1. An image of a distant object
formed by a single converging
lens ____.
•
•
•
•
•
A. can be focused on a screen.
B. can be projected on a wall.
C. is upside down.
D. is real.
E. all of the above.
1. An image of a distant object
formed by a single converging
lens ____.
•
•
•
•
•
A. can be focused on a screen.
B. can be projected on a wall.
C. is upside down.
D. is real.
E. all of the above.
2. An image formed by a single
diverging lens ____.
•
•
•
•
•
A. is upside down.
B. is larger than the object.
C. can be projected on a wall.
D. is virtual.
E. all of the above.
2. An image formed by a single
diverging lens ____.
•
•
•
•
•
A. is upside down.
B. is larger than the object.
C. can be projected on a wall.
D. is virtual.
E. all of the above.
3. A magnifying glass is usually
a ____.
• A. converging lens.
• B. combination of diverging and
converging lenses.
• C. diverging lens.
3. A magnifying glass is usually
a ____.
• A. converging lens.
• B. combination of diverging and
converging lenses.
• C. diverging lens.
4. Which instrument is a human
eye most similar to?
•
•
•
•
A. camera
B. microscope
C. telescope
D. slide projector
4. Which instrument is a human
eye most similar to?
•
•
•
•
A. camera
B. microscope
C. telescope
D. slide projector
1. An object is placed 6.0 cm in
front of a converging lens of
focal length 4.0 cm. Draw a ray
diagram to show the location
and orientation of the image
formed. What is its type and
orientation?
• First draw a line parallel to the principle
axis which refracts through the focal point.
• First draw a line parallel to the principle
axis which refracts through the focal point.
• Then draw a line through the optical center
of the lens. Where they cross is the image.
• This image is real and inverted (case 4).
We use the equations to find the actual
distance and size of the image.
2. The focal length of a double
convex lens is 4.0 cm. An object
2.0 cm high is 10.0 cm from the
lens. Draw a ray diagram of this
situation. Calculate the location
and size of the image. What is
its type and orientation?
• First draw a line parallel to the principle
axis which refracts through the focal point.
• First draw a line parallel to the principle
axis which refracts through the focal point.
• Then draw a line through the optical center
of the lens. Where they cross is the image.
• This image is real and inverted . We use the
equations to find the actual distance and size of
the image.
1
-----f
1
= -----do
1
+ -----di
1
-----4
1
= -----10
1
+ -----di
di = 6.67 cm
hi
------ho
hi
------2 cm
=
di
------do
=
6.67 cm
----------10 cm
hi = 1.3 cm, mag = 0.667
3. A double convex lens has a
focal length of 5.0 cm. An object
0.75 cm high is 3.0 cm from the
lens. Draw a ray diagram of this
situation. Calculate the location
and size of the image. What is
its type and orientation?
• First draw a line parallel to the principle
axis which refracts through the focal point.
• First draw a line parallel to the principle
axis which refracts through the focal point.
• Then draw a line through the optical center
of the lens. Where they cross is the image.
• Image is virtual, upright, and larger.
• Case 6, a magnifying glass.
1
-----f
1
= -----do
1
+ -----di
1
-----5
1
= -----3
1
+ -----di
di = -7.5 cm
hi
------ho
hi
--------0.75 cm
=
di
------do
=
-7.5 cm
----------cm
hi = 1.88 cm, mag = -2.5
4. A concave lens has a focal
length of 6.0 cm. An object is
placed 7.0 cm from the lens.
Draw a ray diagram of this
situation. Calculate the location
and size of the image. What is
its type and orientation? What is
the magnification of this lens?
• First draw a line parallel to the principle
axis which refracts through the focal point.
• First draw a line parallel to the principle
axis which refracts through the focal point.
• Then draw a line through the optical center
of the lens. Where they cross is the image.
• This image is virtual, upright, and smaller. We
use the equations to find the actual distance and
size of the image.
1
-----f
1
= -----do
1
+ -----di
1
------6
1
= -----7
1
+ -----di
di = -3.2 cm
hi
------ho
hi
--------ho
=
di
------do
=
-3.2 cm
----------7 cm
mag = 0.46
5. What type of lenses are used
to correct for farsightedness and
nearsightedness?
• Farsightedness is corrected using a
converging lens (convex).
• Nearsightedness is corrected using a
diverging lens (concave).