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6 Image formation by thin lenses

6.1 Definition of the equation and the respective variables

Optical lenses form images of objects by means of light refraction. Light rays passing through

a lens are refracted at both boundaries of the lens since the indices of refraction are different

for the lens material and the surrounding medium. Depending on the position of the object

and the type of the lens images can be either real or virtual. The image is formed in that

location where the refracted rays (or their extensions in the case of virtual images, see later)

intersect.

A thin lens produces the sharp image of the object, if the reciprocal of its focal distance equals

to the sum of the reciprocals of image and object distances

(thin lens equation):

1 1 1

f o i

where f, o and i are the focal, object and image distances,

respectively (the SI unit is meter for all the quantities, but

for practical reasons, they are mostly expressed in cm or

mm).

The equation above is the common Gaussian form of the lens equation. In the case of real

objects, the o object distance is positive. The focal length is positive (f > 0) for converging

lenses, whereas it is negative (f < 0) for diverging lenses. The i image distance is positive (i >

0) for a real image, whereas it is negative for a virtual one (i < 0).

Related definitions:

Optical lens: A lens is a transparent optical device that focuses or disperses a light beam by

means of refraction. Their two surfaces are parts of the surfaces of spheres. Both surfaces can

be either convex or concave, and one of the surfaces can be planar (flat) as well. A thin lens is

a lens with a thickness that is negligible compared to the radii of curvature of the lens

surfaces.

Convex lens: thicker at the middle than at the edges. If the lens is made of an optically denser

material than the surrounding medium (typically this is the case) a collimated beam of light

passing through the lens converges at one point (focus or focal point). In this case, the lens is

called a positive or converging lens.

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Concave lens: thinner at the middle than at the edges. If the lens is made of an optically

denser material than the surrounding medium, a collimated beam of light passing through the

lens is diverged (spread); the lens is thus called a negative or diverging lens.

Principal plane: optical planes are stretched surfaces given by the interception of beams

entering to the lens parallel to the axis and the extension of beams leaving the system. Lenses

have two principle planes: on the side of the object and on side of the image. For thin lenses,

the location of the principle planes is practically the same (i.e. the two planes overlap with

each other).

Optical axis: the axis, which passes through the optical center of the lens, perpendicular to

the principal plane(s).

Focal point: The point, where the converging lens focuses the collimated beams of light or

the point where the beams of light starts from, in the case of a diverging lens. All kind of

lenses have two focal points located on the two sides of the lens. For symmetric lenses the

two focal points are located at the same distances measured from the center of the lens. (We

focus on symmetric lenses here.)

Focal length/distance (f): the distance from the principal plane(s) to the corresponding focal

points of the lens (F and F′). Its SI unit is m. (Although it is often given in cm or mm.)

Real image: Outgoing rays from a point converge at a real location. The image of an object is

formed by the interception of rays on the opposite side of the lens. Real images can be

projected onto a screen.

Virtual image: Outgoing rays from a point diverge; therefore not the refracted rays

themselves but their backward extensions will intercept and form the image on the object’s

side. These images cannot be projected onto a screen.

Object and image distances: the distance of the object or the image from the principal plane

of the lens. Their SI unit is m. (They are often given in cm or mm.)

6.2 Further relationships

Linear magnification:

The ratio of image and object sizes (I and O, respectively), which is equal to the absolute

value of the ratio of image and object distances.

I iM

O o

The magnification is often calculated using distances with their actual sign as follows:

iM

o

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If M is negative, as it is for real images, the image is inverted (upside-down) with respect to

the object. For virtual images M is positive, so the image is upright.

Dependence of the focal length on the properties of the lens and the medium:

1 2

1 1 1( 1)( )n

f R R

where n is the relative index of refraction of the lens with respect to the medium, R1 and R2 are

radii of spheres surrounding the lens. (In case of diverging lens these values are negative.) For

a symmetric lens R1 and R2 are equal.

Refracting power of the lens (D, diopter):

reciprocal of the focal length measured in meter. (Unit: diopter, m-1)

6.3 Image formation by thin converging lenses

6.3.1 Graphical representation of the thin lens equation

Let’s examine the dependence of image distance on the object distance. Rearrange the thin

lens equation and plot the image distance as a function of object distance:

The obtained curve is a hyperbola, which approaches asymptotically the o=f and i=f dashed

lines. The curve illustrates the characteristics of image formation (see 6.3.2):

(o i )fo

i x yo f

1D

f

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if o<f (the object is inside the focal length), i is negative; therefore a virtual image is

formed on the front side of the lens (i.e. on the same side where the object is located);

if o >f (the object is outside the focal length), i is positive; therefore a real image is

formed on the opposite side of the lens;

if the object approaches the focal point from either directions, the image distance

increases (for virtual images it will be a more negative value);

if the object is located at the focal point (o=f), the image would be formed at infinity,

which practically means, that no image is formed;

for an object located at infinity (o=), the image is formed in the focal point (i=f);

It can be also seen from the graph that the location of the image is more sensitive to

the position of the object in the vicinity of the focal point than farther away from it.

If we plot the reciprocal of the image distance as a function of the reciprocal of the object

distance, a straight line is obtained:

The slope of the straight line is -1 and it crosses the x-axis at x=1/f. The y-interception of the

straight line is 1/f as well.

Characteristics of image formation can be illustrated using this plot as well (see 6.3.2).

If the object is located inside the focal length (o<f, i.e. 1/o>1/f), the image distance is

negative; i.e. a virtual image is formed. Approaching the object to the lens (1/o

increases), the image distance decreases (the absolute value of 1/i increases).

If the object is located exactly at the focal point (o=f, i.e. 1/o = 1/f), the image would

be formed at infinity (i=, i.e. 1/i approaches zero).

If the object is located beyond the focal point (o>f, i.e. 1/o<1/f), the image distance is

positive, thus, the image is real. If the object is moving away from the focal point (o

increases, 1/o decreases), the image distance decreases (1/i increases). If the object is

1 1 1 1 1 (y x )

i f o i o

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twice the focal length away from the lens, the image is formed at the same distance

(i.e. twice the focal length) on the opposite side of the lens.

If the object is at infinity (o=), the image would be formed at the focal point on the

opposite side of the lens (i=f, i.e. 1/i=1/f).

From the graphs above one can estimate the magnification as well using the corresponding o

and i (or 1/o and 1/i) values.

6.3.2 Construction of images – ray diagrams of thin converging lenses

Characteristics of image formation by lenses can be represented using the so-called ray

diagrams as well. The image formed by a single lens can be located and sized with the three

principal rays given below:

any ray that enters parallel to the optical axis on one side of the lens proceeds

towards the focal point on the other side;

any ray that arrives at the lens after passing through the focal point on the front

side, comes out parallel to the axis on the other side;

any ray that passes through the center of the lens will not change its direction.

Intersection of any two rays out of the three mentioned above is enough to construct the

image. In the cases of simple objects (such as an arrow, or a section; see below) it is enough

to use the beams starting from their two end points in order to construct the image.

Consequently, if one of the end points is located on the optical axis, the corresponding image

point will be located on the optical axis as well.

As we could see in 6.3.1, the properties of the image (position, size) are determined by the

location of the object. The ray diagrams below show construction of images for specific

object locations. Since two rays are enough for proper image construction, only rays traveling

in parallel to the optical axis and rays passing through the center of the lens will be shown.

The propagation directions of rays are indicated by the blue, dashed lines, whereas the object

and the image are colored in black and red, respectively. O and I are the object and image

sizes, respectively.

(i) The object is located inside the focal length:

virtual (only the extended imaginary lines give intersection), upright and magnified

image is formed (O < I);

image formation by a simple magnifier or the eyepiece of a light microscope are

typical examples of this situation;

if the object approaches the focal point (i.e. the object distance increases), the image

distance and the size of the image will increase as well.

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(ii) The object is located between 2F and F (distance from the lens is less than twice the focal

length, but greater than the focal length):

a real, inverted, magnified image (O < I) is formed on the opposite side of the lens

outside twice of the focal length;

the objective of the light microscope is a typical example;

if the object approaches 2F, the image distance and the image size will decrease; if the

object is located exactly at a distance twice the focal length, the object and image sizes

are equal (O=I), and the image is located at the same distance (i.e. twice the focal

length) on the opposite side of the lens.

(iii) The object is located beyond 2F:

a real, inverted image is formed, which is reduced in size (O > I) and located in

between 2F’ and F’;

the longer the object distance, the closer the image is to the focal point (F’);

when the object is at infinity, the image is formed exactly at the focal point (F’)

behind the lens;

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image formation by the human eye is a good example here.

6.4 Exercises with solutions

6.4.1 Exercise

Construct the image of the object (indicated by the green arrow) formed by a thin converging

lens.

Solution:

Draw any two of the three principal rays outgoing from the two end points of the object: here

we use rays traveling in parallel to the optical axis and those passing through the center of the

lens.

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Please keep in mind that the object is located inside the focal length; therefore an upright,

virtual and magnified image should be obtained.

6.4.2 Exercise

The figure below depicts the image of an object generated by a thin converging lens. Draw the

object in the appropriate position.

Solution:

In order to introduce the third principal ray of image formation, instead of the ray entering the

lens parallel to the optical axis, the ray passing through the focal point on the front side of the

lens (and then coming out parallel to the axis on the other side) will be used to find the

location of the object.

Please note, that the image is located at a distance between the focal length and twice the

focal length (between F’ and 2F’), indicating that we have an inverted and reduced image. It

means that the object size should be greater than the image size, and the object should be

located at a distance greater than twice the focal length in front of the lens. Furthermore the

object points “downwards”, i.e. to the opposite direction, than the image does.

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6.4.3 Exercise

The figure below depicts an object and its image generated by a thin converging lens. Indicate

the focal points on both sides of the lens.

Solution:

In order to solve the problem, we should use two of the principal rays:

outgoing rays from the object entering the lens parallel to the optical axis cross the

focal point on the opposite side of the lens;

rays passing through the object’s side focal point will leave the lens parallel to the

optical axis.

Thus, F’ can be positioned by drawing the ray outgoing from the object parallel to the optical

axis, whereas F can be located using the ray ingoing to the image parallel to the optical axis.

6.4.4 Exercise

The table shows the focal length of three thin converging lenses along with corresponding

values of object and image distances. Complete the table with the missing data. Use the

appropriate sign when giving the magnification. Do not forget the units.

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lens #1 lens #2 lens #3

focal distance (f) 10 cm 5 cm

object distance (t) 4 cm 5 cm

image distance (i) 8 cm 5 cm

refractive power (D)

magnification (M)

Solution:

Missing data can be calculated using the equations shown below. Please keep in mind, that the

focal length should be expressed in meter when it is used to determine the refractive power.

lens #1 lens #2 lens #3

focal distance (f) 10 cm 5 cm 2.5 cm

object distance (o) 4 cm 13.3 cm 5 cm

image distance (i) - 6.67 cm 8 cm 5 cm

refractive power (D) 10 diopters 20 diopters 40 diopters

magnification (M) 1.67 -0.6 -1

Lens #1: the refractive power can be calculated from the focal length. Image distance can be

determined from the focal and object distances. If the image distance is calculated then we

can determine the magnification.

1 1 1 1 0.1 0.25 6.67cm

10cm 4cm

1D= 10 diopters

0.1m

6.67M= 1.67

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ii i

Image distance has a negative sign, thus the image is virtual, as it was already suggested by

the object and focal distances (o<f). Consequently, the magnification is positive, i.e. the

image is upright.

Lens #2:

1 1 1 1 0.2 0.125 13.3cm

5cm 8

1D= 20 diopters

0.05m

8M= 0.6

13.3

oo cm o

1 1 1 1 D= M=-

i

f o i f o

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Lens #3:

1 1 1 1 0.4 2.5cm

5cm 5cm

1D= 40 diopters

0.025m

5M= 1

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ff f

As it is indicated by the object and image distances, in the case of lens #2 and #3 the images

are real and inverted; it is reflected by the negative magnification values as well.

6.4.5 Exercise

The table shows the refractive power of three converging thin lenses along with

corresponding values of object and image parameters. Complete the table with the missing

data. Indicate the units as well.

lens #1 lens #2 lens #3

refractive power (D) 25 diopters 10 diopters 40 diopters

image distance (i) 10 cm

object distance (o) 5 cm 10 cm

image size (I) 3 cm 4 cm

object size (O) 5 cm

Solution:

1 1 1 1 D= M= =

i I

f o i f o O

In order to calculate the missing data, we should use the equations shown above. Please keep

in mind, that the focal length is expressed in meter when it is used to determine the refractive

power of lenses. In the case of magnification we can use the absolute values.

lens #1 lens #2 lens #3

refractive power (D) 25 diopters 10 diopters 40 diopters

image distance (i) 10 cm - 10 cm 3.33 cm

object distance (o) 6.67 cm 5 cm 10 cm

image size (I) 3 cm 10 cm 4 cm

object size (O) 2 cm 5 cm 12 cm

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Lens #1:

125 0.04m 4cm

1 1 1 1 0.25 0.1 6.67cm

4cm 10cm

3cm 10cm 2cm

6.67cm

ff

oo o

M OO

Lens #2:

110 0.1m 10cm

1 1 1 1 0.1 0.2 10cm

10cm 5cm

10cm 10cm

5cm 5cm

ff

ii i

II

Lens #3:

140 0.025m 2.5cm

1 1 1 1 0.4 0.1 3.33cm

2.5cm 10cm

4cm 3.33cm 12cm

10cm

ff

ii i

OO

6.5 Exercises

1. The focal distance of a thin converging lens is 10 cm. At what distance the object

should be placed to get a magnified, virtual image?

(a) 20cm (b) 12cm (c) 5cm

2. The focal distance of a thin converging lens is 8 cm. At what distance the object

should be placed to get a real, magnified image?

(a) 18cm (b) 13cm (c) 5cm

3. An illuminated object is placed at 10cm from a converging thin lens. The sharp image

of the object can be observed on a screen placed at a distance of 25cm from the lens.

What is the diopter of the lens?

(a) 14 (b) 0,14 (c) 7

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4. What is the focal distance of a thin lens with a refractive power of 16 diopters?

(a) 0.0625 cm (b) 6.25 cm (c) 6.25 m

5. An illuminated object is placed at a distance of 20 cm from a converging thin lens. A

sharp image is formed on a screen placed at 25 cm from the lens. What is the

magnification?

(a) 1.25 (b) 0.8 (c) 500

6. An illuminated object is placed at 20 cm from a converging thin lens. The sharp image

of the object is formed at a distance of 25cm. What is the focal length of the lens?

(a) 11cm (b) 0.09 cm (c) 45 cm

7. A 5-cm tall illuminated object is placed at a distance of 15 cm from a converging thin

lens. The image of the object is formed at a distance of 30 cm from the lens. What is

the image size?

(a) 10 cm (b) 2.5 cm (c) 3 cm

8. The focal distance of a converging lens is 10 cm. What is the magnification if the

object is placed at 8 cm from the lens?

(a) 5 (b) 0.2 (c) 1.25

9. The focal distance of a converging lens is 5 cm. The image size for an object placed at

a distance of 8 cm from the lens is 10 cm. What is the object size?

(a) 16.7cm (b) 6 cm (c) 13.3 cm

10. The refractive power of a converging lens is 25 diopters. What is the focal distance?

(a) 0.04 cm (b) 4 cm (c) 25 cm

11. The refractive power of a converging lens is 20 diopters. What kind of image is

generated if the object is placed in 7.5 cm distance from the lens?

(a) real, reduced (b) real, magnified (c) virtual, magnified

12. The refractive power of a converging lens is 20 diopters. At what distance should the

object be placed in order to receive an inverted, reduced image?

(a) 15 cm (b) 7.5 cm (c) 3 cm

13. A converging lens forms a 3× magnified real image of an object placed at a distance of

12 cm from the lens. What is the image distance?

(a) 15 cm (b) 4 cm (c) 36 cm

14. A converging lens forms a 3× magnified virtual image of an object placed 8 cm from

the lens. What is the image distance?

(a) 24 cm (b) -24 cm (c) -2,6 cm

15. A converging lens forms a 5× magnified real image of an object. If the image distance

is 10 cm, what is the object distance?

(a) 2 cm (b) 50 cm (c) -2 cm

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16. A, B and C are optical lenses of the same size but of different materials. Which one

has the longest focal length, if the order of their absolute indices of refraction is the

following: nA<nB<nC.

(a) A (b) B (c) C