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Concepts and Math!
Convex Lenses: Summary
Object Location
Image Orientation
Image Size
Image Type
Beyond c
Inverted
Reduced
Real
At c
Inverted
Same as object
Real
Between c and f
Inverted
Enlarged
Real
At f
No image
No image
No image
Closer than f
Upright
Enlarged
Virtual
This is the same as for a concave mirror!
This is why convex lenses held close to an object make good
magnifying glasses!
Concave Lenses: Summary
Object Location
Image Orientation
Image Size
Image Type
Anywhere
Upright
Reduced
Virtual
This is the same as for a convex mirror!
Converging lenses and converging mirrors have the same image properties.
Diverging lenses and diverging mirrors have the same image properties.
Return of the Whiteboard!
A large lens is used to focus an image of an
object onto a screen. If the left half of the lens is
covered with a dark card, which of the following
occurs?
(A) The left half of the image disappears.
(B) The right half of the image disappears.
(C) The image becomes blurred.
(D) The image becomes dimmer.
(E) No image is formed.
Although principal rays help guide us to locate the image, we cannot forget
the important fact that each point on the object emits rays in all directions.
The lens is completely filled with rays from every point of the object!
(The image is also formed by infinite rays from the
middle of the object, the bottom of the object, etc.)
So, if we cover half of the lens...
The entire image would still exist!
However, less light would be forming the image.
Therefore the image would be dimmer.
Which three of the glass lenses above, when placed
in air, will cause parallel rays of light to converge?
(A) I, II, and III
(B) I, III, and V
(C) l, IV, and V
(D) II, III, and IV
(E) II, IV, and V
I, III, and V are more convex than concave –
they will cause light rays to converge.
II and IV are more concave than convex –
they will cause light rays to diverge.
Geometric Optics Equation #1
1 1 1
+ =
di do f
Applies to both mirrors and lenses.
di is the distance from the image to the mirror or lens
do is the distance from the object to the mirror or lens
f is the focal length of the mirror or lens
Focal length can be positive or negative!
Mirrors and lenses that have a focal point will have a
positive focal length.
Mirrors and lenses that have an antifocal point will
have a negative focal length.
Converging lenses and mirrors (concave mirrors and
convex lenses) have a positive focal length.
Diverging lenses and mirrors (convex mirrors and
concave lenses) have a negative focal length.
A Point of Possible Confusion
• Real images have a positive di
• Virtual images have a negative di
If you end up with a negative di when you do
the calculations, it means that a virtual image
is produced!
This applies to both mirrors and lenses, and
you must be consistent!
Whiteboard Problem Solving I
A postage stamp is placed 30 centimeters to the left
of a converging lens of focal length 60 centimeters.
Where is the image of the stamp located?
(A) 60 cm to the left of the lens
(B) 20 cm to the left of the lens
(C) 20 cm to the right of the lens
(D) 30 cm to the right of the lens
(E) 60 cm to the right of the lens
Whiteboard Problem Solving II
A concave mirror with a radius of curvature
of 1.0 m is used to collect light from a
distant star. The distance between the
mirror and the image of the star is nearly
(A) 0.25 m (B) 0.50 m (C) 0.75 m
(D) 1.0 m (E) 2.0 m
Solution!
A star is so far away that we can comfortably use
the approximation do = ∞!
This gives
1 1 1
+ =
di ¥ f
Since 1/∞ = 0, this results in di = f
Geometric Optics Equation #2
hi
di
=ho
do
hi is the height of the image
ho is the height of the object
A negative image height means that the image is inverted.
Memorize both of these equations
(don’t forget the negative sign!)
Attack of the Whiteboard
15 cm
5 cm
f’
f’
10 cm
10 cm
Where will the image be located and what will it look like?
do = 15 cm
f = -10 cm
1 1
1
+ =
di 15 -10
di = -6 cm
ho = 5 cm
hi
-6cm
=5 cm
15 cm
hi = -2 cm
f’
(Virtual image
6 cm from
lens)
(Inverted and
reduced image)
f’
Deep Whiteboard Thoughts!
do
ho
An object of height h0
is located at a distance
do from a plane mirror.
1 1 1
+ =
Using the mathematical models
and
di do f
hi
di
=- ,
ho
do
determine the location and height of the
image formed by the plane mirror!
Hint: What is the radius of curvature of a plane mirror?
do
fplane mirror = ∞
ho
1 1 1
+ = = 0!
di do ¥
di = -do
This means that the mirror will
produce a virtual image
(negative image distance) that
is equidistant from the mirror.
hi
di
=ho
do
Since di = -do 
-di / do = 1
Therefore hi / ho = 1
hi = ho
The image is the same height as
the object!