Chapter 7

Chapter 7: Geometrical Optics

7.1 Reflection at a Spherical Surface

L.O 7.1.1 State laws of reflection

L.O 7.1.2 Sketch and use ray diagrams to determine the characteristics of image

formed by spherical mirrors

L.O 7.1.3 Use

for real object only

A spherical mirror is a reflecting surface with spherical geometry.

Two types:

i. Convex, if the reflection takes place on the outer surface of the spherical shape.

ii. Concave, if the reflecting surface is on the inner surface of the sphere.

C : centre of curvature of the surface mirror.

P : centre of the surface mirror (vertex or pole).

Line CP : principal or optical axis.

AB : aperture of the mirror.

F : focal point of the mirror.

f : focal length (FP, distance between focal point and the centre of the mirror).

r : radius of curvature of the mirror.

Laws of reflection state:

The incident ray, the reflected ray and the

normal all lie in the same plane.

The angle of incidence, i equal the angle

of reflection, r as shown in figure below.

Chapter 7

Ray diagrams for spherical mirrors:

Ray diagram is defined as the simple graphical method to indicate the positions of the

object and image in a system of mirrors or lenses.

Table below shows the conditions for ray diagram:

Ray 1 Ray 2 Ray 3

A ray parallel to the

principal axis passes

through or diverges from

the focal point F after

reflection.

A ray which passes through or

is directed towards the focal

point F is reflected as a ray

parallel to the principal axis.

A ray which passes through or

is directed towards the centre

of curvature C will be

reflected back along the same

path.

At least any

two rays for

drawing the

ray diagram.

Chapter 7

Concave Mirror

Object

Distance u

Ray Diagram Image Property

u > r

Real

Inverted

Diminished

Between point C and F

u = r

Real

Inverted

Same size

Formed at point C

(Application: Overhead

projector)

f < u < r

Real

Inverted

Magnified

Formed at a distance

greater than C

u = f

Real or virtual

Formed at infinity

(Application: Torchlight, car

spotlight)

u < f

Virtual

Upright

Magnified

Formed at the back of

the mirror

(Application: Make-up

mirror, shaving mirror,

mirror used by dentist)

Chapter 7

u at

infinity

Real

Inverted

Diminished

Formed at point F

(Application: Reflection

telescope)

Convex Mirror

Object

Distance u

Ray Diagram Image Property

Any position

in front of

the convex

mirror

Virtual

Upright

Diminished

Formed at the back of

mirror

(Application: rear-mirror and

side-mirrors of cars, at sharp

corners of a road, and the

corners in a supermarket

because it has a wide field of

view and providing an

upright image.)

Equation (formula) of spherical mirror:

vuf

111

Relationship between focal length, f and radius of curvature, r:

ΔFCM is isosceles (FC = FM).

Consider ray AM is paraxial (parallel

and very close to the principal axis).

2

CP2

1FP

FP FM

rf

For spherical

mirror only

Chapter 7

Linear magnification of the spherical mirror, m is defined as the ratio between image height,

hi and object height, ho.

u

v

h

hm i

0

m is a positive value if the image formed is upright and it is negative if the image formed is

inverted.

Table below shows the sign convention for spherical mirror’s equation:

Physical Quantity Positive sign (+) Negative sign (-)

Object Distance, u Real object

(in front of the mirror)

Virtual object

(at the back of the mirror)

Image Distance, v Real image

(same side of the object)

Virtual image

(Opposite side of the object)

Focal length, f Concave mirror Convex mirror

Linear magnification, m Upright image Inverted image

Object/ Image Height, h Upright image Inverted image

Note:

Real image is formed by the actual light rays that pass through the image.

Real image can be projected on the screen.

Example

Question Solution

A dentist uses a small mirror attached to a

thin rod to examine one of your teeth. When

the tooth is 1.20 cm in front of the mirror, the

image it forms is 9.25 cm behind the mirror.

Determine

a. the focal length of the mirror and state

the type of the mirror used

b. the magnification of the image

Chapter 7

Question Solution

An upright image is formed 20.5 cm from the

real object by using the spherical mirror. The

image’s height is one fourth of object’s

height.

a. Where should the mirror be placed

relative to the object?

b. Calculate the radius of curvature of the

mirror and describe the type of mirror

required.

c. Sketch and label a ray diagram to show

the formation of the image

A mirror on the passenger side of your car is

convex and has a radius of curvature 20.0

cm. Another car is seen in this side mirror

and is 11.0 m in front of the mirror (behind

your car). If this car is 1.5 m tall, calculate

the height of the car image.

A person of 1.60 m height stands 0.60 m

from a surface of a hanging shiny globe in a

garden.

a. If the diameter of the globe is 18 cm,

where is the image of the

b. person relative to the surface of the

globe?

c. How tall is the person’s image?

d. State the characteristics of the person’s

image.

Chapter 7

Exercise

Question

An object is placed 10 cm in front of a concave mirror whose focal length is 15 cm.

Determine

a. the position of the image.

b. the linear magnification and state the properties of the image.

Answer: v = ‒30 cm; m = 3

A concave mirror forms an image on a wall 3.20 m from the mirror of the filament of a

headlight lamp. If the height of the filament is 5.0 mm and the height of its image is 35.0 cm,

calculate

a. the position of the filament from the pole of the mirror.

b. the radius of curvature of the mirror.

Answer: 4.57 cm; 9.01 cm

a. A concave mirror forms an inverted image four times larger than the object. Find the

focal length of the mirror, assuming the distance between object and image is 0.600 m.

b. A convex mirror forms a virtual image half the size of the object. Assuming the distance

between image and object is 20.0 cm, determine the radius of curvature of the mirror.

Answer : 160 mm, -267 mm

If a concave mirror has a focal length of 10 cm, find the two positions where an object can

be placed to give, in each case, an image twice the height of the object.

Answer: 15 cm, 5.0 cm

A convex mirror of radius of curvature 40 cm forms an image which is half the height of the

object. Find the object and image position.

Answer: 20 cm, 10 cm behind the mirror

A 1.74 m tall shopper in a department store is 5.19 m from a security mirror. The shopper

notices that his image in the mirror appears to be only 16.3 cm tall.

a. Is the shopper’s image upright or inverted? Explain.

b. Determine the radius of curvature of the mirror.

Answer: u think, 1.07 m

A concave mirror of a focal length 36 cm produces an image whose distance from the mirror

is one third of the object distance. Calculate the object and image distances.

Asnwer: 144 cm, 48 cm

Chapter 7

7.2 Refraction at a Plane and Spherical Surfaces

L.O 7.2.1 State and use the laws of refraction (Snell’s Law) for layers of materials

with different densities

Refraction is defined as the changing of direction of a light ray and its speed of

propagation as it passes from one medium into another.

Refraction at a plane

Laws of refraction state:

The incident ray, the refracted ray and the normal all lie in the same plane.

For two given media,

constant sin

sin

1

2 n

n

r

i or rnin sin sin 21

where n1 : refractive index of medium 1 (medium containing incident ray)

n2 : refractive index of medium 2 (medium containing refracted ray)

n1 < n2

(Medium 1 is less dense medium 2)

n1 > n2

(Medium 1 is denser than medium 2)

The light ray is bent toward the normal

The light ray is bent away from the normal

Special cases:

Snell’s Law

i = 0°

i = critical angle i > critical angle

(Must be from denser to less dense medium)

Chapter 7

Refractive index (index of refraction)

Definition – is defined as the constant ratio r

i

sin

sin for the two given media.

The value of refractive index depends on the type of medium and the colour of the light.

It is dimensionless and its value greater than 1.

Consider the light ray travels from medium 1 into medium 2, the refractive index can be

denoted by

If medium 1 is vacuum, then the refractive index is called absolute refractive index,

written as:

v

cn

mediumin light ofvelocity

in vacuumlight ofvelocity

Relationship between refractive index and the wavelength of light:

As light travels from one medium to another, its wavelength, changes but its frequency,

f remains constant.

The wavelength changes because of different material. The frequency remains constant

because the number of wave cycles arriving per unit time must equal the number

leaving per unit time so that the boundary surface cannot create or destroy waves.

By considering a light travels from medium 1 (n1) into medium 2 (n2), the velocity of light

in each medium is given by

11 fv and 22 fv

The refractive index can be written as:

0

mediumin light of wavelength

in vacuumlight of wavelengthn

Other equation for absolute refractive index

in term of depth is given by

depthapparent

depth realn

(Medium containing

the incident ray) (Medium containing

the refracted ray)

2

121

2 mediumin light ofvelocity

1 mediumin light ofvelocity

v

vn

Chapter 7

L.O 7.2.2 Use

for spherical surface

Equation of spherical refracting surface:

r

nn

v

n

u

n 1221

Table below shows the sign convention for spherical refracting surface’s equation:

Physical Quantity Positive sign (+) Negative sign (-)

Object Distance, u Real object

(the object is on the same side of the

spherical surface from where the light

is coming)

Virtual object

(the object is on the opposite side of

the spherical surface from where the

light is coming)

Image Distance, v Real image

(the image is on the opposite side of

the spherical surface from where the

light is coming)

Virtual image

(the image is on the same side of the

spherical surface from where the light

is coming)

Radius of

curvature, r

Convex surface

(the center of curvature is on the the

opposite side of the spherical surface

from where the light is coming)

Concave surface

(the center of curvature is on the the

same side of the spherical surface

from where the light is coming)

If the refraction surface is flat (plane), r = ∞

The equation (formula) of linear magnification for refraction by the spherical surface is

given by

un

vn

h

hm

o

i

2

1

Chapter 7

Example

Question Solution

A fifty cent coin is at the bottom of a

swimming pool of depth 3.00 m. The refractive

index of air and water are 1.00 and 1.33

respectively. Determine the apparent depth of

the coin.

A cylindrical glass rod in air has a refractive

index of 1.52. One end is ground to a

hemispherical surface with radius, r = 3.00 cm

as shown in figure below.

Calculate,

a. the position of the image for a small object

on the axis of the rod, 10.0 cm to the left of

the pole as shown in figure.

b. the linear magnification.

(Given the refractive index of air , na= 1.00)

Figure below shows an object O placed at a

distance 20.0 cm from the surface P of a glass

sphere of radius 5.0 cm and refractive index of

1.63.

Chapter 7

Exercise

Question

We wish to determine the depth of a swimming pool filled with water by measuring the width

(x = 5.50 m) and then noting that the bottom edge of the pool is just visible at an angle of

14.0 above the horizontal as shown in figure below.

Calculate the depth of the pool. (Given nwater = 1.33 and

nair = 1.00)

Answer : 5.16 m

A light beam travels at 1.94 x 108 m s

-1 in quartz. The wavelength of the light in quartz is 355

nm.

a. Find the index of refraction of quartz at this wavelength.

b. If this same light travels through air, what is its wavelength there?

(Given the speed of light in vacuum, c = 3.00 x 108 m s

-1)

Answer: 1.55; 550 nm

A pond with a total depth (ice + water) of 4.00 m is covered by a transparent layer of ice of

thickness 0.32 m. Determine the time required for light to travel vertically from the surface of

the ice to the bottom of the pond. The refractive index of ice and water are 1.31 and 1.33

respectively.

(Given the speed of light in vacuum is 3.00 108 m s

-1.)

Answer: 1.77×10-8

s

A small strip of paper is pasted on one side of a glass sphere of radius 5 cm. The paper is

then view from the opposite surface of the sphere. Find the position of the image.

(Given refractive index of glass =1.52 and refractive index of air=1.00)

Answer : 20.83 cm in front of the concave surface (second refracting surface)

A point source of light is placed at a distance of 25.0 cm from the centre of a glass sphere of

radius 10 cm. Find the image position of the source.

(Given refractive index of glass =1.50 and refractive index of air=1.00)

Answer : 28 cm at the back of the concave surface (second refracting surface)

Chapter 7

7.3 Thin lenses

L.O 7.3.1 Sketch and use ray diagrams to determine the characteristic of image

formed by concave and convex lenses

Thin lens is defined as a transparent material with two spherical refracting surfaces whose

thickness is thin compared to the radii of curvature of the two refracting surfaces.

Thin lenses are divided into two types:

i. Convex (Converging) lens which are thicker at the centre than the edges

ii. Concave (Diverging) lens which are thinner at the centre then at the edges

Both convex and concave lenses have two refractive surfaces of radii of curvature r1 and r2

with their respective centre of curvatures at C1 and C2:

Centre of curvature (point C1 and C2) is defined as the centre of the sphere of which

the surface of the lens is a part.

Radius of curvature (r1 and r2) is defined as the radius of the sphere of which the

surface of the lens is a part.

Principal (Optical) axis is defined as the line joining the two centres of curvature of a

lens.

Optical centre (point O) is defined as the point at which any rays entering the lens

pass without deviation.

Chapter 7

Table below shows the conditions for ray diagram:

Ray 1 Ray 2 Ray 3

Ray which is parallel to the

principal axis will be

deflected by the lens towards/

away from the focal point F.

Ray passing through the

optical centre is un-deflected.

Ray which passes through the

focal point (convex lens) or

appear to converge to the

focal point (concave lens)

becomes parallel to the

principal axis after emerging

from lens.

At least any

two rays for

drawing the

ray diagram.

Chapter 7

Converging Lens

Object

Distance u

Ray Diagram Image Property

u > 2f

Real

Inverted

Diminished

Between F2 and 2F2 at the

back of the lens

(Application: Camera lens,

human eye lens)

u = 2f

Real

Inverted

Same size

Formed at point 2F2 at

the back of the lens

(Application: Photocopier)

f < u < 2f

Real

Inverted

Magnified

Formed at a distance

greater than 2F2 at the

back of the lens

(Application: Projector lens,

objective lens of microscope)

u = f

Real or virtual

Formed at infinity

(Application: Eyepiece of

astronomical telescope)

u < f

Virtual

Upright

Magnified

Formed in front of the

lens

(Application: Magnifying

lens)

Chapter 7

u = ∞

Real

Inverted

Diminished

Formed at F2

(Application: Objective lens

of a telescope)

Diverging (Concave) lens

Object

Distance u

Ray Diagram Image Property

Any position

in front of

the

diverging

lens

Virtual

Upright

Diminished

Formed in front of the

lens

L.O 7.3.2 Use thin lens equation for real object only

L.O 7.3.3 Use lens maker’s equation

L.O 7.3.4 Use the thin lens formula for a combination of two convex lenses

Thin lens equation for real object only:

fvu

111

Lens maker’s equation:

21medium

material 111

1

rrn

n

f

where f : focal length

r1 : radius of curvature of first refracting surface

r2 : radius of curvature of second refracting surface

nmaterial : refractive index of lens material

nmedium : refractive index of medium

The radius of curvature of flat refracting surface is infinity, r = ∞.

Chapter 7

If the medium is air, then the lens maker’s equation can be written as

21

111

1

rrn

f

Linear magnification of the spherical mirror, m is defined as the ratio between image height,

hi and object height, ho.

u

v

h

hm i

0

m is a positive value if the image formed is upright and it is negative if the image formed is

inverted.

Since fvu

111 , the linear magnification equation can be written as

vfvu

111 1

f

vm

Table below shows the sign convention for both lens maker’s equation and thin lens equation:

Physical Quantity Positive sign (+) Negative sign (-)

Object Distance, u Real object

(the object is on the same side of the

spherical surface from where the light

is coming)

Virtual object

(the object is on the opposite side of

the spherical surface from where the

light is coming)

Image Distance, v Real image

(the image is on the opposite side of

the spherical surface from where the

light is coming)

Virtual image

(the image is on the same side of the

spherical surface from where the light

is coming)

Radius of

curvature, r

Convex surface

(the center of curvature is on the the

opposite side of the spherical surface

from where the light is coming)

Concave surface

(the center of curvature is on the the

same side of the spherical surface

from where the light is coming)

Focal length, f Converging lenses Diverging lenses

Object/ Image

Height, h

Upright image Inverted image

Chapter 7

Combination of Two Convex Lenses

Many optical instruments, such as microscopes and telescopes, use two converging

lenses together to produce an image.

In both instruments, the 1st lens (closest to the object)is called the objective and the 2

nd

lens (closest to the eye) is referred to as the eyepiece or ocular.

The image formed by the 1st lens is treated as the object for the 2

nd lens and the final

image is the image formed by the 2nd

lens.

The position of the final image in a two lenses system can be determined by applying the

thin lens formula to each lens separately.

The overall magnification of a two lenses system is the product of the magnifications of

the separate lenses.

21mmm

where

m1 : magnification due to the first lens

m2 : magnification due to the second lens

Chapter 7

Example

Question Solution

A biconvex lens is made of glass with

refractive index 1.52 having the radii of

curvature of 20 cm respectively. Calculate the

focal length of the lens in

a. water,

b. carbon disulfide.

(Given nw = 1.33 and nc = 1.63)

A person of height 1.75 m is standing 2.50 m

in from of a camera. The camera uses a thin

biconvex lens of radii of curvature 7.69 mm.

The lens made from the crown glass of

refractive index 1.52.

a. Calculate the focal length of the lens.

b. Sketch a labeled ray diagram to show the

formation of the image.

c. Determine the position of the image and

its height.

d. State the characteristics of the image.

An object is placed 90.0 cm from a glass lens

(n = 1.56) with one concave surface of radius

22.0 cm and one convex surface of radius

18.5 cm. Determine

a. the image position.

b. the linear magnification.

Chapter 7

Question Solution

An object is 15.0 cm from a convex lens of

focal length 10.0 cm. Another convex lens of

focal length 7.5 cm is 40.0 cm behind the

first. Find the position and magnification of

the image formed by

a. the first convex lens

b. both lenses

Exercise

Question

A converging lens with a focal length of 90.0 cm forms an image of a 3.20 cm tall real object

that is to the left of the lens. The image is 4.50 cm tall and inverted. Find

a. the object position from the lens.

b. the image position from the lens. Is the image real or virtual?

Answer: 154 cm; 217 cm

A thin plano-convex lens is made of glass of refractive index 1.66. When an object is set up

10 cm from the lens, a virtual image ten times its size is formed. Determine

a. the focal length of the lens,

b. the radius of curvature of the convex surface.

Answer: 11.1 cm, 7.33 cm

The objective and eyepiece of the compound microscope are both converging lenses and have

focal lengths of 15.0 mm and 25.5 mm respectively. A distance of 61.0 mm separates the

lenses. The microscope is being used to examine a sample placed 24.1 mm in front of the

objective.

a. Determine

i. the position of the final image,

ii. the overall magnification of the microscope.

b. State the characteristics of the final image.

Answer: -129 mm, -9.9, u think

A converging lens with a focal length of 4.0 cm is to the left of a second identical lens. When

a feather is placed 12 cm to the left of the first lens, the final image is the same size and

orientation as the feather itself. Calculate the separation between the lenses.

Answer: 12.0 cm