4 Optical Instruments
The Eye
The light is focused by the cornea, with
assistance from the lens, on the retina. When
the object is far away, system focal length is
maximum, about 2.5 cm (distance from the
cornea to retina). For closer objects, the
curvature of the lens increase thus decreasing
the focal length so that image is still focused
on the retina – accommodation. In the lens
equation for the eye
1
s
+ =
1
s'
1
f
the distance to the image s’ is fixed and changing s requires changing f. The
closest point for which the lens can still focus the image is called the near
point. This distance changes with age and from person to person; the
“standard” value is defined as 25 cm.
26
Farsightedness
Nearsightedness
The size of the image is
y ' = − ys '/ s
Since s’ = 2.5 cm and does not change,
the bigger the distance to the object s,
the smaller the image.
27
Example 12. The near-point distance of a person is 75 cm. The reading
glasses, that are very close to the eye (a compound lens) correct it to 25 cm.
That is, the object put 25 cm in front of eye glasses forms a virtual image 75 cm
from the eye. Describe the glasses.
The lens equation for the glasses is
1
s
+ s1' =
1
f
=
1
25 cm
+ −751cm =
2
75 cm
; f = 37.5 cm = 0.375 m;
P = 2.67 D; m = −75 / − 25 = 3
Such glasses not only change the near point, but also magnify the object (the
fact well-known to everyone who has them)!
A simple magnifier – a converging lens put close to an eye if the object is
28
closer to the lens than its focal length.
The (compound) microscope
is used to look at very small objects at small distances.
The lens close to the object –
objective – forms a real image.
The eyepiece is a simple
magnifier. The eyepiece is placed
so that the image formed by the
objective is in its first focal plane
(viewing the image at infinity). The
distance between the second focus
of the objective and the first focus
of the eyepiece is the tube length
(about 16 cm).
The magnification of the objective is
tan β = y / f o = − y '/ L ⇒ mo = y '/ y = − L / f o
29
The magnification of the eyepiece (simple magnifier) is
θ o = y / xnp , θ = y / f e ,
M e = θ / θ o = xnp / f e
where xnp is the near point distance.
The overall magnification is
M = m0 M e = − Lxnp / f o f e
30
Example 13. The focal lengths of an objective and an eyepiece are 1.0 and 1.8
cm. The near point of the viewer is 25 cm. Find the magnifying power and the
position of the object if the final image is viewed at infinity. The distance
between the lenses is 22 cm.
The tube length is the separation
between the lenses minus the focal
distances,
L = (22 – 1.0 - 1.8) cm = 19.2 cm
The magnification is
M = m0 M e = − Lxnp / f o f e
= −(19.2 ⋅ 25) / (1.0 ⋅1.8 ) = −267
For the final image to be viewed at infinity, the image for the objective should be
in the focal plane of the eyepiece. Then, from the figure, the image distance for
the objective is s ' = f 0 + L = 1.0 cm + 19.2 cm = 20.2 cm
The lens equation for the objective
object:
1/ s + 1/ s ' = 1/ f o gives the position of the
1/ s = 1/ (1.0 cm ) − 1/ ( 20.2 cm ) = 0.95 cm −1 ; s = 1.05 cm
31
Telescope
is used to view large objects very far away by creating a smaller, but much
closer real image of the object.
The purpose of the objective is not to magnify, but to produce the image that is
close to us. This image is in the focal plane of the eyepiece so that it is viewed
at infinity. Since this image is in the second focal plane of the objective (the
object is at infinity!), the separation between the lenses is f o + f e
The magnifying power is the angular magnification M = θ e / θ o
According to the figure,
tan θ 0 = sy =
− y'
fo
≈ θ o ; tan θ e =
M = θe / θ0 = − fo / fe
y'
fe
≈ θe ;
32
Example 14. The distance between the lenses is 40 cm. The focal length of the
eyepiece is 4 cm. Find the magnifying power.
The distance between the lenses is the sum of focal lengths. Therefore,
f o = 40 cm − 4 cm = 36 cm
and the magnifying power is
M = − f o / f e = −36 / 4 = 9
.
33
Review of Chapter 32
An image is real if actual light rays converge to the image point (in front of the
mirror or on refracted-light side of a refractor). An image is virtual if only
extensions of the actual rays converge to it.
An object is real if actual light rays diverge from it (actual object or a real
image). An object is virtual if only extensions of the actual rays diverge from it.
Mirror equation
1/ s + 1/ s ' = 1/ f ;
f = r/2
where the focal length is the image distance when the object is at infinity
Lateral magnification M = y '/ y = − s '/ s
Images can be easily located using the ray diagram for three principal rays:
the parallel ray, the central (radial) ray, and the focal ray.
Sign conventions: s is positive if the object is on the incident side of the mirror;
s’ is positive if the image is on the reflected-light side; r and f are positive if the
mirror is concave (the center is on the reflected-light side).
34
Refraction. At a single surface
n1
s
+ ns2' =
n2 − n1
r
, m = y '/ y = −n1s '/ n2 s
Sign conventions: s is positive if the object is on the incident side; s’ is positive
if the image is on the refracted-light side; r is positive if the center is on the
refracted-light side.
Thin lenses. Thin lens and lens-maker’s equations
1
s
+ s1' =
n −1
r1
+ 1r−2n ≡ 1f ,
1
f
= ( n − 1)  r11 − r12 
Incident rays parallel to the axis emerge directed towards or away from the first
focal point; incident rays directed towards or away from the second focal
point emerge parallel with the axis.
Power and magnification
P = 1/ f ; m = y '/ y = − s '/ s
Images can be easily located using the ray diagram for two of three principal
rays: the parallel ray, the central (radial) ray, and the focal ray.
Sign conventions: s > 0 if the object is on the incident side; s’ > 0 if the image is
on the refracted-light side; f > 0 for converging lens, f < 0 for diverging lens.
35
The Eye. Light is should be focused on the retina, about 2.5 cm from the
cornea. F for the lens changes depending on the distance from the object; the
closest distance for which the focusing is still possible – the near point, about
25 cm. The closer the object, the larger is the image on the retina.
The simple magnifier. A lens with a focus length that is smaller than the
near point, M = θ / θ = x / f
e
o
np
The compound microscope has two lenses – the objective and the eyepiece.
The object is outside the focal point of the objective with an image in the focal
plane of the eyepiece. The eyepiece is a simple magnifier.
The magnifying power
M = m0 M e = − Lxnp / f o f e
where the tube length is the distance between the second focal point of the
objective and the first of the eyepieces.
The telescope also has two lenses which are separated by a distance equal to
sum of their focal lengths. The objective creates a smaller image than the
object but at a much closer distance. The eyepiece is a simple magnifier.
The magnifying power
M = θ e / θ 0 = − fo / fe
36