Unit 18
SUBJECT
REFRACTION AT A SINGLE CURVED SURFACE,
VERGENCE
The section of the Core Competency that you will be
CORE
COMPETENCIES working towards in this unit is:
Foundation knowledge for:
4. Optical Appliances
The ability to dispense an appropriate optical appliance
WORK SET
Optics
ADDITIONAL
READING
ABDO College
Introductory Mathematics
Units 7, 13 & 15
LEARNING
OUTCOMES
After completing the work set you should be able to:
A H Tunnacliffe & J G Hirst
Sections 2.2.3 – 2.2.5 & 7.3.6.1
•
Define:
(a) converging and diverging spherical refracting
surfaces
(b) the terms vertex, centre of curvature, principal
axis of a surface
(c) vergence
(d) linear magnification in terms of reduced
vergences
•
Recall the sign convention applied to distances and
angles associated with refraction at a single
spherical surface
•
Explain what is meant by spherical aberration
•
Derive:
(a) the fundamental paraxial equation for refraction
at a single spherical surface
(b) Newton’s equation for a single refracting
surface.
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Unit 18
LEARNING
OUTCOMES
1
•
Calculate:
(a) the powers of convex and concave surfaces
(b) image position and size by means of the
fundamental paraxial equation and
magnification formulae
•
Draw diagrams illustrating the first and second focal
lengths of converging and diverging surface
•
Recall the construction rays necessary to draw ray
diagrams (to scale) to show the formation of images
produced by a single convex refracting surface and
by a single concave refracting surface, and to use
these rays to produce diagrams for the images
produced by all possible positions of the object.
YOUNG’ S CONSTRUCTION FOR A CURVED SURFACE
The graphical construction for the passage of a ray through a plane surface at an
oblique angle was undertaken in the Assignment for Unit 14.
(Note that this should be referred to as Dowell’s construction for a plane surface.)
Now we draw Young’s Construction for a curved (spherical) surface. Fig.7.31 of
Tunnacliffe & Hirst has been redrawn in Fig.1 in a little more detail.
In addition to the refracting surface itself, Young’s Construction requires that you draw
to scale two arcs of circles centred at C;
1.
With radius
=
2.
With radius
=
 n′ 
 ×r
n

n

 ′×r
n

An incident ray which strikes the refracting surface of radius r at D, also strikes the
construction arc of radius (n′ /n × r) at T1 (by continuation if necessary). Join T1C.
T1C intersects the construction arc of radius (n/n′ × r) at T2. Join DT2 (and continue if
necessary). The proof below shows that the path DT2 is the path of the refracted ray.
Applying the sine rule to triangle DT1C:
sin i sin CT1D
=
CT1
CD
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Refraction at a Single Curved Surface, Vergence
 n′ 
But CT1 =   r and CD = r.
n
Hence, n sin i = n′ sin CT1D
Since, n sin i
= n′ sin i′,
it is clear that angle i′ = angle CT1D.
Thus the two triangles, DT1C and DT2C are equi-angular and must be similar.
Hence angle CDT2 must equal angle CT1D = angle i′.
Thus the line DT2 represents the direction of the refracted ray.
Fig.1a Convex Surface
N
Convex
Surface
i
D
i′
n
A
n′
i
T2
C
N′
3
T1
Unit 18
Fig.1b Concave Surface
T1
i
T2
N′
2
N
Concave
Surface
C
i′
i′
D
n′
n
A
VERGENCE
The term “vergence” is used very extensively in practical optics, usually in connection
with convergence and divergence. It refers to the shape of wave-fronts.
Fig.2
Convergent light has wave-fronts
which are convex to the incident
direction.
Divergent light has wave-fronts
which are concave to the
incident direction.
4
Refraction at a Single Curved Surface, Vergence
3
REDUCED DISTANCE AND VERGENCE
Tunnacliffe & Hirst 2.2.3.2 refers to reduced vergence and uses a ‘bar’ notation for the
symbols.
As you will see from the description there is no difference between this ‘reduced’
′
n
vergence and the vergence as defined by or .

n′
Note that when light travels through a denser medium, it is slowed down.

In the same time that light travels a distance  in air, it will travel
( or  ) in a
n
refractive index n.

This distance is called the “reduced distance” or “equivalent air distance”.
n
Vergence is also referred to as ‘dioptric distance’.
‘Reduced’ distances were popular when calculators were not generally available and
accurate calculations had to be performed with 7 figure tables. Today they offer no
advantage and the omission of the refractive indices causes problems with the
construction of ray diagrams.
Fig.3a
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Unit 18
Imagine wave-fronts drawn through crests of waves emanating from O in Fig.3a.
If the distance from A to O is –50cm (minus because the direction is always measured
from the lens or surface or wave-front) then the vergence of the light arriving at A is
n
1
L= =
= –2.00D , i.e. it is diverging and of vergence –2.00D.
 –0.5
Fig 3b
Now let us imagine that in Fig.3b, the medium through which light travels is denser,
say n = 1.5. There would be 1½ times as many wave-fronts on arrival at A.
Hence L =
n 1.5
=
= −3.00D
 −0.5
Note that in Fig 3b the wavelength of the waves is
4
2
3
of those in Fig 3a.
USE OF THE CONJUGATE FOCI RELATIONSHIP
The fundamental paraxial equation or the conjugate foci relationship, is derived in
paragraph 2.2.3 of Tunnacliffe & Hirst. The fundamental paraxial equation or the
conjugate foci relationship is derived in paragraph 2.2.3 of Tunnacliffe & Hirst.
Alternatively, a video lecture can be accessed by visiting goo.gl/XAXo67 or by
scanning the QR code below.
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Refraction at a Single Curved Surface, Vergence
Use of this relationship enables us to find the position and size of the image of an
object which is formed by a refracting surface.
Note that at this stage it is very useful to imagine that the curved refracting surface is
ground on one end of a long glass or plastics rod so that the image is formed within the
material upon which the surface has been ground.
Having read paragraphs 2.2.3 to 2.2.5 in Tunnacliffe & Hirst you should work
carefully through the following examples.
Example 1
A convex surface on the end of a plastics rod of refractive index 1.498, lying in air,
has a radius of curvature of 83mm.
a)
What is the curvature of the surface?
b)
What is the power of the surface?
c)
What would be the power of the surface if the rod is placed in a tank containing
water of refractive index 1.332?
a)
The curvature of the surface, R =
1
when r is expressed in metres,
r
1000
when r is expressed in millimetres.
r
1000 1000
Hence=
R = = +12.05m −1
r
+83
or
b)
The power of the surface, F, is given by F =
in millimetres.
Here, n′ = 1.498 and n = 1, so F =
c)
1000 ( n′ − n )
r
when r is expressed
1000 (1.498 − 1)
= +6.00D
+83
When the rod is placed in water, its radius does not change but, since n is now
equal to 1.332,
its power under water
becomes F =
1000 ( n′ − n ) 1000 (1.498 − 1.332 )
=
= +2.00D
+83
r
which is one-third of its power in air!
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Unit 18
Example 2
The rod described in Example 1 is set up in air, 50cm from a 1cm high, luminous
object. Find the position and size of the image formed in the rod.
Fig.4
n=1
h
A
B
B′
ℓ = –50 cm
ℓ′
The refraction is illustrated in Fig.4.
We are given, ℓ = –50cm, n = 1.0 (the object lies in air), h = +1cm, n′ = 1.498,
and from Example 1, F = +6.00D.
Using the conjugate foci relationship
L′ = L + F
n
n′
where,
and F = +6.00D
L′ = , L =

′
100n′
100n
or when distances are substituted in cm: L′ =
and L =
'


100
we have L =
= −2.00D and since L′ = L + F = –2.00 + 6.00 = +4.00D
−50
Since
′ =
100n′
149.8
we get ′ =
= +37.45cm
L′
+4.00
So the image is formed 37.45cm to the right of A, the vertex of the surface.
The magnification of the image, m is given by
so m =
h′ L
=
h L′
h′ L −2
=
=
= −0.5 and h′ = m × h = –0.5 × 1.0 = –0.5cm.
h L′ +4
Thus the rod forms a real inverted image of the object.
The image lies 37.45cm to the right of the vertex (the image distance ℓ′ has a plus
sign) and is of height 0.5cm and is inverted (since the image height, h′, carries a minus
sign).
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Refraction at a Single Curved Surface, Vergence
Example 3
The object and the rod described in Example 2 above are now placed under water (the
refractive index of the water is 1.332). What is the new position and size of the image?
We are given, ℓ = –50cm, n = 1.332 (the object now lies in water), h = +1cm,
n′ = 1.498, and from Example 1 the power of the surface, F, in water is +2.00D.
Using the conjugate foci relationship
where,
L′ =
n′
,
′
L=
n

L′ = L + F
and F = +2.00D
or when distances are substituted in cm L′ =
100n′
100n
and L =
'


133.2
= −2.664D
−50
and since L′ = L + F = –2.664 + 2.00 = –0.664D
we have
L=
Since
′ =
100n′
149.8
we get ′ =
= −225.6cm
−0.664
L′
So the image is formed –225.6cm to the left of the vertex of the surface, A.
The magnification of the image, m is given by
so m =
h′ L
=
h L′
h′ L −2.664
=
=
= +4.01 and h′ = m × h = +4.01 × 1.0 = +4.01cm.
h L′ −0.664
Thus the rod now forms a virtual erect image of the object.
The image lies –225.6cm to the left of the vertex (the image distance ℓ′ has a minus
sign) and is of height +4.01cm and is erect (since the image height, h′, carries a plus
sign). The image formation in this case is illustrated in Fig.5. Note that the diagram is
not to scale.
Fig.5
n = 1.332
h′
h
B
B′
A
ℓ = –50 cm
ℓ′
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Unit 18
5
NOTE ABOUT NEUTRALISATION
Remember that neutralisation forms part of your practical examination; you are
required to neutralise a pair of lenses by hand. Continue to practice hand neutralisation
from now until the examination. If you consistently neutralise two pairs of spectacles
daily in your practice and check by focimeter, this should provide you with sufficient
practise.
10
Refraction at a Single Curved Surface, Vergence
Assignment 18
Remember that marks are allocated for diagrams and so marks are deducted
where relevant diagrams are omitted.
1.
2.
3.
a)
What is the curvature of a surface whose radius is 83.33mm?
b)
What is the sag of this surface over a diameter of 60mm?
[5]
[15]
A pencil of light is made to converge to a point 40cm beyond a lens. Find the
curvature of the wavefront:
a) on leaving the lens,
[5]
b)
5cm from the lens,
[5]
c)
39.9cm from the lens,
[5]
d)
41cm from the lens.
[5]
a)
An object 1cm high is placed 20cm in front of a convex spherical surface of
25cm radius of curvature which separates air from glass of refractive index
1.60. Find the position and size of the image.
[40]
(Note: a diagram is needed for this question.)
b)
What are the positions of the principal foci of this surface?
[20]
4.
An object 2cm high is placed 10cm in front of a convex spherical surface of
6.64cm radius of curvature which separates air from plastic of refractive index
1.498. Find the position and size of the image.
[40]
(Note: a diagram is needed for this question.)
5.
An object 1cm high is placed 30cm in front of a concave spherical surface of
60mm radius of curvature which separates air from glass of refractive index 1.90.
Find the position and size of the image.
[40]
(Note: a diagram is needed for this question.)
Please make sure you show your Supervisor your completed assignment so they can
ensure it is complete and to the required standard, and remember you can contact
your Tutor for support if you have any issues.
11