1. Phy2005 Applied Physics II Spring 2017 Announcements:
Today: catch up w/ syllabus and practice
Exam 3 April 10 in-class

2. Q2. A parallel beam of light is sent through an aquarium.
If a convex lens is held in the water, it focuses the beam (…….. ……………………. ) than outside the water
nair = 1, nwater = 1.33
closer to the lens
(b) at the same position as
(c) farther from the lens

3. off the optical axis as a point light source.
Rules for Images
Trace principal beams considering one end of an object
off the optical axis as a point light source.
A beam passing through the focal point runs parallel to
the optical axis after a lens.
A beam coming through a lens in parallel to the optical
axis passes through the focal point.
A beam running on the optical axis remains on the optical
axis.
A beam that pass through the geometrical center of
a lens will not be bent.
Find a point where the principle beams or their imaginary
extensions converge. That’s where the image of the point source
will be formed.

4. two focal points: f1 and f2 Parallel beams: no image!!

5. Virtual image
Magnifying glass
object
object
Virtual image

6. Glass lens (nG = 1.52)
screen D x
A candle is a distance D from a screen. A converging lens
of focal length f is placed in between.
There are 2 positions x of the lens for which a real image will be
formed on the screen. Where are they?
b) What is the ratio of the image sizes to the two lens positions?

7. screen D
D-x ho f f hi x
similar
ho+hi ho
divide: f x

8. hi ho f D
screen x
D-x
Another similarity:
quadratic equation x2-xD+DF=0 so

9. hi ho f D
screen x
D-x
Magnification
so plug in values of x for each solution

10. New definitions: we found: object distance p = D-xhi ho f D
screen x
D-x
we found:
New definitions:
object distance p = D-x
image distance q = x
magnification M = hi/ho
Rewrite:
Or:

11. Exactly the same as the mirror eq.!!! Now let’s think about the sign.
Lens equation and magnification
1/p + 1/q = 1/f
M = -q/p
Exactly the same as the mirror eq.!!!
Now let’s think about the sign.
positive
negative p
real object
virtual object
(multiple lenses) q
real image
(opposite side of object)
virtual image
(same side of object) f
for converging lens
for diverging lens M
erect image
inverted image

12. Ex. 1 An object is placed at a distance of 80 cm from a thin
lens along the axis. If a real image forms at a distance 40 cm
from the lens, on the opposite side from the object, what is
the focal length of the lens?
26.6 cm
(2) 45.3 cm
(3) 90.9 cm
(4) cm
(5) 21.4 cm

13. Ex. 2 It is desired to use a 60-cm focal length diverging lens to form a virtual image of an object. The image is to
be one- third as large as the object. Where should the object be placed and what will be the image distance in cm?
(1) (55, 22.5)
(2) (-165, -45.2)
(3) (55, -22.5)
(4) (-155, 41.3)
(5) (120, -40)

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15. Q1 In a slide projector, the slide is 12cm from the lens
Q1 In a slide projector, the slide is 12cm from the lens. The image appears on the screen 4.0 m away.
What is the focal length of the lens?
(1) 40 cm (2) 31 cm (3) 1.3 m (4) 11.7 cm (5) 0 cm

16. Q2 A diverging lens has a focal length of -12 cm
Q2 A diverging lens has a focal length of -12 cm. An object of height ho = 4 cm is placed 4 cm from the lens. Which is closest to the size of the image?
(1) 4 cm (2) 3 cm (3) 1 cm (4) 2 cm (5) 0 cm