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Physics 6C
Geometric Optics
Mirrors and Thin Lenses
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We have already learned the basics of Reflection and Refraction.
Reflection - angle of incidence = angle of reflection
Refraction - light bends toward the normal according to Snell’s Law
Now we apply those concepts to some simple types of mirrors and lenses.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We have already learned the basics of Reflection and Refraction:
Reflection - angle of incidence = angle of reflection
Refraction - light bends toward the normal according to Snell’s Law
Now we apply those concepts to some simple types of mirrors and lenses.
Flat Mirror
This is the simplest mirror – a flat reflecting surface. The light rays bounce off and you see an
image that seems to be behind the mirror. This is called a VIRTUAL IMAGE because the light rays
do not actually travel behind the mirror. The image will appear reversed, but will be the same size
and the same distance from the mirror. A typical light ray entering the eye of the viewer is shown.
The object distance is labeled S and the image distance is labeled S’.
Real Object
Virtual Image
S
S’
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Spherical Mirrors
For curved mirrors we will assume that the shape is spherical (think of a big shiny ball, and slice off
any piece of that – there’s your spherical mirror). This will make our math relatively simple, with
only a couple of formulas. The hard part will be to get the negative signs correct.
The radius of curvature describes the shape of the mirror. This is the same as the radius of the big
shiny ball that the mirror was cut from.
We will have two types of mirrors, depending on which direction they curve:
CONCAVE mirrors curve toward you, and have POSITIVE R (like the inside of the sphere).
CONVEX mirrors curve away from you, and have NEGATIVE R (think of the outside of the ball).
There is a point called the FOCAL POINT which is halfway between the mirror and the center.
C
R
R
C
Shiny side
Shiny side
Concave Mirror – R is positive
Convex Mirror – R is negative
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
Ray 1 through
the center
Optical Axis
Focal
Point
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
Ray 1 through
the center
Ray 1 reflects
directly back
Optical Axis
Focal
Point
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
Optical Axis
Ray 2 through
the focal point
Focal
Point
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
Ray 2 reflects
parallel to axis
Optical Axis
Ray 2 through
the focal point
Focal
Point
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
Focal
Point
Optical Axis
Ray 3 comes in
parallel to axis
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
Ray 3 reflects
through focal point
Focal
Point
Optical Axis
Ray 3 comes in
parallel to axis
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
We will learn 2 techniques for dealing with mirrors (and lenses):
• Graphical – draw the light rays and the image is at their intersection.
• Formula – use a couple of formulas to locate and describe an image.
First the Graphical Method:
For a spherical mirror there are 3 basic rays that you can draw:
1) Any ray that goes through the CENTER of the circle reflects directly back to the light source.
2) Any ray that goes through the FOCAL POINT is reflected back PARALLEL to the optical axis.
3) Any ray that starts parallel to the optical axis is reflected back through the focal point.
(opposite of 2)
All 3 rays shown with the
image at their intersection
Optical Axis
Image
Object
1
2
3
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Before we answer this let’s look at a few basic formulas for spherical mirrors.
1) The focal length is half the radius.
f
R
2
Remember the sign convention – if the mirror is
concave R is positive. If convex, R is negative.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Before we answer this let’s look at a few basic formulas for spherical mirrors.
1) The focal length is half the radius.
f
R
2
Remember the sign convention – if the mirror is
concave R is positive. If convex, R is negative.
2) This formula relates the object (S) and image
(S’) positions to the focal length (f) of the mirror.
1 1 1
 
f S S
Here S is always positive for mirrors, and
S’ is positive if the image is on the same
side as the object (a REAL image).
To remember this, just follow the light – a
real (positive) image will have light rays
passing through it.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Before we answer this let’s look at a few basic formulas for spherical mirrors.
1) The focal length is half the radius.
f
R
2
Remember the sign convention – if the mirror is
concave R is positive. If convex, R is negative.
2) This formula relates the object (S) and image
(S’) positions to the focal length (f) of the mirror.
1 1 1
 
f S S
3) The magnification (m) of the image is related
to the relative positions of the object and image.
m
y
S

y
S
Here S is always positive for mirrors, and
S’ is positive if the image is on the same
side as the object (a REAL image).
To remember this, just follow the light – a
real (positive) image will have light rays
passing through it.
Don’t forget the negative sign in this
formula. The sign of m tells you if the
image is upright (+) or inverted (-)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
R  0.5m  f  0.25m
S  0.2m
object distance
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
Now we can use formula 2 to locate the image (S’)
focal length
R  0.5m  f  0.25m
S  0.2m
object distance
1 1 1
 
f S S
1
1
1


 S  1m
0.25 0.2 S
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
R  0.5m  f  0.25m
S  0.2m
Now we can use formula 2 to locate the image (S’)
object distance
1 1 1
 
f S S
1
1
1


 S  1m
0.25 0.2 S
This means the image will be located 1m BEHIND the mirror.
This is a VIRTUAL image.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
R  0.5m  f  0.25m
S  0.2m
Now we can use formula 2 to locate the image (S’)
object distance
1 1 1
 
f S S
1
1
1


 S  1m
0.25 0.2 S
This means the image will be located 1m BEHIND the mirror.
This is a VIRTUAL image.
For the magnification, just use formula 3.
m
S
 1m

 5
S
0.2m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
OK, back to the problem. We have given information:
focal length
R  0.5m  f  0.25m
S  0.2m
Now we can use formula 2 to locate the image (S’)
object distance
1 1 1
 
f S S
1
1
1


 S  1m
0.25 0.2 S
This means the image will be located 1m BEHIND the mirror.
This is a VIRTUAL image.
For the magnification, just use formula 3.
m
S
 1m

 5
S
0.2m
So the image is upright (+) and 5 times as large as
the object.
We could also draw the ray diagram…
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example using the Formula Method:
A concave makeup mirror with radius of curvature 0.5m is held 0.2m from a woman’s face.
Where is her image and how large is it?
Notice the 3 rays in the diagram. They all start at the object and go toward the mirror. Ray 1
through the center is easy to draw. So is ray 2, which starts out flat, then bounces off the mirror
and goes through the focal point (f).
Ray 3 is the tricky one. Since the object is inside the focal point (closer to the mirror, or S<f) we
can’t draw the ray through the focal point. Instead we pretend the ray came from the focal point
and passed through the object on its way to the mirror, then bounced off flat.
The outgoing rays do not intersect! So we have to trace them backwards to find their intersection
point behind the mirror. This is what your brain does for you every time you look in a mirror. The
virtual image appears at the point where the outgoing light rays seem to be coming from.
3
Object
1
f
Image
S
S’
2
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Convex Mirrors
These will work the same way as concave, but R and f are negative. Take
a look at where the center of the sphere is – it is behind the mirror. There
are no light rays there. This is why the radius is negative. Because the
light rays do not go there.
The 3 typical light rays are shown.
•Ray 1 points toward the center and bounces straight back.
•Ray 2 starts flat and bounces off as if it is coming from the focal point.
•Ray 3 starts toward the focal point and bounces off flat.
object
R
f
3
C
Image (this is a virtual image behind
the mirror, so S’ is negative)
1
2
Convex Mirror – R is negative
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SPHERICAL MIRROR EQUATIONS
AND SIGN CONVENTION
Concave Mirror Illustrated
Light In Side
S < 0 Virtual Object
S > 0 Real Object
F
C
V
Light Out Side
S’ > 0 Real Image
C This Side, R > 0
Optic Axis
S’ < 0 Virtual Image
C This Side, R < 0
C – Center of Curvature
R – Radius of Curvature
F – Focal Point (Same Side as C)
V – Vertex
Equations: Paraxial Approximation
f
R
2
m
y
S

y
S
1 1 1


S S f
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REFRACTION AT SPHERICAL INTERFACE
BETWEEN TWO OPTICAL MATERIALS
Light In Side
S > 0 Real Object
S’ < 0 Virtual Image
C This Side, R < 0
na – Index of
Refraction
Light Out Side
S < 0 Virtual Object
S’ > 0 Real Image
C This Side, R > 0
nb – Index of
Refraction
Illustrated Interface Has C, Center of Curvature,
On The Light Out Side, Thus R > 0
A Flat Interface Has R = ∞
na nb nb  na


S
S
R
m
y
n S
 a
y
nbS
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part a) consider the fish to
be the light source, and
calculate the image position
for light rays exiting the bowl.
We will be using this formula:
n a nb nb  n a


S S'
R
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part a) consider the fish to
be the light source, and
calculate the image position
for light rays exiting the bowl.
We will be using this formula:
n a nb nb  n a


S S'
R
Here is the given information:
na  1.33;nb  1;S  14cm;R  14cm
This radius is negative
because the center of the
bowl is on the same side as
the light source (the fish)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part a) consider the fish to
be the light source, and
calculate the image position
for light rays exiting the bowl.
We will be using this formula:
n a nb nb  n a


S S'
R
Here is the given information:
na  1.33;nb  1;S  14cm;R  14cm
1.33 1 1  1.33


 S  14cm
14cm S  14cm
This radius is negative
because the center of the
bowl is on the same side as
the light source (the fish)
A negative value for S’ means the image is on the same side of the
interface as the object (i.e. inside the bowl in this case). So the
observer will see a virtual image of the fish at the center of the bowl.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part a) consider the fish to
be the light source, and
calculate the image position
for light rays exiting the bowl.
Magnification can be found
from this formula:
n S
m a
nbS
We will be using this formula:
n a nb nb  n a


S S'
R
Here is the given information:
na  1.33;nb  1;S  14cm;R  14cm
1.33 1 1  1.33


 S  14cm
14cm S  14cm
This radius is negative
because the center of the
bowl is on the same side as
the light source (the fish)
A negative value for S’ means the image is on the same side of the
interface as the object (i.e. inside the bowl in this case). So the
observer will see a virtual image of the fish at the center of the bowl.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part a) consider the fish to
be the light source, and
calculate the image position
for light rays exiting the bowl.
We will be using this formula:
n a nb nb  n a


S S'
R
Here is the given information:
na  1.33;nb  1;S  14cm;R  14cm
1.33 1 1  1.33


 S  14cm
14cm S  14cm
Magnification can be found
from this formula:
n S
m a
nbS
m
1.33(14cm)
 1.33
1(14cm)
The fish appears larger
by a factor of 1.33
This radius is negative
because the center of the
bowl is on the same side as
the light source (the fish)
A negative value for S’ means the image is on the same side of the
interface as the object (i.e. inside the bowl in this case). So the
observer will see a virtual image of the fish at the center of the bowl.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part b) the light source is the sun, which is really
far away (i.e. object distance is infinity). This is what
they mean by “parallel rays from the sun”.
The focal point will be where the sun’s
rays converge, so we need to find the
sunlight
image distance S’.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part b) the light source is the sun, which is really
far away (i.e. object distance is infinity). This is what
they mean by “parallel rays from the sun”.
The focal point will be where the sun’s
rays converge, so we need to find the
sunlight
image distance S’.
Our given information becomes:
na  1;nb  1.33;S  ;R  14cm
This radius is positive because the
center of the bowl is on the opposite
side as the light source (the sun)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 24.23
A small tropical fish is at the center of a water-filled (n=1.33) spherical fishbowl 28cm in diameter.
a) Find the apparent position and magnification of the fish to an observer outside the bowl.
b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish,
which might swim into the focal point of the parallel rays from the sun. Is the focal point
actually within the bowl?
For part b) the light source is the sun, which is really
far away (i.e. object distance is infinity). This is what
they mean by “parallel rays from the sun”.
The focal point will be where the sun’s
rays converge, so we need to find the
sunlight
image distance S’.
Our given information becomes:
na  1;nb  1.33;S  ;R  14cm
1 1.33 1.33  1


 S  56cm

S
 14cm
Focal Point
This radius is positive because the
center of the bowl is on the opposite
side as the light source (the sun)
This image is beyond the other side of the bowl (28cm away), so the fish will be safe.
Prepared by Vince Zaccone
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Assistance Services at UCSB
THIN LENS EQUATIONS
AND SIGN CONVENTION
Surface 1
Surface 2
Light In Side
Light Out Side
S > 0 Real Object
S < 0 Virtual Object
S’ < 0 Virtual Image
S’ > 0 Real Image
C1 This Side, R1 <
C1 This Side, R1 > 0
0
C2 This Side, R2 > 0
C2 This Side, R2 <
0
n – Index of Refraction
C1 – Center of Curvature, Surface 1
C2 – Center of Curvature, Surface 2
Illustrated Lens is Double Convex Converging
With C1 on the Light Out Side and C2 on the Light In Side
Equations:
 1
1
1 

 (n 1)

f
 R1 R 2 
1 1 1


S S f
m
y
S

y
S
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Two Basic Types of Lenses
CONVERGING
Focal Point
•f is positive
•Thicker in middle
•Object outside focal
point = real image
•Object inside focal
point = virtual image
DIVERGING
•f is negative
•Thinner in middle
•Real object always
gives a virtual image
Focal Point
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
Radius=20cm
Radius=15cm
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
To find the focal length we use
the thin lens equation:
Radius=20cm
Radius=15cm
 1
1
1 

 (n 1)

f
 R1 R 2 
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
To find the focal length we use
the thin lens equation:
 1
1
1 

 (n 1)

f
 R1 R 2 
R1=+20cm
R2=+15cm
Light traveling
this direction
The difficult part is to get the signs correct for the
radii. We can suppose the light is coming from the
left, so the light encounters the 20cm side first.
Since the center of that 20cm-radius circle is on the
other side (where the light rays are going to end up)
we call this radius positive – so R1=+20cm.
Similarly, the 15cm-radius circle has its center on
the other side, so this is also positive: R2=+15cm
Your basic rule of thumb is this: follow the
light rays – they end up on the positive side.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
To find the focal length we use
the thin lens equation:
 1
1
1 

 (n 1)

f
 R1 R 2 
R1=+20cm
R2=+15cm
Light traveling
this direction
1
1 
 1
 (1.6  1)

  f  100cm
f
 20cm 15cm 
For extra bonus fun, try calculating the focal length when the light
comes from the other side – so the 15cm radius is encountered first.
You ought to get the same answer for the focal length.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
f=-100cm
S=+50cm
For part b) we can use the formula:
1 1 1


S S f
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
f=-100cm
S=+50cm
S’=-33.3cm
For part b) we can use the formula:
1 1 1


S S f
1
1
1


 S  33 13 cm
50cm S  100cm
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
f=-100cm
S=+50cm
S’=-33.3cm
For part b) we can use the formula:
1 1 1


S S f
1
1
1


 S  33 13 cm
50cm S  100cm
The height of the image comes
from our magnification formula:
 33 13 cm
y
S
y
m
 

 y  8cm
y
S
12cm
50cm
The image is virtual, upright, and 8cm tall.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
f=-100cm
S=+50cm
S’=-33.3cm
For part b) we can use the formula:
1 1 1


S S f
1
1
1


 S  33 13 cm
50cm S  100cm
The height of the image comes
from our magnification formula:
 33 13 cm
y
S
y
m
 

 y  8cm
y
S
12cm
50cm
The image is virtual, upright, and 8cm tall.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
f=-100cm
S=+50cm
S’=-33.3cm
For part b) we can use the formula:
1 1 1


S S f
1
1
1


 S  33 13 cm
50cm S  100cm
The height of the image comes
from our magnification formula:
 33 13 cm
y
S
y
m
 

 y  8cm
y
S
12cm
50cm
The image is virtual, upright, and 8cm tall.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem
a) Find the focal length of the thin lens shown. The index of refraction is 1.6.
b) A 12cm-tall object is placed 50cm away from this lens. Find the image location and height. Is
this image real or virtual? Draw the ray diagram.
f=-100cm
S=+50cm
S’=-33.3cm
For part b) we can use the formula:
1 1 1


S S f
1
1
1


 S  33 13 cm
50cm S  100cm
The red ray in our diagram is
initially headed for the focal point
on the other side of the lens at
x=+100cm.
The lens deflects it parallel to the
axis, and we trace it back to find
the image (at the intersection
with the other 2 rays)
The height of the image comes
from our magnification formula:
 33 13 cm
y
S
y
m
 

 y  8cm
y
S
12cm
50cm
The image is virtual, upright, and 8cm tall.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB