CHAPTER 7: OPTICS

CHAPTER 7

OPTICS

KMM

CHAPTER 7 1
Optics

The study of light based on the assumptions :
1. Light travels in a fixed direction in a straight line as it

passes through a uniform medium.
2. It changes its direction when it meets the surface of a

different medium.

7.1 Reflection at a spherical surface
7.2 Refraction at spherical surfaces
7.3 Thin lenses

7.1 Reflection at a spherical surface (1 hour)

At the end of this topic, students should be able :

a)State radius of curvature, R =2f for spherical mirror

a)Sketch and use ray diagrams with a minimum of two rays to
determine the characteristics of image formed by spherical mirrors.

c) Use mirror equations for real object only.

d) Define and use magnification,

2

Spherical Mirror
• A spherical mirror is a mirror with a curved reflecting surface.
• It can be made of glass or any polished metal surfaces.

• Two types of spherical mirror :
i. Concave mirror (converging), if the reflecting surface is inside.
ii. Convex mirror (diverging), if the reflecting surface is outside.

3

CONCAVE MIRROR Imaginary spherical CONVEX MIRROR
AA

CF PP FC

BB

ff
C → centre of curvature of the surfarce mirror. r
P → centre of the surface mirror (vertex or pole).
Line CP → principal or optical axis
AB → aperture of the mirror.
F → focal point (focus) of the mirror.
f → focal length (FP, the distance between focal point, F and the centre, P of the
mirror.
r → radius of curvature of the mirror.
4

Focal point, F (focus)

Concave mirror Convex mirror

Focal length, f Focal length, f

▪ For concave mirror – is defined ▪ For convex mirror – is defined as a
as a point where the incident point where the incident parallel
parallel rays converge after rays seem to diverge from a point
reflection on the mirror. behind the mirror after reflection.

5

7.1 a) State radius of curvature R= 2f for spherical mirror

• A ray parallel to the principal axis is incident on the surface
of a spherical mirror at point X.

• A radial line from the center of of curvature through point X
is normal to the mirror.

• The angle of incidence is equal to the angle of reflection :

• By alternate interior angles, Concave mirror
• Triangle XFC is isosceles since it has two equal angles,

therefore,

• Since the incident ray is close to the principal axis, is small.
As a result,

where,

Convex mirror 66

7.1 b) Sketch Ray Diagram (minimum 2 rays) to determine
image formed
Ray 1 : A ray drawn from the top of the object parallel to the

principal axis and is reflected through the focal point F.

11

7

Ray 2 : A ray drawn from the top of the object through the focal
point F and is reflected parallel to the principal axis.

2
2

8

Ray 3 : A ray drawn from the top of the object through the
center of curvature C and is reflected back along its
incoming path.

3
3

9

Images Formed By Concave Mirror

Object distance, u Ray diagram Image characteristic

u>r C F P Real
Front back Inverted
Diminished
u=r F P Formed between
C point C and F.
10
Front back Real
Inverted
Same size
Formed at point C.

10

Object distance, u Ray diagram Image characteristic

f<u<r C P Real
F back Inverted
Magnified
Front Formed at a distance
greater than CP.
u=f CF P
Virtual
11 Front back Upright
Magnified
Formed at infinity.

11

Object distance, u Ray diagram Image characteristic

u<f Virtual
Upright
F P Magnified
C back Formed at the back
of the mirror
Front

F o Real
o Inverted
o Diminised

o12 Formed at F

12

“SOITA” = SKETCH OBJECT AND IMAGE THROUGH ARROW

1

5

2 OBJECT IMAGE

3 6
4 56

∞C F ∞

1
2
3

4 IMAGE

In front mirror Behind mirror

13

Image formed by a convex mirror using ray diagram

Image Characteristics :
▪ virtual
▪ upright
▪ diminished
▪ formed at the
▪ back of the mirror
(behind the mirror)

14

Shaving /make-up mirrors are
used to see the face.

Why ?

Because they provide upright diminished Because they provide upright
images of object and also have a wider and magnified images.
field of view than plane mirrors.
15

7.1 c) Use mirror equation for real object only

1. Spherical mirror’s equation

Or Where

2. Lateral magnification, m is defined as the ratio between the image
height to the object height
where u = object distance

v = image distance

r = radius of curvature

f = focal length

ho= object height 16
hi= image height

7.1 d)Apply magnification

Lateral magnification, m is defined as the ratio between the image
height to the object height

where u = object distance
v = image distance
r = radius of curvature
f = focal length
ho= object height
hi= image height

If > 1.0, image is larger than object

If < 1.0, image is smaller than object

If = 1.0, image is same size as the object

17

Sign conventions for Spherical Mirror

No. Quantity Positive sign (+) Negative sign (-)

1u Real in front of the mirror Virtual behind the mirror
Object distance object object

v Real Inverted In front of Virtual Upright behind the
2 Image distance image image the mirror image image mirror

3 m , magnification Upright behind the mirror Inverted In front of the mirror
hi, height image image image

4r Concave mirror Convex mirror
Radius of curvature (converging) (diverging)

5f Concave mirror Convex mirror
Focal length

Image RIVU: Real, Inverted, Virtual, Upright NOTE : Flat surface, r1=8 ∞
characteristics

Real Image Note : Image

• Real image is one where the rays actually pas through the image.
(location where the light actually converges)

• It can be projected on the screen.

Virtual Image

• Virtual image is one where the rays do not actually pass through
the image. (location from where light appears to have converged)

• It cannot be formed on the screen. 19

RIVU: Real, Inverted, Virtual, Upright

7.2 Refraction At Spherical Surface

At the end of this chapter, students should be able to:

a) Use for spherical surface.

20

7.2 a) Use Equation of spherical refracting surface
Refraction At A Spherical Surface

Equation of
spherical refracting

surface

Where
u = object distance from pole
v = image distance from pole
n1 = refractive index of medium 1 (medium containing the incident ray)
n2 = refractive index of medium 2 (medium containing the refracted ray)

21

In front Behind

In front Behind

Convex spherical surface, pole
thus r is positive (+)

Flat surface, thus r
is infinity (∞)

In front Behind

Concave spherical surface,
thus r is positive (-)

22

Sign Conventions For Refracting Surface

No Quantity Positive sign (+) Positive sign (−)

1 u Real object in front of Virtual behind the
the surface object surface
Object distance

v Real image behind the Virtual in front of
surface image the surface
2 Image distance

inverted upright

3 r Convex surface Concave surface

Radius of curvature

Image RIVU: Real, Inverted, Virtual, Upright NOTE : Flat surface ,2r3 = ∞
characteristics

7.3 Thin Lenses
At the end of this chapter, students should be able to:

a) Use thin lens equation, for real object only.

b) Determine the focal length of a convex lens (Experiment 5: Geometrical Optics)

c) Use lens maker’s equation

d) Apply magnification, respectively.

e) Use the thin lens formula for a combination of two convex lenses

Thin Lens

• 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.

• Two types of lens:
i) Convex (Converging)
ii) Concave (Diverging)

Lenses

• Lenses are made of transparent materials, like glass Convex (Converging)
Lens
or plastic, that typically have an index of refraction
greater than that of air.

• Each of a lens’ two faces is part of a sphere and can Concave (Diverging)
be convex or concave (or one face may be flat). Lens

• If a lens is thicker at the center than the edges, it is
a convex, or converging, lens since parallel rays will
be converged to meet at the focus.

• A lens which is thinner in the center than the edges
is a concave, or diverging, lens since rays going
through it will be spread out.

26

Lenses: Focal Length

• Like mirrors, lenses have a principal axis perpendicular to their surface
and passing through their midpoint.

• Lenses also have a vertical axis, or principal plane, through their middle.

• They have a focal point, F, and the focal length is the distance from the
vertical

axis to F.
• There is no real center of curvature, so 2F is used to denote twice the foc2a7 l

Terms of Thin Lens O

O

• A lens has 2 spherical surfaces with radii of curvature r1 & r2 .
• Radius of curvature (r1 and r2) - the radius of each spherical surface.
• Center of the lens , O - a point at where all rays pass through this point

will continue in a straight line.
• Principal axis - a line through the center of curvature C1 & C2 .

Focal Point (focus, F) and Focal Length, f

Convex (Converging lens)
F – a point where the incident parallel
rays converge after passing through
f the lens
Concave (Diverging lens)
F – a point where the incident parallel
rays seem to diverge from after
passing through the lens

Focal length, f – distance between focal
f point , F and center of the lens.

7.3 a) Use thin lens equation for real object
c) Use len’s maker equation
Thin lens formula

Lens maker’s equation

Where f : focal length

r1 : radius of curvature for 1st refracting surface 30
r2 : radius of curvature for 2nd refracting surface
nmedium : refractive index of the medium
nmaterial : refractive index of the lens material

30

• If the medium is air (n1= nair=1) thus the lens maker’s
equation can be written as

7.3 d) Apply magnification
Linear magnification, m
• is defined as the ratio between image height, hi and object
height, ho.

31 31

Sign Conventions For Thin Lens

No Quantity Positive sign (+) Negative sign (-)

1 u Real object Virtual object
Object distance (in front of the lens) (behind the lens)

2 v Real image, inverted Virtual image, upright
Image distance (behind the lens) (in front of the lens)

3 r Convex surface Concave surface
Radius of curvature

4 f Convex lens Concave lens
Focal length

5 m, magnification Upright (in front of lens) Inverted (behind the lens)
hi, image height

RIVU: Real, Inverted, Virtual, Upright NOTE : Flat surface , r = ∞ 32

Converging (convex) Lenses

r1 r2 r1 r2 r1 r2

(+ve) (-ve) (+ve) ( ∞) (+ve) ( +e)

Biconvex Plano-convex Convex meniscus

Diverging (concave) Lenses

r1 r2 r1 r2 r1 r2

(−ve) (+ve) (−ve) ( ∞) (+ve) (+ve)

Biconcave Plano-concave Concave meniscus

7.3 e) Use the thin lens formula for a 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 2nd 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 2nd 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.

M= M₁M₂

Example 7.1.1
An object 2.0 cm high is placed 30 cm from a concave mirror with a
radius of curvature of 10 cm. Find the location and its characteristics.
Solution

Given : h = 2.0 cm, u = +30 cm, r = +10 cm, f = r/2= +5 cm

Characteristics :
1) diminished
2) in front of the mirror/real image
3) inverted

35

Example 7.1.2

A woman who is 1.50 m is located 3.0 m from an anti shoplifting mirror. The
focal length of the mirror is 0.25 m. Find :
(a) the position of her image
(b) the magnification
(c) the height of the image.

Solution Given : u = 3.0 m, f = -0.25 m (convex mirror)

(a) (c)

(b)

36

Example 7.1.3
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. Find
(a)focal length of the mirror,
(b)the magnification of the image.
Solution
Given : u = +1.20 cm, v = -9.25 cm (behind the mirror/ virtual image)

(a) (b)

37 37

Example 7.1.4

A shaving or makeup mirror forms an image of a light bulb on a
wall of a bathroom that is 3.50 m from the mirror. The height of the
bulb is 8.0 mm and the height of its image is 40 cm.
a) Sketch a labeled ray diagram to show the formation of the bulb’s

image.
b) Calculate

i. the position of the bulb from the pole of the mirror,
ii. the focal length of the mirror.

Solution 7.1.4 8 mm P
(a) CF

40 cm

3.50 m

(b) i.

The position of the bulb is 7.0 cm in front of the mirror.

(b) ii.

40

Exercise :
1. 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

person relative to the surface of the globe?
b. How large is the person’s image?
c. State the characteristics of the person’s image.

(Virtual, upright, diminished, formed
behind the reflecting surface)

2. a. A concave mirror forms an inverted image four times larger than the object.
Calculate 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. (ANS. :160 mm ; 267 mm)

3. a. 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.
i. Is the shopper’s image upright or inverted? Explain.
ii. Determine the radius of curvature of the mirror.

b. 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. (ANS. : u think, 1.07 m ; 144 cm, 48 cm)

4. 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. ( 15cm, 5.0cm )

5. 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.

( 20cm,10cm behind the mirror )

6. What are the nature, size, and location of the image formed when a 6 cm tall

object is located 15 cm from a spherical concave mirror of focal length 20 cm ?
(virtual, upright, -60 cm, + 24 cm)

7. An object is placed 15 cm from a

a) concave mirror

b) convex mirror

of radius of curvature 20 cm. Calculate the image position and magnification in

each case. ( +30 cm,-2, -6 cm, 2/5) 43

8. 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? (16.4 cm)
b. Calculate the radius of curvature of the mirror and describe the type of
mirror required. (10.9 cm)
c. Sketch and label a ray diagram to show the formation of the image.

9. A spherical mirror forms a real image 18 cm from the surface. The image is
twice as large as the object. Find the location of the object and the focal length
of the mirror. ( 9.0 cm, 6.0 cm)

10. The magnification of a mirror is -0.333. Where is the object if its image is
formed on a card 540 mm from the mirror? What is the focal length?
(1.62 m, -405 mm)

44

Example 7.2.2

A cylindrical glass rod has an index of refraction 1.50. One end is ground to a
hemispherical surface with radius r = 20 mm. A point object on the axis of the
rod, 80 mm to the left of the vertex. The rod is in air.
Calculate the image distance.
Solution

n1 = 1 , n2 = 1.50 , r =+20 mm , u = + 80 mm

+ve (to the right of vertex → behind the surfa45ce)

Example 7.2.3

A set of coins is embedded in a spherical plastic paper-weight having a radius
of 3.0 cm. The index of refraction of the plastic is n1 = 1.50. One coin is
located 2.0 cm from the edge of the sphere. Find the position of the image of
the coin.

Solution

v

v = −1.7 cm

46

Example 7.2.4 Normal line

Refer to FIGURE 22.2 A, v
calculate the image distance.
Solution 8 cm

u = real/actual depth FIGURE 22.2 A 47
v = apparent depth

Exercise

1. A student wishes 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 1.

FIGURE 1

Calculate the depth of the pool. ANS. : 5.16 m
(Given nwater = 1.33 and nair = 1.00)

48

2. A goldfish is swimming inside a spherical plastic bowl of water, with an index of
refraction of 1.33. If the fish is 20.0 cm from the wall of the 40.0 cm radius bowl, where
does it appear to an observer outside the bowl ?
(17.2 cm from the wall , inside the bowl of water)

3. 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. Determine the position of the image.
(Given nglass = 1.52, nair = 1.00)
(20.83 cm in front of the 2nd refracting surface)

4. A point source of light is palced at a distance of 25.0 cm from the center of the glass sphere
of radius 10 cm. Determine the image position of the source.
(Given nglass = 1.52, nair = 1.00)
(25.2 cm at the back of the 2nd refracting surface)