CHAPTER 32

OPTICAL IMAGES

• Introduction to ray diagrams

• Reflection from mirrors

• Plane mirrors• Spherical mirrors• Sign convention

• Lenses

• Converging Lenses• Diverging Lenses• Chromatic aberration

Introduction to ray diagrams:

Rays emerge from all points on an object. Only some of those from the top of the head are shown here. Note: the image does not really exist ... you cannot display it on a screen, for example! It is called a virtual image.

** Note that only 2 rays are required to locate the image position **

object image

*

*

Geometry of image formation (plane mirror):

ΔOAB and ΔOPQ are congruent triangles,i.e., OA = OP and AB = PQ

Lateral magnification m =

PQAB

( = 1).

Also, object distance (OA) = image distance (OP)

i.e., s = ′ s .Image is virtual and upright.

OA

B

P

Qs

1

2

′ s

θ

θ

θ

θθ

Question 32.1: You are 1.62 m tall and want to be able to see your full image in a vertical plane mirror.

(a) What is the minimum height of the mirror that will meet your needs? (b) How far above the floor should the bottom of the mirror in (a) be placed, assuming that the top of your head is 14 cm above your eye level? (c) If you move closer to the mirror, describe what you would see in the mirror.

Use ray diagrams to explain your answers.

A ray diagram showing rays from your feet and the top of your head reaching your eyes is shown above. (a) Therefore, the mirror must be half your height, i.e.,

0.81 m.

(b) The top of the minimum height mirror must be halfway between your eyes and the top of your head, i.e., 1.55 m from the floor. Therefore, the bottom of the mirror should be

1.55 m − 0.81 m = 0.74 m,above the floor.

1.62 m

(c) You move closer to the mirror. A ray diagram showing rays from your feet and the top of your head reaching your eyes is shown above, with the same size mirror. Notice you can still see all your body; it makes no difference where you stand!

Reflection in a plane mirror produces:lateral inversion

x

y

z

RH set of axes(object)

LH set of axes(image)

′ x ′ y

′ z

Formation of an image with two plane mirrors.

The rays from the object into the eye satisfy the law of

reflection. We see that the image at P1 due to mirror 1

acts as an object for mirror 2 . The final image appears

at P2. Note that both images are virtual.

IMPORTANT CONCEPT: Even though the image at

P1 is virtual, it can act as an object for mirror 2 .

1

2

P1 P2

Even though a virtual image does not actually exist, you can see it and photogragh it because it can act as an object for another lens or mirror and produce a real image.

PS: Ever done a selfie in a mirror?

Question 32.2: When two plane mirrors are parallel, such as on opposite walls in a hairdressing salon, multiple images arise because each image in one mirror serves as an object for the other mirror. An object is placed between parallel mirrors separated by 3.00 m. The object is 1.00 m in front of the left mirror and 2.00 m in front of the right mirror.

(a) What is the distance from the left mirror to the first four images in that mirror?

(b) What is the distance from the right mirror to the first four images in that mirror?

(a) For the left hand mirror (red) the images are 1.0 m; 5.0 m; 7.0 m; 11.0 m

behind the mirror.

(b) For the right hand mirror (blue) the images are2.0 m; 4.0 m; 8.0 m; 10.0 m

behind the mirror.

1.0 m

5.0 m

7.0 m

2.0 m

4.0 m

10.0 m 11.0 m

8.0 m

3.0 m

See the notes I wrote on

“retroreflection”

available on the useful notes link on the web-site.

REFLECTION FROM CURVED SURFACES

1. Concave mirror

Using simple geometry we have:β = α + θ and γ = α + 2θ.

Eliminate θ and we get: α + γ = 2βIf α, β and γ are small then:

α ≈ℓs

β ≈ℓr

γ ≈ℓ′ s

i.e., 1s

+1′ s

=2r

•α β γφ

θθ

ℓ

1

2

s ′ s

r image object C

opticalaxis

opticalaxis

•

s ′ s

r image object

NOTE: The image produced in this example is real; it actually exists, i.e., we could see it on a screen.

My convention is that if the rays forming the image are solid lines, it is a real image. Dashed lines are rays that do not actually exist ... they show paths that rays appear to follow, for example, behind a mirror.

1s

+1′ s

=2r

If s = ∞ then ′ s =

r2

= f ,

where F is the principal focus and f is the focal length.

∴

1s

+1′ s

=1f

.

Note, it is a real focus.

Can you think of an example?

•

s = ∞ ′ s

Beam parallel to the optical axis (i.e., from a very distant object)

• F C

To produce a parallel beam of light put the source at the principal focus.

Can you think of an example?

•

′ s = ∞ s = f

C

r

• Object distance ( s): if the object is on the same side of the reflecting or refracting surface as the incoming light, s > 0.

• Image distance ( ′ s ): if the image is on the same side of the reflecting or refracting surface as the outgoing light, ′ s > 0.

• Radius of curvature (r): if the center of curvature is on the same side as the outgoing light, r > 0.

Tee, hee, hee ...there’s a sign conventionfor mirrors and lenses!

Image formation in a concave mirror

I. Object distance: s > r

s > 0 ′ s > 0 r > 0The lateral (or linear) magnification is defined as

m = ′ y

y .

But y

s = − ′ y ′ s . ∴m = − ′ s

s (⇒ negative).

Definition of magnification.

Note in this case that m < 0 and m < 1. The image is real, inverted and reduced.

• image

object

•

1

2

C F

s

′ s

y

′ y

Image formation in a concave mirror

II. Object distance: r2 < s < r (i.e., f < s < r)

s > 0 ′ s > 0 r > 0The lateral (or linear) magnification is

m = ′ y

y = − ′ s s (⇒ negative).

Note that m < 0 and m > 1. The image is real, inverted and enlarged.

•

image

object

1

2

C F

s

′ s

y

′ y •

Image formation in a concave mirror

III. Object distance: s < f (i.e., s < r2)

s > 0 ′ s < 0 r > 0The lateral (or linear) magnification is

m = ′ y

y = ′ s s (⇒ positive).

Note that m > 1. The image is virtual, upright and enlarged.

•

image

object

1 2

C F

s

′ s

′ y

• y

• When s > r: then m < 0 and m < 1. The image is real, inverted and reduced.

• When f < s < r: then m < 0 and m > 1. The image is real, inverted and enlarged.

• When s < f : then m > 1. The image is virtual, upright and enlarged.

positive m ( m > 0) ⇔ virtual image (upright)negative m ( m < 0) ⇔ real image (inverted)

Here’s a summary!For a concave mirror ...

f > 0

m = − ′ s s .

CF

r

Question 32.3: A concave mirror has a radius of curvature of 24.0 cm. Find the image position for objects that are placed

(a) 55.0 cm, (b) 24.0 cm, (c) 12.0 cm, and (d) 8.00 cm from the mirror.

For each case give the magnification and state whether the image is real or virtual; upright or inverted.

Given r = 0.24 m. Since the mirror is concave then r and

f ( =r2 = 0.12 m) are both positive.

(a) For s = 0.55 m and using 1s

+1′ s

=1f

, we have:

1′ s

=1

0.12 m−

10.55 m

= 6.52 m−1, i.e., ′ s = 0.15 m.

∴m = − ′ s s = −0.15 m

0.55 m = −0.27.

Since ′ s > 0, m < 0 and m < 1, the image is on the same side as the object, it is real, inverted and reduced.

(b) For s = 0.24 m:

1′ s

=1

0.12 m−

10.24 m

= 0.417 m−1, i.e., ′ s = 0.24 m.

∴m = − ′ s s = −0.24 m

0.24 m = −1.00.

Since ′ s > 0, m < 0 and m = 1, the image is on the same side as the object, it is real, inverted and the same size as the object.

(c) s = 0.12 m:

1′ s

=1

0.12 m−

10.12 m

= 0, i.e., ′ s = −

10

= ∞.

So, the emerging rays are parallel and do not form an image.

(d) s = 0.08m:

1′ s

=1

0.12 m−

10.08 m

= −4.17 m−1, i.e., ′ s = −0.24 m.

∴m = − ′ s s = 0.24 m

0.08 m = 3.0.

Since ′ s < 0, m > 0 and m > 1, the image is on the opposite side to the object, it is virtual, upright and enlarged.

2. Convex mirror:

Using simple geometry:θ = α + β and 2θ = α + γ

Eliminate θ and we get: γ − α = 2β

i.e., ℓ′ s −ℓs

=2ℓr

so, 1′ s −

1s

=2r.

But, using the sign convention we find:

s > 0 ′ s < 0 r < 0,

∴

1′ s

+1s

=2r,

which is the same result obtained for a concave mirror!

• C rβγ

θ

1

2

θ

′ s

α

s

θ

image object

• C rβγ

θ

1

2

θ

′ s

α

s

θ

image object

1′ s

+1s

=2r.

When s = ∞, i.e., an incoming parallel beam of light, we

have: 1′ s

=2r

=1f

i.e., f =

r2

(virtual focus),

so 1′ s

+1s

=1f

.

** Note that by the sign convention, r < 0, and so

f < 0 for a convex lens **

Image formation in a convex mirror

By simple geometry the lateral magnification of a convex mirror is:

m = ′ y

y = − ′ s s ,

but ′ s < 0, so m > 0.

Note also, 0 < m < 1 always and so the image is upright, virtual and reduced in size.

Can you think of an example ?

• C

1

2 ′ s

s

object • F image

Question 32.4: A convex mirror has a radius of curvature of 24.0 cm. Find the image position for objects that are placed

(a) 55.0 cm, (b) 24.0 cm, (c) 12.0 cm, and (d) 8.00 cm from the mirror.

For each case give the magnification and state whether the image is real or virtual; upright or inverted.

Given r = 0.24 m. Since the mirror is convex then r and

f ( =r2 = 0.12 m) are negative.

(a) s = 0.55 m:

1′ s

= −1

0.12 m−

10.55 m

= −10.2 m−1,

i.e., ′ s = −0.10 m.

∴m = − ′ s s = 0.10 m

0.55 m = +0.18.

Since ′ s < 0, m > 0 and m < 1, the image is on the opposite side as the object, it is virtual, upright and reduced.

(b) s = 0.24 m:

1′ s

= −1

0.12 m−

10.24 m

= −12.5 m−1,

i.e., ′ s = −0.08 m.

∴m = − ′ s s = 0.08 m

0.24 m = +0.33.

Since ′ s < 0, m > 0 and m < 1, the image is on the opposite side as the object, it is virtual, upright and reduced.

(c) s = 0.12 m:

1′ s

= −1

0.12 m−

10.12 m

= −16.7 m−1,

i.e., ′ s = −0.06 m.

∴m = − ′ s s = 0.06 m

0.12 m = +0.50.

Since ′ s < 0, m > 0 and m < 1, the image is on the opposite as the object, it is virtual, upright and reduced.

(d) s = 0.08 m:

1′ s

= −1

0.12 m−

10.08 m

= −20.83 m−1,

i.e., ′ s = −0.048 m.

∴m = − ′ s s = 0.048 m

0.08 m = +0.60.

Since ′ s < 0, m > 0 and m > 1, the image is on the opposite side to the object, it is virtual, upright and reduced.

Question 32.5: A dentist wants a small mirror that will produce an upright image that has a magnification of 5.50 when the mirror is located 2.10 cm from a tooth.

(a) Should the mirror be concave or convex? (b) What should the radius of curvature of the

mirror be?

(b) The mirror cannot be convex because a convex mirror always produces a diminished virtual image. Therefore, it must be concave.

(a) s = 2.10 cm. The image is upright and so it is virtual, i.e., behind

the mirror, so ′ s < 0.

The magnification: m = − ′ s s = 5.50

∴ ′ s = −5.50 × 2.10 cm = −11.6 cm.

∴

1s

+1′ s

=1f

=2r

=1

2.10 cm−

111.6 cm

= 0.39 cm−1

i.e., r =

20.39 cm−1 = 5.13 cm.

∴f = r2 = 2.56 cm,

so s < f .

s ′ s

See the notes on the web-site for chapter 32 on how to solve mirror problems using a scale drawing ...

Types of Lenses ...

Cross-sections of various types of lenses. The upper group are converging lenses, the lower group are diverging lenses.

Planoconvex Biconvex** Positive Meniscus

ConvergingLenses

Planoconcave Biconcave** Negative Meniscus

DivergingLenses

Diverging lens(concave)

Converging lens (convex)

Focal length(f)

Parallelbeam

Real focus

Principalfocus

Convex lens ...

5.0 cm

Focal length(f)

Parallelbeam

Real focus

Principalfocus

Principalfocus

Parallelbeam

Virtual focus

Focal length(f)

Lens-maker’s formula:

Optometrists who prescribe lenses and opticians who make them do not normally specify the focal length of a lens but its refractive power, in units called diopters.

Refractive power =

1f (in meters)

(diopters)

Converging lenses have f > 0 but diverging lenses have f < 0.

1f

=n2n1

−1⎛

⎝ ⎜

⎞

⎠ ⎟

1r1−

1r2

⎛

⎝ ⎜

⎞

⎠ ⎟

n2 n1

r1

r2

Sign convention: If the center of curvature is on the same side as the outgoing light r > 0 .

Incoming light Outgoing light

r2

r1

r1 > 0 r2 < 0

• •

Incoming light Outgoing light r2

r1 r1 < 0 r2 > 0

••

Incoming light Outgoing light r1

r2

r1 > 0 r2 > 0

••

Question 32.6: Place the following in order of increasing focal length, i.e., from shortest to longest:

r1

1

r1

r1

2

r1

r2 r2 > r1

3

r1 > 0 both positive both positive

r2 = ∞

1:

1f1

= (n −1) 1r1

. ∴f1 =

r1(n −1)

.

2:

1f2

= (n −1) 1r1−

1r1

⎛

⎝ ⎜

⎞

⎠ ⎟ = 0. ∴f2 = ∞.

3:

1f3

= (n −1) 1r1−

1r2

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = (n −1) r2 − r1

r1r2

⎛ ⎝ ⎜ ⎞

⎠ ⎟ .

∴f3 =

r1(n −1)

r2(r2 − r1)

.

∴f1 < f3 < f2.

** See also lens-maker’s formula under “useful notes” **

> 1

r1

1

r1

r1

2

r1

r2 r2 > r1

3

Note, if n1 = 1 (air), the lens maker’s formula becomes:

1f

= (n2 −1) 1r1−

1r2

⎛

⎝ ⎜

⎞

⎠ ⎟ .

But n2 varies with wavelength (λ) ... dispersion.

Therefore, the focal length, f, varies with light color. Since, for glass, n2 increases for decreasing λ (i.e., from

red light to blue light ), the focal length for blue light is less than for red light. So, blue light is “bent” more than red light.

** See the articles on the web-site I wrote about aberration **

blurred focus“chromatic aberration”

Question 32.7: Find the focal length of a lens that has an index of refraction of 1.62, a convex surface with a radius of curvature of r1 = 0.40 m and a concave

surface with a radius of curvature of r2 = 1.00 m.

r1

r2

• •

Using the sign convention r1 > 0 and r2 > 0. We have

1f

=n2n1

−1⎛ ⎝ ⎜ ⎞

⎠ ⎟ 1

r1−

1r2

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ,

where n2 = 1.62 and n1 = 1.00.

∴

1f

= 1.62 −1( ) 10.40 m

−1

1.00 m⎛ ⎝

⎞ ⎠ = 0.93 m−1,

i.e., f = 1.08 m.

r1

r2

Incoming light Outgoing light

••

For a convex lens we have (see “useful notes”):

1s

+1′ s

=1f

and m = ′ y

y = − ′ s s .

(Also, by substitution: m = − ′ s − f

f= −

fs − f

.)

If s > 2f then m < 1, i.e., an inverted and reduced image.

Object Image

′ y

′ s

y

s

f f

Convex lens ...

When:• s > f then m < 0 (real, inverted)• s = 2f then m = 1 (real, inverted)• s = f then m = ∞• s < f then m > 1 (virtual, upright)

Interesting case !!

F F

s

f f

0

real invertedimage

object distance (s)

′ s < 0

′ s > 0

virtual uprightimage

m = − fs − f

s > 0 f > 0

s = f s < f s > f

m = −

fs− f

If s < f , then m > 1. Also,

since 1s

>1f

, then 1′ s

< 0,

i.e., ′ s < 0.

So, to use a convex lens as a magnifying glass make

s < f , i.e., object distance less than the focal length of the lens. The image is on the same side as the object and it is enlarged, upright and virtual.

f f

Object

Image

s ′ s

Question 32.8: An object that is 0.03 m high is placed 0.25 m in front of a thin lens that has a power equal to 10.0 D.

(a) Find the position and the size of the image. (b) Repeat part (a) if the object is placed 0.20 m in

front of the lens. (c) Repeat part (a) if the object is placed 0.05 m in

front of the lens.

The power of a lens is given by P =

1f

.

∴f = P−1 = 0.10 m.

(a) With s = 0.25 m. The lens formula is 1s

+1′ s

=1f

.

∴

1′ s

=1f−

1s

=s − f

fs, i.e.,

′ s =

fss − f

.

With s = 0.25 m and f = 0.10 m, we find

′ s =

(0.10 m)(0.25 m)(0.25 m − 0.10 m)

= 0.167 m.

∴m = − ′ s

s= −

0.167 m0.25 m

= −0.667.

The object height is y = 0.03 m, so the image height is

′ y = my = (−0.667)(0.03 m) = −0.02 m.Since ′ s > 0 the image is real. Since ′ y < y and ′ y < 0, the image is reduced and inverted.

(b) With s = 0.20 m, we find

′ s =

(0.10 m)(0.20 m)(0.20 m − 0.10 m)

= 0.02 m.

∴m = − ′ s

s= −

0.20 m0.20 m

= −1.00,

and the image height is

′ y = my = (−1.00)(0.03 m) = −0.03 m.Since ′ s > 0 the image is real. Since ′ y = y and ′ y < 0, the image is the same size as the object and inverted.

(c) With s = 0.05 m, we find

′ s =

(0.10 m)(0.05 m)(0.05 m − 0.10 m)

= −0.10 m.

∴m = − ′ s

s=

0.10 m0.05 m

= +2.00,

and the image height is

′ y = my = (2.00)(0.03 m) = 0.06 mSince ′ s < 0 the image is virtual. Since ′ y > y and

′ y > 0, the image is enlarged and upright.

Question 32.9: If the rays all start from the object at the same time, which arrives at the image first or do they arrive at the same time?

A B

C

Remember Fermat’s principle? It states that a light ray traveling between two points will take the path with the

least time. So, if one of these light ray paths took less time than any of the others, by Fermat’s principle they would all take that path! But we

know that light travel along A, B and C ... so they must all take the same time (even though the distances traveled are not the same)! Although path C is the shortest, the ray travels further through glass where the velocity is less.

A B

C

Concave lens ...

Note for a concave lens: s > 0, ′ s < 0, f < 0 and ′ s < s .

Since m = − ′ s s then 0 < m < 1, always.

So, a diverging lens always produces a virtual image of a real object no matter where it is positioned. The image is upright and smaller relative to the object

A B C D E F A B C D E F

A

Object Image

F ′ s

s

A f

Question 32.10: A double concave lens* that has an index of refraction of 1.435 has radii whose magnitudes are 0.03 m and 0.25 m. An object is positioned 0.80 m to the left of the lens. Find

(a) the focal length of the lens,(b) the position of the image, and(c) the magnification of the image.(d) Is the image real or virtual, upright or inverted?

* Both sides of the lens are concave.

r1 = −0.30 m (opposite side of outgoing light).

r2 = 0.25 m (same side of outgoing light).

n1 = 1.00 and n2 = 1.45

(a)

1f

=n2n1

−1⎛

⎝ ⎜

⎞

⎠ ⎟

1r1−

1r2

⎛

⎝ ⎜

⎞

⎠ ⎟

=

1.451.00

−1⎛ ⎝

⎞ ⎠

1−0.30 m

−1

0.25 m⎛ ⎝

⎞ ⎠ = −3.30 m−1

∴f = −0.303 m (diverging lens).

(b) s = 0.80 m and 1s

+1′ s

=1f

.

∴

10.80 m

+1′ s

=1

−0.303 m

i.e., 1′ s

=1

−0.303 m−

10.80 m

= −4.55 m−1

∴ ′ s = −0.22 m(opposite side to outgoing light)

r1 r2Incoming light Outgoing light

(c) m = − ′ s

s= −

−0.22 m0.80 m

= +0.275.

(d) Since m > 0 the image is virtual and upright. Since

m > 1 the image is diminished.

See useful notes on web-site to see how to solve this problem using a scale drawing.

Note: the image is not necessarily on the same side as the outgoing light!

Incoming light Outgoing light

Object

Image

FRESNEL LENS:

Only the radii of the front and back surfaces of a lens determine the focal length, not the thickness in between! So, much weight can

be saved with this plano-convex lens by removing the “unused” part of a lens!

Lighthouse lens

Compared with regular lenses, Fresnel lenses produce a poorer image quality, so they tend to be used only where quality is not critical or where the bulk of a solid lens would be prohibitive. Cheap Fresnel lenses can be stamped or moulded out of transparent plastic and are used in overhead projectors, projection televisions, and hand-held sheet magnifying glasses. Fresnel lenses have been used to increase the visual size of CRT displays in pocket televisions, and they are also used in traffic lights.