Slide 1
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Slide 2
___________________________________
Multiple Lenses
An object 6 cm tall is placed 40 cm from a thin converging (double convex) lens
of f = 8 cm. A second converging lens of 12 cm focal length is placed 20 cm from
the first lens. Determine the position, size and character of the final image.
___________________________________
___________________________________
___________________________________
___________________________________
-40 cm
-30 cm
-20 cm
-10 cm
+10 cm
+30 cm
___________________________________
f = +12
f = +8
___________________________________
The image of the first lens can be used as the
object of the second lens
Slide 3
___________________________________
Multiple Lenses
An object 6 cm tall is placed 40 cm from a thin converging (double convex) lens
of f = 8 cm. A second converging lens of 12 cm focal length is placed 20 cm from
the first lens. Determine the position, size and character of the final image.
___________________________________
___________________________________
___________________________________
___________________________________
-40 cm
-30 cm
The first lens
is a double
convex lens
s  40
s  ?
f 8
-20 cm
-10 cm
1 1 1
 
s s f
+10 cm
___________________________________
f = +8
s
M 
s
s„ is positive,
10
Object is in front

therefore image
is
1
1
1
of lens, so
40
 sis
lens
s 8 40 behind the
positive
 0.25
1 1 1
 
s f s
s  10
+30 cm
f = +12
Since the magnification is
negative, the image is inverted
at (0.25)(6cm)=1.5 cm tall
___________________________________
Slide 4
___________________________________
Multiple Lenses
An object 6 cm tall is placed 40 cm from a thin converging (double convex) lens
of f = 8 cm. A second converging lens of 12 cm focal length is placed 20 cm from
the first lens. Determine the position, size and character of the final image.
___________________________________
___________________________________
___________________________________
___________________________________
-40 cm
-30 cm
The second
lens is a
double
convex lens
-20 cm
-10 cm
+10 cm
1 1 1
 
s s f
s  10
Object is 10 cm
1 of 1second
1
in front
 
lens,sso
is 10
 s12
positive
s  ?
f  12
s  60
Slide 5
___________________________________
f = +12
f = +8
1 1 1
 
s f s
+30 cm
___________________________________
Therefore the final
s
M 
image
is inverted is
at 9
Since the
magnification
s
cmthe
tallimage
and isislocated
positive,
not inverted
s„ is negative,,
60
cm to the
left
oftall
  image is
and is 40
(1.5cm)(6)
=9
cm
therefore
10
the left most lens.
60 cm in front of

6
the lens
___________________________________
Multiple Lenses
An object is placed 20 cm from a thin diverging (double concave) lens of f = 10 cm.
A second lens (converging) of 20 cm focal length is placed 10 cm to the right of the
first lens. Determine the position, size and character of the final image.
___________________________________
___________________________________
___________________________________
___________________________________
-20 cm
-10 cm
20 cm
f = - 10
30 cm
40 cm
50 cm
60 cm
f = +20
___________________________________
___________________________________
The image of the first lens can be used as the
object of the second lens
Slide 6
___________________________________
Multiple Lenses
An object is placed 20 cm from a thin diverging (double concave) lens of f = 10 cm.
A second lens (converging) of 20 cm focal length is placed 10 cm to the right of the
first lens. Determine the position, size and character of the final image.
___________________________________
___________________________________
___________________________________
___________________________________
-20 cm
-10 cm
The first lens
is a double
concave lens
s  20
s  ?
f  10
20 cm
f = - 10
40 cm
50 cm
f = +20
1 1 1
 
s f s
Object is in front
1
1
1
of lens,so s is
s 10 20
positive
s  6.7
30 cm
s
s
Since the magnification is
6.67

positive, the image is upright
20
s„ is negative,
1 image is
therefore

3 of the lens
in front
M 
60 cm
___________________________________
___________________________________
Slide 7
___________________________________
Multiple Lenses
An object is placed 20 cm from a thin diverging (double concave) lens of f = 10 cm.
A second lens (converging) of 20 cm focal length is placed 10 cm to the right of the
first lens. Determine the position, size and character of the final image.
___________________________________
___________________________________
___________________________________
___________________________________
-20 cm
-10 cm
The second
lens is a
double convex
lens
s  16.67
s  ?
f  20
30 cm
20 cm
60 cm
f = +20
f = - 10
M 
1 1 1
 
s f s
___________________________________
s
s
s  100
___________________________________
Multiple Lenses

___________________________________
Since the magnification is
100
positive, the image is upright

s„ isnegative,
16.67
therefore
image
is
6
in front of the lens
Object is in front
1 so
1
1
of lens,
 sis
s 20 16.67
positive
Slide 8
50 cm
40 cm
We determine the effect of a system of lenses by considering the
image of one lens to be the object for the next lens.
___________________________________
___________________________________
___________________________________
+1
0
-1
+2
+3
+4
___________________________________
In front
behind
\
1 1 1
1 1
   1

s1' f1 s1
1.5 3
s1 = +1.5, f1 = +1
s1'  3
In front
For the second lens:
\
f = -4
f = +1
For the first lens:
s2'  0.8
Slide 9
'
1
Insfront
m1  
 2
s1
'
2
s
4

s2
5
m  m1m2  
___________________________________
8
5
___________________________________
Multiple Lenses

___________________________________
1 1 1
1 1
5
  
 
s2'
f 2 s2 4 1
4
s2 = +1, f2 = -4
m2  
+6
+5
Objects of the second lens can be virtual. Let‟s move the second
lens closer to the first lens (in fact, to its focus):
___________________________________
___________________________________
+1
0
-1
___________________________________
+2
+3
+4
+6
+5
___________________________________
f = +1
For the first lens:
f = -4
s1 = +1.5, f1 = +1
1
'

1

1
 1
1

1
sfor
f1 second
s1 lens
1.5is VIRTUAL.
3
TheObject
objectopposite
1 the
s'
'
we will use the BST (Burns
m1   1  2 Therefore
side
as
light,
\ s1  3
s1
Schlueter
Theorem)
for ray tracing. The
therefore
1negative.
1 1
1
1 1of its
lens will pretend
For the second lens: s2 = -2, f2 = -4
 to havethe negative


'
focal lengthsand
thus
f 2 sopposite
4 properties.
2 4
2
2
s2'
The diverging lens will now pretend to be a
'
\ s2  4 m2   s  2 converging lens.
2
m  m1m2  4
Note the negative object distance for the 2nd lens.
___________________________________
___________________________________
Slide
10
___________________________________
Problem 1

___________________________________
Suppose we interchange the converging and diverging lenses in the
preceding case.

1A
What is the relation of the new magnification m‟
to the original magnification m ?
(a) m’ < m
___________________________________
(c) m’ > m
(b) m’ = m
___________________________________
___________________________________
___________________________________
-1
1B
0
f = -4
+2
+3
+4
+5
+6
___________________________________
• What is the nature of the final image?
(a) real
Slide
11
+1
f = +1
(b) virtual
___________________________________
Problem 1

1A
___________________________________
Suppose we interchange the converging and diverging lenses in the
preceding case.

What is the relation of the new magnification m‟
to the original magnification m ?
___________________________________
___________________________________
___________________________________
-1
0
f = -4
+1
f = +1
+2
+3
+4
+5
+6
• Since the formula for the magnification is equal to the product of the
magnifications of each lens (m = m 1 m 2), you might think that interchanging the
lenses does not change the overall magnification.
• This argument misses the point that the magnification of a lens is not a property
of the lens, but depends also on the object distance!
• Consider the ray shown which illustrates that the magnification must be < 1!
Slide
• What is the nature of the final image?
(a) real
___________________________________
___________________________________
Problem 1
12
___________________________________
___________________________________
(b) virtual
___________________________________
1B
___________________________________
___________________________________
-1
0
f = -4
+1
f = +1
+2
+3
+4
+5
+6
___________________________________
___________________________________
• The ray used in part A actually shows that the image is real and inverted.
• The equations:
1 1 1
1 1 11
   

s1' f1 s1
4 1.5 12
1 1 1
11 12
   1 
s2'
f 2 s2
23 23
Slide
___________________________________
The Microscope
13
___________________________________
___________________________________
M angular  mobjective M eyepiece
 25cm  L 



f

 eyepiece  f objective 
___________________________________
___________________________________
___________________________________
L
Slide
14
___________________________________
Microscope Question
A microscope has an objective of focal length 0.300 cm and a eyepiece focal
length of 2.00 cm.
a) Where must the image formed by the objective be for the eyepiece to
produce a virtual image 25.0 cm in front of the eyepiece?
b) If the lenses are 20.0 cm apart, what is the distance of the objective from the
object on the slide?
c) What is the total magnification of the microscope?
d) What distance would the object have to be from a single lens that gave the
same magnification? What would its focal length have to be?
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Slide
15
Microscope Question
A microscope has an objective of focal length 0.300 cm and a eyepiece focal
length of 2.00 cm.
a) Where must the image formed by the objective be for the eyepiece to
produce a virtual image 25.0 cm in front of the eyepiece?
b) If the lenses are 20.0 cm apart, what is the distance of the objective from the
object on the slide?
c) What is the total magnification of the microscope?
d) What distance would the object have to be from a single lens that gave the
same magnification? What would its focal length have to be?
1
1
1


se 25cm 2cm
1
1
1


se 2cm 25cm
1 1 1
 
se s f
1
25cm  2cm

se  2cm  25cm 
50
cm
23
 2.17cm
se 
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Slide
16
Microscope Question
A microscope has an objective of focal length 0.300 cm and a eyepiece focal
length of 2.00 cm.
a) Where must the image formed by the objective be for the eyepiece to
produce a virtual image 25.0 cm in front of the eyepiece?
b) If the lenses are 20.0 cm apart, what is the distance of the objective from the
object on the slide?
1
1
1
1 1 1


 
so 17.8cm 0.3cm
s s f
o
o
1
1
1


so 0.3cm 17.8cm
so  se  20cm
1 17.8cm  0.3cm

s  0.3cm 17.8cm 
so  20cm  se
 20cm  2.17cm
s  0.305cm
 17.8cm
Slide
17
Microscope Question
A microscope has an objective of focal length 0.300 cm and a eyepiece focal
length of 2.00 cm.
a) Where must the image formed by the objective be for the eyepiece to
produce a virtual image 25.0 cm in front of the eyepiece?
b) If the lenses are 20.0 cm apart, what is the distance of the objective from the
object on the slide?
c) What is the total magnification of the microscope?
d) What distance would the object have to be from a single lens that gave the
same magnification? What would its focal length have to be?
M  m1m2
M angular  mobjective M eyepiece
M  m1m2
s s
 o e
so se
17.8cm 25cm

0.305cm 1.85cm
 788.65
Slide
18
 25cm  L 


 f


 eyepiece  f objective 
 25cm  20cm 

 2.00cm  0.300 
A microscope has an objective of focal length 0.300 cm and a eyepiece focal
length of 2.00 cm.
a) Where must the image formed by the objective be for the eyepiece to
produce a virtual image 25.0 cm in front of the eyepiece?
b) If the lenses are 20.0 cm apart, what is the distance of the objective from the
object on the slide?
c) What is the total magnification of the microscope?
d) What distance would the object have to be from a single lens that gave the
same magnification? What would its focal length have to be?
s
s
25cm
788
 0.031cm
s
1 1 1
 
so so f
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
 833.33
Microscope Question
m
___________________________________
1
1


0.031cm 25cm
25cm  0.031cm

 0.031cm  25cm 
1
f
1
f
0.775
cm
25.031
 0.031cm
f 
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Slide
The Telescope
19
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
M angular  
Slide
f objective
f eyepiece
Telescope Question
20
An astronomical telescope has an objective of 50 cm focal length. The
eyepiece has a focal length of 3.5 cm. How far must these lenses be
separated when viewing and object 200 cm from the objective?
1 1 1
 
so si f
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
1
1
1
 
200cm si 50cm
1
1
1


si 50cm 200cm

___________________________________
200cm  50cm
 50cm  200cm 
Therefore the eyepiece must be placed
so that the principal focus is at to
location of the objective‟s image, to for a
virtual image at infinity. Thus the
separation of the two lenses will be:
66.67 cm+ 3.5 cm = 70.2 cm
___________________________________
si  66.67cm
Slide
Amazing Eye
21


One of first organs to develop.
100 million Receptors
200,000 /mm2
 Sensitive to single photons!


Candle from 12 miles
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Ciliary Muscles
___________________________________
Slide
22
The Physics of Focusing the Eye
___________________________________
___________________________________
Ciliary Muscles
___________________________________
Cornea n= 1.38
Lens
n = 1.4
___________________________________
Vitreous n = 1.33
___________________________________
___________________________________
___________________________________
Which part of the eye does most of the light bending?
1) Lens
2) Cornea
3) Retina
4) Cones
Lens and cornea have similar shape, and index of refraction. Cornea has
air/cornea interface 1.38/1, 70% of bending. Lens has Lens/Vitreous interface
1.4/1.33. Lens is important because it can change shape.
Slide
___________________________________
Eye (Relaxed)
23
___________________________________
25 mm
___________________________________
___________________________________
___________________________________
Determine the focal length of your eye when looking at an object far away.
1 1 1
 
s s f
Object is far away:
so  
Image at retina:
si  25mm
Slide
___________________________________
1
1
1


 25 mm f
f relaxed  25 mm
___________________________________
Eye (Tensed)
24
___________________________________
25 cm
___________________________________
25 mm
___________________________________
___________________________________
___________________________________
Determine the focal length of your eye when looking at an object up close (25 cm).
1 1 1
 
s s f
Object is up close:
Want image at retina:
so  25cm  250mm
si  25mm
Recall:
1
1
1


250 mm 25 mm f
ftense  22.7 mm
f relaxed  25 mm
___________________________________
___________________________________
Slide
___________________________________
Near Point, Far Point
___________________________________
Eye‟s lens changes shape (changes f )
___________________________________
25




Can only change shape so much
“Near Point”




26
Closest do where image can be at retina
Normally, ~25 cm (if far-sighted then further)
___________________________________
___________________________________
___________________________________
___________________________________
“Far Point”

Slide
Object at any do can have image be at retina (di =
approx. 25 mm)
Furthest do where image can be at retina
Normally, infinity (if near-sighted then closer)
If you are nearsighted...
(far point is too close)
Too far for near-sighted eye to focus
do
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
dfar
Near-sighted eye can focus on this!
Contacts form virtual image at far point –
becomes object for eye.
___________________________________
___________________________________
Want to have (virtual) image of distant object, d o = , at the far point, di = -dfar.
flens = -dfar
Slide
27
___________________________________
Refractive Power of Lens
___________________________________
___________________________________
Diopter = 1/f = POWER
where f is focal length of lens in meters.
Person with far point of 5 meters, would need contacts with focal
length –5 meters.
Doctor‟s prescription reads:
1/(-5m) = –0.20 Diopters
___________________________________
___________________________________
___________________________________
___________________________________
Slide
___________________________________
If you are farsighted...
28
___________________________________
(near point is too far)
Too close for far-sighted eye to focus
___________________________________
___________________________________
do =25cm
dnear=50 cm
___________________________________
Far-sighted eye can focus on this!
___________________________________
1
1
1


d0 d near flens
ContactsWant
formthe
virtual
nearimage
point at
to near
be at do.
point – becomes object for eye.
When object is at do, lens must
create an (virtual) image at -dnear.
Slide
29



1
1
1


25cm 50cm flens
f  50cm
___________________________________
The Eye
The “Normal Eye”
___________________________________
Far Point  distance that relaxed eye can focus onto retina = 
Near Point  closest distance that can be focused on to the retina
1 1 1
1
   0
f s s'
2.5 cm
f  2.5 cm
___________________________________
___________________________________
___________________________________
___________________________________
2.5cm
1 1 1 1
1
   
f s s ' 25 2.5
f  2.3 cm
___________________________________
___________________________________
25cm
Therefore the normal eye acts as a lens with a focal length which can vary from 2.5 cm
(the eye diameter) to 2.3 cm which allows objects from 25 cm to infinity to be focused
on the retina
This is called “accommodation”
Diopter: 1/f where f is in metres
Slide
30
An intuitive way to view eye corrections
Near-sighted eye is elongated, image of distant object forms in front of retina
___________________________________
___________________________________
___________________________________
___________________________________
Add diverging lens, image forms on retina
Far-sighted eye is short, image of close object forms behind retina
___________________________________
___________________________________
___________________________________
Add converging lens, image forms on retina
Slide
31
DEFECTS OF VISION
Myopia – Short sightedness
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Due to enlarged size of the eye ball, the images of
distant objects will be focused in front of the retina. This defect is
known as Myopia.
www.jnvkannur.temple.at
Remedy is the use of a concave
lens as shown.
Slide
32
___________________________________
Due to the reduced size of the eye ball, the images of nearby objects
will be focused behind the retina. This defect is known as
Hypermetropia.
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Remedy is the use of a convex lens as shown.
Slide
This little Piggy
33
___________________________________
___________________________________
In The Lord of the Flies,
Piggy‟s glasses are used to
focus the Sun‟s rays and
start a fire. What type of lens
do you need for this?
___________________________________
Convex lens
___________________________________
___________________________________
Later in the novel, Piggy‟s
glasses are broken, and
poor Piggy has a hard
time seeing because he is
nearsighted. What type of
lenses where in his
glasses?
___________________________________
Concave lens
Remember, do your research if you are going to be an author.
___________________________________
Slide
___________________________________
Understanding
34
___________________________________
An arrow-shaped object is placed in front of a plane mirror as shown below.
The image would look like:
___________________________________
___________________________________
a)
___________________________________
b)
___________________________________
c)
___________________________________
d)
e)
Slide
___________________________________
Understanding
35
___________________________________
An illuminated arrow is placed 2 cm in front of a diverging lens with focal
length of -6 cm. The image is:
a)
b)
c)
d)
e)
___________________________________
real, inverted, smaller than the object
Virtual, inverted, larger than the object
Virtual, upright, larger than the object
Real, upright, larger than the object
Virtual, upright, smaller than the object
___________________________________
___________________________________
A diverging lens (has a negative focal length) will always create an
upright virtual image in front of the object. Since the image distance
is smaller than the object distance, the image will be smaller as well.
Slide
36
___________________________________
___________________________________
___________________________________
Understanding
An object is placed in front of three different optical devices, two lenses and a
mirror, with focal points as shown in the figure. Which will produce real
images?
___________________________________
___________________________________
___________________________________
___________________________________
a)
b)
c)
d)
e)
I only
II only
III only
I and II
II and III
I
II
III
A single concave lens produces only virtual images. An object
placed inside the focal length of a convex lens will result in a virtual
image. This eliminates I and II. An object outside the focal length of
a concave mirror will produce an inverted, real image
___________________________________
___________________________________
Slide
37
___________________________________
Understanding
A concave mirror with a radius of curvature 1.5 m is used to collect light from a
distant source. The distance between the image formed and the mirror is
closest to:
a)
b)
c)
d)
e)
0.75 m
1m
1.5 m
2m
3m
38
___________________________________
A student sets up an optics experiment with a converging lens of focal length
10 cm. He places an illuminated arrow 2 cm high at 15 cm from the lens axis.
The size of the image:
0.5 cm
1 cm
2 cm
3.5 cm
4 cm
m
1 1 1
 
so si f
si
s0
30cm
15cm
 2
1
1
1
 
15cm si 10cm

1
1
1


si 10cm 15cm
si  30cm
39
Understanding
An object is placed in front of a convex mirror. The location of the image is
closest to:
a)
b)
c)
d)
e)
A
B
C
D
E
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
Since the magnification is -2 times. The size of the image is 2 cm x 2
which is 4cm and the image is inverted.
Slide
___________________________________
___________________________________
Understanding
a)
b)
c)
d)
e)
___________________________________
___________________________________
Since the object is distant, then the light rays that approach the
mirror are parallel. The focus is r/2 where r is the radius of
curvature. In this example, r=1.5 m, so f =0.75 m
Slide
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
object
C
D
___________________________________
E
___________________________________
focus
___________________________________
A
B
A convex (diverging) mirror will produce an upright, smaller, virtual
image.
Slide
40
___________________________________
Understanding
For which of the cases will the image of the arrow be virtual and smaller than
the object:
___________________________________
___________________________________
___________________________________
___________________________________
a)
b)
c)
d)
e)
I only
II only
III only
I and II
I and III
III
II
I
___________________________________
___________________________________
Diverging elements like I and III will always produce smaller virtual
images. II will produce a virtual image, but it will be larger. This is
basically a magnifying glass.
Slide
41
___________________________________
Free Response Problem
The figure above shows an enlarged portion of the glass wall of a fish tank,
currently empty so that air (n=1) is on either side of the glass (n=1.5). The glass is
0.5 cm thick. A ray R is incident on the glass at a 300 angle with the normal as
shown.
a) On the left figure, continue the ray, showing qualitatively what happens at the
next interface.
b) At what distance above the normal line N will the transmitted ray emerge out of
the glass?
c) Determine the incident angle at the second interface that will ensure total
internal reflection. Could the initial ray R have its incident angle adjusted to
make this happen?
d) Suppose the tank is filled with water (n=1.33) as on the right figure. Show
qualitatively what happens at the glass water interface
glass
air
air
air
N
R
42
water
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
N
300
300
Slide
glass
___________________________________
R
0.5cm
0.5cm
___________________________________
Free Response Problem
The figure above shows an enlarged portion of the glass wall of a fish tank,
currently empty so that air (n=1) is on either side of the glass (n=1.5). The glass is
0.5 cm thick. A ray R is incident on the glass at a 300 angle with the normal as
shown.
a) On the left figure, continue the ray, showing qualitatively what happens at the
next interface.
air
300
300
R
glass
air
300
air
N
water
N
300
19.50
0.5cm
glass
R
0.5cm
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
The reflected light will also have an angle of reflection of 30 0
ni sin i   nr sin  r 
ni sin i   nr sin  r 
n
sin  r   i sin i 
nr
n
sin  r   i sin i 
nr
 1 

 sin  30  
  1.5 

  1.5 

 sin 19.5  
 1 

 r  sin 1  
 r  sin 1  
 19.5
 30
Thus the ray
exiting back into
air is parallel to
the original ray
___________________________________
Slide
43
___________________________________
Free Response Problem
The figure above shows an enlarged portion of the glass wall of a fish tank,
currently empty so that air (n=1) is on either side of the glass (n=1.5). The glass is
0.5 cm thick. A ray R is incident on the glass at a 300 angle with the normal as
shown.
b) At what distance above the normal line N will the transmitted ray emerge out of
the glass?
air
300
300
R
air
glass
300
d
air
N
glass
water
N
300
19.50
R
0.5cm
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
0.5cm
___________________________________
We can determine d from the geometry
d
0.5cm
d  0.5cm tan 19.5 
tan 19.5  
 0.18cm
Slide
44
The figure above shows an enlarged portion of the glass wall of a fish tank,
currently empty so that air (n=1) is on either side of the glass (n=1.5). The glass is
0.5 cm thick. A ray R is incident on the glass at a 300 angle with the normal as
shown.
c) Determine the incident angle at the second interface that will ensure total
internal reflection. Could the initial ray R have its incident angle adjusted to
make this happen?
air
300
300
R
air
glass
glass
water
N
300
R
0.5cm
d) Suppose the tank is filled with water (n=1.33) as on the right figure. Show
qualitatively what happens at the glass water interface
300
300
R
The ray exiting
into water is
NOT parallel to
the original ray,
with the angle of
refraction now
being:
air
air
glass
N
0.5cm
R
ni sin i   nr sin  r 
n
sin  r   i sin i 
nr
  1.5 

 sin 19.5  
  1.33 

 22.1
 r  sin 1  
___________________________________
19.50
___________________________________
___________________________________
___________________________________
___________________________________
water
N
300
19.50
___________________________________
___________________________________
The figure above shows an enlarged portion of the glass wall of a fish tank,
currently empty so that air (n=1) is on either side of the glass (n=1.5). The glass is
0.5 cm thick. A ray R is incident on the glass at a 300 angle with the normal as
shown.
glass
___________________________________
___________________________________
Free Response Problem
air
___________________________________
0.5cm
Since the ray that exits into the air has to
exit at 900 to be total reflected, and that we
have the incoming ray parallel to the
outgoing ray. We would need to incoming
ray have an incident angle of 900, thus
indicting that it would not enter the glass,
so we could not make it happen.
n 
 c  sin  r 
 ni 
1  1 
 sin 

 1.5 
 41.8
1
45
air
N
19.50
We require the critical angle
Slide
___________________________________
Free Response Problem
___________________________________
___________________________________
0.5cm
___________________________________
Slide
___________________________________
Free Response Problem
46
A converging lens with focal length 4 cm has an object placed 6 cm in front of
it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first
lens.
a)
b)
c)
d)
e)
f)
Determine the position of the image formed by the first lens.
Draw a ray diagram needed to display the image from first lens.
What is the magnification of the image?
Determine the position of the image formed by the second lens.
Draw a ray diagram needed to display the image from the second lens.
Determine the overall magnification and image orientation of final image
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
0
Slide
47
16
4
28
48
___________________________________
Free Response Problem
A converging lens with focal length 4 cm has an object placed 6 cm in front of
it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first
lens.
___________________________________
___________________________________
a) Determine the position of the image formed by the first lens.
___________________________________
___________________________________
0
16
4
28
48
___________________________________
1 1 1
 
so si f
1
1
1
 
6cm si 4cm
___________________________________
1
1
1


si 4cm 6cm
6cm  4cm
24cm 2
si  12cm
Therefore the image is at the
12 cm + 8 cm = 20 cm position

Slide
48
___________________________________
Free Response Problem
A converging lens with focal length 4 cm has an object placed 6 cm in front of
it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first
lens.
b) Draw a ray diagram needed to display the image from first lens.
___________________________________
___________________________________
___________________________________
___________________________________
0
4
16
28
48
___________________________________
___________________________________
Therefore the image is at the
12 cm + 8 cm = 20 cm position
Slide
49
___________________________________
Free Response Problem
A converging lens with focal length 4 cm has an object placed 6 cm in front of
it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first
lens.
___________________________________
___________________________________
c) What is the magnification of the image?
___________________________________
___________________________________
0
16
4
m
m
si
so
12cm
6cm
 2

Slide
50
28
48
si
so
___________________________________
___________________________________
Therefore the image is twice
as big and is inverted.
___________________________________
Free Response Problem
A converging lens with focal length 4 cm has an object placed 6 cm in front of
it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first
lens.
___________________________________
___________________________________
d) Determine the position of the image formed by the second lens.
___________________________________
___________________________________
0
16
4
1
1
1
 
16cm si 8cm
16cm  8cm
128cm 2
si  5.33cm
___________________________________
___________________________________
Therefore the virtual image is located
at 36 cm - 5.33 cm = 30.67 cm

51
48
1 1 1
 
so si f
1
1
1


si 8cm 16cm
Slide
28
___________________________________
Free Response Problem
A converging lens with focal length 4 cm has an object placed 6 cm in front of
it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first
lens.
___________________________________
___________________________________
e) Draw a ray diagram needed to display the image from the second lens.
___________________________________
___________________________________
0
4
16
28
48
___________________________________
___________________________________
Therefore the final virtual
image is inverted.
Slide
___________________________________
Free Response Problem
52
A converging lens with focal length 4 cm has an object placed 6 cm in front of
it. A diverging lens with focal length – 8 cm is placed 28 cm behind the first
lens.
___________________________________
___________________________________
f) Determine the overall magnification and image orientation of final image
___________________________________
___________________________________
0
16
4
 s 
m   2    i 
 so 
 s 
m   2    i 
 so 
 5.33cm 
  2   

 16cm 
 0.667
Slide
48
28
Magnification
from first lens
___________________________________
Therefore the image is 2/3 the
original size and is inverted.
___________________________________
Free Response Problem
53
Two thin converging lenses of focal lengths 10 cm and 20 cm are separated
by 20 cm. An object is placed 15 cm in front of the first lens.
a)
b)
c)
d)
e)
f)
___________________________________
Determine the position of the image formed by the first lens.
What is the magnification of the image?
Draw a ray diagram needed to display the image from first lens.
Determine the position of the image formed by the second lens.
Determine the overall magnification and image orientation of final image.
Draw a ray diagram needed to display the image from the second lens.
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
0
5
Slide
10
20
25
30
40
45
50
55
60
___________________________________
Free Response Problem
54
Two thin converging lenses of focal lengths 10 cm and 20 cm are separated
by 20 cm. An object is placed 15 cm in front of the first lens.
___________________________________
a) Determine the position of the image formed by the first lens.
b) What is the magnification of the image?
c) Draw a ray diagram needed to display the image from first lens.
___________________________________
1 1 1
 
so si f
1
1
1
 
15cm s1i 10cm
M1  
1
1
1


s1i 10cm 15cm
___________________________________
30cm
15cm
 2
M1  
si
so
___________________________________
___________________________________
s1i  30cm
___________________________________
0
5
10
20
25
30
40
45
50
55
60
Slide
___________________________________
Free Response Problem
55
Two thin converging lenses of focal lengths 10 cm and 20 cm are separated
by 20 cm. An object is placed 15 cm in front of the first lens.
___________________________________
d) Determine the position of the image formed by the second lens.
e) Determine the overall magnification and image orientation of final image.
f) Draw a ray diagram needed to display the image from the second lens.
___________________________________
1
1
1


10cm s2i 20cm
1 1 1
 
so si f
M2  
1
1
1


s2i 20cm 10cm
2
s2i  6 cm
3
0
5
10
20
25
si
so
M f  M1M 2
30
40
45
2
6 cm
M2   3
10cm
2

3
4
2
M f   2     
3
3
50
55
___________________________________
___________________________________
___________________________________
___________________________________
60
This line is back projected through the centre of lens 2 (which
would have come from lens 1 back to the object)
Slide
___________________________________
Free Response Problem
56
A converging lens with focal length 25 cm has an object placed 150 cm in
front of it. A diverging lens with focal length – 15 cm is placed 20 cm behind
the first lens.
a)
b)
c)
d)
e)
f)
Determine the position of the image formed by the first lens.
What is the magnification of the image?
Draw a ray diagram needed to display the image from first lens.
Determine the position of the image formed by the second lens.
Determine the overall magnification and image orientation of final image.
Draw a ray diagram needed to display the image from the second lens.
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
0
20
Slide
40
60
80
100
110
140
220
240
___________________________________
Free Response Problem
57
A converging lens with focal length 25 cm has an object placed 150 cm in
front of it. A diverging lens with focal length – 15 cm is placed 20 cm behind
the first lens.
a) Determine the position of the image formed by the first lens.
b) What is the magnification of the image?
c) Draw a ray diagram needed to display the image from first lens.
1 1 1
 
so si f
1
1
1
 
150cm s1i 25cm
M1  
1
1
1


s1i 25cm 150cm
si
so
20
40
60
80
100
___________________________________
___________________________________
30cm
150cm
 0.2
M1  
s1i  30cm
0
___________________________________
___________________________________
___________________________________
___________________________________
110
140
220
240
Slide
___________________________________
Free Response Problem
58
A converging lens with focal length 25 cm has an object placed 150 cm in
front of it. A diverging lens with focal length – 15 cm is placed 20 cm behind
the first lens.
d) Determine the position of the image formed by the second lens.
e) Determine the overall magnification and image orientation of final image.
f) Draw a ray diagram needed to display the image from the second lens.
1 1 1
 
so si f
0
20
1
1
1


10cm s2i 15cm
M2  
si
so
M2  
30cm
10cm
3
1
1
1


s2i 15cm 10cm M  M M
f
1
2 M f   0.2  3  0.6
s2i  30cm
40
60
80
100
110
140
220
240
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________
___________________________________