Download Lecture 29:lens formula practice, microscopes

Phy2005
Applied Physics II
Spring 2017
Announcements:
• Today: compound lenses, micro-, telescope
• Exam 3 April 10 in-class
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Last time
D-x
D
ho
hi
f
x
Definitions:
object distance p
image distance q
magnification M = hi/ho
Lens equation:
q
1 1 1
 
and M  
p q f
p
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
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#8
S11 Exam 3. 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 of 40 cm from the lens, on the
opposite side from the object, what is the focal length
of the lens in cm?
(1) 26.6
(2) 45.3
(3) 90.9
(4) 120.0
(5) 21.4
Optical Instruments
Eye Glasses
Perfect Eye
Nearsighted
Nearsighted can be corrected with a diverging lens.
 A far object can be focused on retina.
Farsighted
A
Power of lens: diopter = 1/f (in m)
(+) diopter  converging lens
(-) diopter  diverging lens
Larger diopter
 Stronger lens (shorter f)
Material
n
Cornea
1.38
Aqueous
Humor
Lens
1.331.34
1.411.45
1.34
Vitreous
Humor
Air
Water
1.00
1.33
Compound Microscope
(two converging lenses)
objective
po
qo
eyepiece
pe
qe
Magnified inverted virtual image
Real image formed by the objective lens  an object for the eyepiece lens
Each set follows lens
Equations!!!
fo
po
qo
Mo
fe
pe
qe
Me
M = MoMe = (-qo/po)(-qe/pe)
= (qo/po)(qe/pe)
M = MoMe = (-qo/po)(-qe/pe)
= (qo/po)(qe/pe)
Magnification becomes larger qo >> po:
1/po + 1/qo = 1/fo
1/po ≈ 1/fo
The object should be put near the focal point of
the objective lens.
Q. In a compound microscope, the obj. lens has a focal length of 8 mm,
And the eyepiece has a focal length 40 mm. The distance between the
lenses is 200 mm. If the object is placed 8.4 mm from the obj. lens,
What would be the magnification?
M = (qo/po)(qe/pe) = (168/8.4)(-160/32) = -100
po = 8.4
qo = ?
fo = 8
D = 200
pe = ?
qe = ?
fe = 40
1/po + 1/qo = 1/fo
1/qo = 1/fo – 1/po = 1/8 – 1/8.4
Inverted (M<0)
but
Virtual image
(qe<0)
qo = 168
pe = D – qo = 200 – 168
= 32
1/qe = 1/fe – 1/pe
= 1/40 – 1/32
qe = -160
Ex. 3 Two thin lenses with 10.0-cm focal lengths are mounted at
opposite ends of a 30.0-cm long tube. An object is located
45.0 cm from one end of the tube. How far from the opposite
end is the final image?
(1) 38.2 cm
(2) 12.8 cm
(3) 25.6 cm
(4) 33.6 cm
(5) 24.0 cm
#7 S11 Exam 3 Two thin lenses with focal lengths 50.0 cm and 60.0
cm are placed in contact in an orientation so that their optic axes
coincide. What is the focal length of the two in combination?
(1) 110 cm (2) 13.6 cm (3) 55.0 cm (4) 27.3 cm (5) 10.0 cm
(see section 27.13: Virtual objects)
1
1 1
 
f1 p1 q1
(1) Object p1 and image q1 for lens 1 alone
1
1
1


f 2 p2 q2
(2) Object p2 and image q2 for lens 1 alone
p2   q1
Image of lens 1 is virtual object for lens 2
1 1
1 1
  
Add (1) and (2): effective focal length is
f1 f 2 p1 q2
1 1 1
 
f
f1 f 2
Telescope: angular magnification