EP025 Chapter 6 Geometrical Optics

CHAPTER 1 : GEOMETRICAL OPTICS
( 5 HOURS )

Terminology

1. Geometrical Optics : Optik Geometri. JOTTING
2. Light : Cahaya. SPACE
3. Ray : Sinar.
4. Beam of light : Alur cahaya.
5. Reflection : Pantulan.
6. Incident ray : Sinar tuju.
7. Plane mirror : Cermin rata.
8. Spherical mirror : Cermin sfera.
9. Radius : Jejari.
10. Curvature : Kelengkungan.
11. Principal axis : Paksi utama.
12. Convex : Cembung.
13. Concave : Cekung.
14. Virtual : Maya.
15. Upright : Menegak.
16. Inverted : Songsang.
17. Laterally reverse : Songsang sisi.
18. Magnification : Pembesaran.
19. Refraction : Pembiasan.
20. Interference : Interferens.
21. Diffraction : Pembelauan.
22. Polarization : Pengutuban.
23. Ray diagram : Rajah sinar.
24. Thin lenses : Kanta nipis.
25. Focal length : Panjang fokus.
26. Device : Peralatan.

Introduction :

1. Geometrical optics describes the phenomena related to the ray of light
(using the principle of trigonometric); however, it is not concerned with its
nature or with the reasons for the phenomena. It is used to determine the
properties of images by using scaled diagram.

2. Properties of a ray of light :
a) A form of energy (known as photon which depends on frequency).
b) Propagates in a straight line.
c) Has constant velocity, c ≈ 3.0x108 ms-1 (1080 million kmh-1).
d) An element of electromagnetic waves.

3. Phenomena related to light :
a) Reflection.
b) Refraction.
c) Interference.
d) Diffraction.

4. Generally, the properties of images formed from reflection and refraction can
be determined by using 3 different methods :
a) Experiment (practical method).
b) Related physics equation (mathematical method) based on the sign
convention.
c) Ray diagram (geometrical method).

| Chapter 1 | Geometrical Optics | page 1 / 22 |

Reflection :

1. Reflection refers to the return of all or part of beam of light when it encounters
the boundary between two media.

2. Images of pictures can be seen in a mirror when the light that reaches the
mirror is reflected to the eyes of the observer.

3. Laws of reflection :
i) The angle of incidence, i is equal to the angle of reflection, r.
∠ θi = ∠ θf
ii) The incident ray, the reflected ray and the normal are all lie in the
same plane.

JOTTING
SPACE

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4. Equation related to reflection (mathematical method) : …..(1.1)
1= 1+1
f uv

where f = focal length (length between the mirror and the focal point).
u = object distance (length between the mirror and the object).
v = image distance (length between the mirror and the image).

Note : Focal point, F is a point where all reflected rays of parallel incident rays
intersect from the mirror.

The rule of sign (Sign Convention) :

+ve -ve
Virtual object
u Real object (behind the mirror)
(in front the mirror) Convex mirror
(Diverging mirror)
Concave mirror Focal point is located
behind the mirror
f (Converging mirror) Virtual image
Focal point is located in (behind the mirror)

front of the mirror (Upright)

Real image

v (in front the mirror)

(Inverted)

Note : Equation (1.1) also can be written as :

v = uf
u−f

Based on the above equation, we can determine where the object
should be placed in order to form the real or virtual image.

| Chapter 1 | Geometrical Optics | page 2 / 22 |

5. Properties (characteristics) of image :
i. Real OR virtual / apparent.
ii. Upright / erect OR inverted.
iii. Enlarge / magnified OR diminished OR same size.

Note : Naturally, real image must be inverted and virtual image must be upright.

Real image → Inverted.
Virtual image → Upright.

Note : Virtual object is generally found if ≥ 2 mirror surfaces are being used.
If the image produced from the first mirror (primary image) is formed
behind the second mirror, the image is then being identified as the
virtual object for the second mirror (secondary object).

6. Linear magnification, M : …..(1.2)
= The ratio of height h of an image to height H of the object.

M= h = v
Hu

Note : If M < 1 then, the image is diminished. JOTTING
If M = 1 then, the image is same size as the object. SPACE
If M > 1 then, the image is enlarged.

Reflection of plane mirror :

1. The focal length of the plane mirror, f → ∞ (all reflected rays of parallel incident
rays from the mirror never intersect each other).

Based on equation (1.1) : ; if f → ∞, then 1 → 0.
1= 1+1 f
f uv

∴ 0= 1+ 1
uv

∴ v = -u

Hence, the image formed in a plane mirror (characteristics of image) :
i. Apparent / virtual (the sign of v is -ve)
ii. Upright / erect.
iii. Same size as object (magnitude of v = magnitude of u, then M = 1)

| Chapter 1 | Geometrical Optics | page 3 / 22 |

2. The properties of the image can be also determined by using a ‘ray diagram’

(geometrical method) based on the law of reflection by choosing any point on

the object : uv

B JOTTING
SPACE
C

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Note : To use the ray diagram, it needs only two incident rays. More incident
rays are needed just only to verify the intersection point.

Note : The normal line is always normal @ perpendicular to the mirror surface.

Steps taken to draw a ray diagram for reflection of plane mirror (use geometry
set) :

(Zahidi’s Rule of Reflection, 2006 : plane mirror)

i. Draw the plane mirror.
ii. Draw the object at distance u from the mirror (the object can be drawn at

the left side or even right side of the mirror).
iii. Choose any one point on the object and label as A (normally as O).
iv. Draw one incident ray which connects point A to any point on the mirror

(label as B). Show the direction of the ray using an arrow.
v. Draw the normal line on the mirror.
vi. Measure the angle of incident, ∠i.
vii. Measure the angle of reflection, ∠r (make sure ∠i = ∠r).
viii. Draw the reflected ray which connects point B through angle ∠r.

Show the direction of the ray using an arrow.
ix. Repeat steps iv to viii for other point on the mirror (label as C).
x. Determine the intersection point between the two reflected rays drawn

before. Label the point as A’ (normally as I). This point actually is the
image of the point object A chosen before.
xi. Measure the distance v from the mirror.
xii. Determine the properties of the image produced.

Note : In the ray diagram, image produced is considered real if the image is
formed in front of the mirror (same side as the real object) and vise versa.

Based on the ray diagram, the characteristics of the image produced by a plane
mirror are always the same (no matter where the object is being located in front
of the mirror), i.e. :
i. Apparent / virtual (it is formed behind the mirror).
ii. Upright / erect.
iii. Same size as object.
iv. Laterally reverse (can be determined only by ray diagram).

| Chapter 1 | Geometrical Optics | page 4 / 22 |

Example :

A boy is standing 50 cm from a long plane mirror. Find the distance from (a) the boy’s
image to the mirror (b) the boy to his image. [Ans: (a) v = -50cm (b) d = 100cm]

Reflection of spherical mirrors :

1. Spherical mirror can be divided into 2 types :
i. Convex spherical mirror (diverging mirror).
ii. Concave spherical mirror (converging mirror).

The location and properties (characteristics) of an image depend on the distance
between the object and the mirror, u.

2. Convex spherical mirror :
= A spherical surface (outer surface of the part of a sphere) that reflects light.
The parallel rays are diverged from the focal point, F.

JOTTING
SPACE

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Formation of images by a convex spherical mirror (based on the ray diagram) :

Position of Ray diagram Properties of
object image

normal
line

Anywhere Virtual
Upright
Diminished

C

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Note : In spherical mirror, the normal line is any line which connects the centre
of the sphere, C to the surface of the mirror (always normal @
perpendicular to the mirror surface).

| Chapter 1 | Geometrical Optics | page 5 / 22 |

Note : Determination of the properties of image of a real object of reflection on
the convex spherical mirror using mathematical method :

Based on equation (1.1) :

1= 1+1
f uv

or v = uf …..(i)
u−f

For convex mirror → -f
For real object → +u

Hence, v = (+u)(−f )
(+u) − (−f )f

= -  uf f  …..(ii)
 u+ 

Since the sign of v is –ve, then the image produced is always virtual
and of course it is upright.

Magnification, M = v …..(iii) JOTTING
u SPACE

Substitute (ii) into (iii), then :

M=  uf f 
 u+ 

u

=  u 2 uf uf 
+

Then, the value of M is always < 1 (diminished) due to the factor of u2.

| Chapter 1 | Geometrical Optics | page 6 / 22 |

3. Concave spherical mirror :
= A spherical surface (inside area of the part of a sphere) that reflects light. The
parallel rays are converged to the focal point.

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Formation of images using a concave spherical mirror :

Position of object Ray diagram Properties of
image

When the object is Virtual JOTTING
nearer than F Upright SPACE
(u < f). Enlarged

When the object is Real
between F and C Inverted
Enlarged
(f < u < 2f).

When the object is at Real
C Inverted
Same size
(u = 2f).

When the object is Real
further than C Inverted
(u > 2f). Diminished

Distance PC = Radius of curvature, r.
Distance PF = Focal length, f.
Distance PFC = Principle axis.

| Chapter 1 | Geometrical Optics | page 7 / 22 |

Relationship between focal length and radius of curvature :

f= r …..(1.3)
2

4. Based on equation (1.1), the equation of reflection of spherical mirror
(mathematical method) :
1= 1+1
f uv

or 2 = 1+ 1 …..(1.4)
r uv

Steps taken to draw a ray diagram for reflection of spherical mirror (use
geometry set):

(Zahidi’s Rule of Reflection, 2006 : spherical mirror)

i. Draw the spherical mirror (concave mirror or convex mirror) of radius of JOTTING
curvature, r. SPACE

ii. Draw the principle axis and label points P, F and C respectively stands
for pole of the mirror, focal point and centre of sphere.

iii Draw the object at distance u from the mirror (normally in the shape
of upright-arrow).

iv. Choose any one point on the object and label as A (normally as O).
v. Draw a normal line which connects the centre of sphere, C to the

object A and the mirror.
vi. Draw two incident rays as well as the respective reflected rays from

the listed pairs below. Show the direction of each ray using an arrow.

a) Convex mirror :

Incident ray Reflected
(connect point A to the mirror) (from the mirror)

Incidents through the normal line Reflects through the normal line
Parallel with the principle axis

Converges the focal point, F Parallel with the principle axis

Parallel with the principle axis Diverges from the focal point, F

Towards the pole of the mirror, P Away from the pole of the mirror,
P with ∠i = ∠r

b) Concave mirror :

Incident ray Reflected
(connect point A to the mirror) (from the mirror)

Incidents through the normal line Reflects through the normal line

Converges the focal point, F Parallel with the principle axis

Parallel with the principle axis Converges the focal point, F
Towards the pole of the mirror, P
Away from the pole of the mirror,
P with ∠i = ∠r

vii. Determine the intersection point between the two reflected rays drawn
before. Label the point as A’ (normally as I). This point actually is the

image of the object of point A chosen before.

| Chapter 1 | Geometrical Optics | page 8 / 22 |

viii. Measure the distance v from the mirror.
ix. Determine the properties of the image produced.

Example :

An object is placed 20 cm from (a) concave mirror (b) convex mirror, of radius of
curvature 25 cm. Calculate the image position and magnification in each case.
[Ans: (a) v = 33.3cm, M = 1.67 (b) v = -7.69cm, M = 0.38]

Refraction :

1. Refraction refers to the change of direction suffered by wave front of light as it
passes obliquely from one medium to another in which its speed of propagation is
changed.

When light moves from one medium to another, it does not propagate along a
straight line. A ray of light deviates from the original direction of propagation.
The way in which a ray propagates through different neighboring media is
described by Snell’s law.

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JOTTING
SPACE
2. Laws of refraction :

i) The incident and refracted rays are on the opposite sides of the
normal at the point of incidence, and all three lie in the same
plane.

ii) Snell’s law :

sin α = nr where n = refractive index (constant).
sin β ni

original
direction

Note : If the incident ray is in air (nair = 1), then :

sin α =n where n = refractive index of the refractive medium.
sin β

3. The refractive index also can be defined as :

= The ratio of the speed of light in vacuum, c to the speed of light in a medium, v.

n = the speed of light in vacuum, c = λ c fc where fc = fv
the speed of light in a medium, v λ v fv

| Chapter 1 | Geometrical Optics | page 9 / 22 |

Note : Refraction of light Reflection of light
Incident angle, θi Weak
0o < θi < θc Strong
Strong
θi = θc Weak (perpendicular Total internal
to the normal line)
θi > θc reflection
None

Refraction of plane surface :

1. Light passing into an optically denser medium is bent towards the normal.
Whereas, light into an optically less dense medium is bent away from the normal.

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JOTTING
SPACE
n2 > n1 n2 < n1

Steps taken to draw a ray diagram for refraction of plane surface
(use geometry set) :

(Zahidi’s Rule of Refraction, 2006 : plane surface)

i. Draw the plane surface of refractive index n2.
ii. Draw an object at distance u from the surface (normally in the form of

point called ‘point object’) in a medium of refractive index n1 and label
as O.
iii. Draw one incident ray which connects point O to any point on the surface
(label as A). Show the direction of the ray using an arrow.
iv. Draw a normal line at A.
v. Measure the angle of incident, ∠α.
vi. Based on Snell’s law, calculate the angle of refraction, ∠β.
vii. Draw the refracted ray which connects point A through angle ∠β.
Show the direction of the ray using an arrow.
viii. Repeat steps iii to vii for other point on the surface (label as B).
ix. Determine the intersection point between the two refracted rays drawn
before. Label the point as I. This point actually is the image of the object O.
x. Measure the distance v from the surface.
xi. Determine the properties of the image produced.

Example :

A light ray israveling in air is incident on a smooth, flat slab of glass at angle of 60.0o to
the plane. Find the angle of refraction in the refractive index of the glass is 1.52.
[Ans: θr = 19.2o]

| Chapter 1 | Geometrical Optics | page 10 / 22 |

Example :

A fifty cent coin is at the bottom of a swimming pool of depth 2.00 m. The refractive
index of air and water are 1.00 and 1.33 respectively. What is the apparent depth of the
coin observed by a man beside the pool? [Ans: d’ = 1.50m]

Refraction of spherical surface :

1. Spherical surface can be divided into 2 types :
i) Convex spherical surface.
ii) Concave spherical surface.

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JOTTING
SPACE
Convex surface Concave surface

2. An object P is placed in a medium of refractive index n’. When light refracts
at a spherical surface of curvature radius r, the formation of an image of the
object depends on the medium of refractive index n.

3. Equation related to refraction of plane and spherical surface (mathematical
method) :

n' + n = n − n' …..(1.5)
uv r

or ni + nr = nr − ni
uv r

where u = object distance.
v = image distance.
r = radius of curvature.

The rule of sign (Sign Convention) :

+ve -ve
Virtual object
u Real object (behind the surface)
(in front the surface) Concave surface
(diverging refracted rays)
Convex surface The centre of curvature is
located in less dense medium.
r (converging refracted rays)
The centre of curvature is Virtual image
(in front the surface)
located in more dense medium
(Upright)
Real image

v (behind the surface)

(Inverted)

| Chapter 1 | Geometrical Optics | page 11 / 22 |

Note : For plane surface, r → ∞ ; if → ∞, then n − n' → 0
r
Then, based on equation (1.5) :
n' + n = n − n'
uv r

∴ n' + n = 0
uv

∴ n' = − n
uv

Steps taken to draw a ray diagram for refraction of spherical surface
(use geometry set) :

(Zahidi’s Rule of Refraction, 2006 : spherical surface)

i. Draw the spherical surface (concave surface or convex surface) of JOTTING
radius of curvature, r and refractive index, n. SPACE

ii. Draw the principle axis and label points P and C respectively stands for
pole of the surface and centre of sphere.

iii Draw the object at distance u from the surface (normally in the form of
point called ‘point object’) on the principle axis and label as O.

iv. Draw one incident ray which connects point O to any point on the
surface (label as A). Show the direction of the ray using an arrow.

v. Draw a normal line which connects point A to the centre of sphere, C.
vi. Measure the angle of incident, ∠α.
vii. Based on Snell’s law, calculate the angle of refraction, ∠β.
viii. Draw the refracted ray which connects point A through angle ∠β.

Show the direction of the ray using an arrow.
ix. Repeat steps iv to viii for other point on the surface (label as B).

x. Determine the intersection point between the two refracted rays drawn
before.Label the point as I. This point actually is the image of the object O.

xi. Measure the distance v from the surface.
xii. Determine the properties of the image produced.

Example :

A point object O is placed inside a glass sphere of radius 20.0 cm. If the refractive index
of glass is 1.5 and the object is at a distance of 9.0 cm from the surface, (a) calculate the
image distance if the object is being observed from the nearest surface (b) state whether
it is real or virtual image. [Ans: (a) -7.06cm (b) virtual]

| Chapter 1 | Geometrical Optics | page 12 / 22 |

Thin lenses :

1. A lens is considered thin if its thickness is much smaller than the radius of
curvature of the surface that forms the boundaries of the lens.

2. A lens is a transparent body comprising two spherical surfaces or a spherical
surface and a flat surface.

3. Several types of lens :

Biconcave lens Biconvex lens

© Physics Teaching Courseware © Physics Teaching CoursewarePlano concave lensPlano convex lens
JOTTING
SPACE© Physics Teaching Courseware

4. Generally, thin lenses can be classified into two types :
i) Convex lens.
ii) Concave lens.

Convex lens

Concave lens

| Chapter 1 | Geometrical Optics | page 13 / 22 |

5. When considering the propagation of light, we can apply the so-called ‘lens

formula’ as follow :

1= 1+1 …..(1.6)
f uv

where f = focal length.
u = object distance.
v = image distance.

The rule of sign (Sign Convention) :

+ve -ve

u Real object Virtual object
(in front the lens) (behind the lens)

Convex lens Concave lens
(the parallel rays are
f (the parallel rays are diverged from the focal
converged to the focal
point).
point). Virtual image
(in front the lens)
Real image
(Upright)
v (behind the lens)

(Inverted)

Note : Equation (1.6) also can be written as : JOTTING
SPACE
v= uf
u−f

Based on the above equation, we can determine where the object should
be placed in order to form the real or virtual image.

6. Linear magnification, M :
= The ratio of height H of an image to height h of the object.

M= h = v …..(1.7)
Hu

| Chapter 1 | Geometrical Optics | page 14 / 22 |

Derivation : Lens Formula

n1 n1
n2

O 1st surface 2nd surface I Io

of r1 of r2

uv

vo

1. Consider a ray from an object O in a medium of refractive index n1 passes
through the first surface of a thin lens of refractive index n2 and form an image Io.

Based on equation (1.5) :

n1 + n2 = n2 − n1 ; r1 = convex surface
u vo + r1

2. Io then acts as a virtual object for the second surface of the lens and finally JOTTING
refracted to form the actual image, I. SPACE

Based on equation (1.5) :

n2 + n1 = n1 − n2 ; r2 = concave surface
− vo v + r2

3. Adding up the two equations :

n1 + n1 = (n2 − n1 ) 1 + 1 
u v r1 r2

4. Dividing the equation by n1 :

1 + 1 =  n2 − 1 1 + 1  …..(i)
u v n1 r1 r2

5. If the object O is at infinity, then u = ∞.
Hence, the image is formed at the focal length, v = f.
Then :

1 + 1 =  n2 − 1 1 + 1  …..(ii)
∞ f n1 r1 r2

6. Comparing equation (i) and (ii) : …..shown.
1+ 1=1
uv f

Note : Equation (ii) is well known as the ‘lens-makers equation’.

| Chapter 1 | Geometrical Optics | page 15 / 22 |

7. Formation of images using a concave lens :

Position of object Ray diagram Properties of
image

Anywhere Virtual
Upright
Diminished

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Note : Determination of the properties of image of a real object of refraction on
the concave lens using mathematical method :

Based on equation (1.6) :

1= 1+1
f uv

or v= uf …..(i) JOTTING
u−f SPACE

For concave lens → -f
For real object → +u

Hence, v = (+u)(−f )
(+u) − (−f )f

= -  uf f  …..(ii)
 u+ 

Since the sign of v is –ve, then the image produced is always virtual
and of course it is upright.

Magnification, M = v …..(iii)
u

Substitute (ii) into (iii), then :

M =  uf f 
 u+ 

u

=  u 2 uf uf 
+

Then, the value of M is always < 1 (diminished) due to the factor of u2.

| Chapter 1 | Geometrical Optics | page 16 / 22 |

8. Formation of images using a convex lens :

Position of object Ray diagram Properties of
image
When the object is
nearer than f Virtual
(u < f). Upright
Enlarge

When the object At ∞
at f Inverted
Enlarge
(u = f).

When the object is Real JOTTING
between f and 2f Inverted SPACE
Enlarge
(f < u < 2f).
Real
When the object Inverted
at 2f Same size

(u = 2f). Real
Inverted
When the object is Diminished
further than 2f
(u > 2f). © Physics Teaching Courseware

| Chapter 1 | Geometrical Optics | page 17 / 22 |

Steps taken to draw a ray diagram for refraction of thin lenses
(use geometry set) :

(Zahidi’s Rule of Refraction, 2006 : thin lenses)

i. Draw the lens (concave lens or convex lens).
ii. Draw the principle axis and label points P, F and 2F respectively stands

for pole of the lens and focal point at both side of the lens.
iii Draw the object at distance u from the lens (normally in the shape of

upright-arrow) and label as O.
iv. Draw two incident rays as well as the respective refracted rays from the

listed pairs below. Show the direction of each ray using an arrow.

a) Concave lens :

Incident ray Reflected
(connect point O to the lens) (from the lens)
Parallel with the principle axis
Diverges from
Towards the pole of the lens, P the incident focal point, F

Away from the pole of the lens
through the same path

b) Convex lens : JOTTING
SPACE
Incident ray Reflected
(connect point O to the lens) (from the lens)
Parallel with the principle axis
Converges the refracted
Towards the pole of the lens, P focal point, F

Away from the pole of the lens
through the same path

Towards the incident focal point, F Parallel with the principle axis

v. Determine the intersection point between the two reflected rays drawn
before. Label the point as I. This point actually is the image of the

object O.
vi. Measure the distance v from the lens.
vii. Determine the properties of the image produced.

| Chapter 1 | Geometrical Optics | page 18 / 22 |

Lensmaker’s equation :

1. Properties of lenses determined by their shape :

Types of lens Diagram Sign of curvature
radius, r
n1
r1 = +ve
Biconvex n2 r2 = +ve
Lens

Biconcave n1 r1 = -ve
Lens n2 r2 = -ve

JOTTING
SPACE

Meniscus n1 r1 = +ve
Convex n2 r2 = -ve

Lens n1 r1 = +ve
n2 r2 = -ve
Meniscus
Concave

Lens

Plano n1 r1 = +ve
Convex n2 r2 = ∞

Lens

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| Chapter 1 | Geometrical Optics | page 19 / 22 |

Plano n1 r1 = ∞
Concave n2 r2 = -ve

Lens

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2. In order to determine the focal length of a lens of refractive index n2 surrounded
by a medium of refractive index n1 and the radius of curvatures of the border
surfaces are r1 and r2 respectively, we can apply the so-called ‘lens-maker’s
equation’ as follow :

1 =  n2 − 1 1 + 1  …..(1.8)
f n1 r1 r2

Note : If the lens is located in air (n1 = 1), then, the equation becomes :

1 = (n − 1) 1 + 1 
f r1 r2

where n = refractive index of the lens used. JOTTING
SPACE

| Chapter 1 | Geometrical Optics | page 20 / 22 |

Example : JOTTING
A plano-concave lens has a curved surface of radius 20.0 cm. If the focal length of this SPACE
lens is 30.0 cm, what is the refractive index of this material? [Ans: n = 1.67]

Example :
A glass lens of refractive index 1.5 has a focal length of 10.0 cm in the air. If it is
immersed in a transparent liquid of refractive index 1.4, what will the focal length be?

[Ans: f = 70cm]

Example :
An object is placed at a distance of 15.0 cm from a convex lens of focal length 20.0 cm.
(a) Calculate the image position (b) Describe the properties of image formed.
[Ans: (a) v = -60cm]

Example :
Find the possible positions where a converging lens of focal length 8.0 cm can
produce image of an object on a screen located 40.0 cm from the object.
[Ans: u = 28.9cm or 11.06cm]

| Chapter 1 | Geometrical Optics | page 21 / 22 |

System of Two Separated Converging Lenses :

Primary lens Secondary lens
(Objective lens) (Eyepiece lens)

Primary Primary Secondary Secondary
object image = object image

up vp JOTTING
fp SPACE

fs
us vs

1. The secondary lens can only be used if the primary image is real no matter it is
located in front or behind the secondary lens.

2. The primary image acts as the secondary object. If the primary image is located
in front of the secondary lens, then it is a real secondary object. Whereas, if the
primary image is located behind the secondary lens, then it is a virtual secondary
object.

3. Since the primary image is also referring as the secondary object, then size of
primary image, Hp and secondary object, hs is the same.

If the size of primary object and secondary image are hp and Hs respectively, the
total magnification of the two lens is :

M= Hs =  Hs x hs  ; hs = Hp
hp hs hp

=  Hs x Hp 
hs hp

= MsMp

| Chapter 1 | Geometrical Optics | page 22 / 22 |