the lens equation
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The Lens Equation
Prepared by:
Princess T. PermolanMS Physics Teaching - DLSU
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Review What is a lens?
What are the different typesof lenses?
What are the type of imagesformed by each type of lens?
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LENS is a piece of transparent
glass that came from theGreek word lentilwhichconverges real and virtual
refracted light rays.
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Diverging Lens/ Convex Lens It is a type of lens
that is thinner at the
middle than on itsedges
It is use to divergelight refracted rays.
The one that is beingconverge at thefocus is the virtualrays.
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Image formed by a
Converging Lens When the object is beyond 2F / 2F
The image formed is real, it is inverted
and reduced When the object is between F & O / F
& O
The image formed is virtual, it is upright
and enlarged When the object is at F
No Image Formation
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Image formed by a
Diverging Lens The image formed is always virtual,
upright and smaller than the object
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Targets: Determine the type of
the image formed, the
location of the image,orientation of the imageand the size of theimage through Lens andmagnification equation
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Targets Solve problems that
are related to imageformation inconverging lens usingLens equation
Show the implication
of the computedvalues through raydiagramming
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Lens Equation
1
f
1
do
1
dif= the focal length of the lens
do = distance of the object from the lens
di distance of the image from the lens
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Sign Convention for
focal length (fL)
+f =focal length of a
converging lens
-f =focal length of a
diverging lens
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Sign Convention for
Image distance
+di =when the image is in theopposite side of the object
(REAL IMAGE)
-di =when the image is on the same
side of the lens (VIRTUALIMAGE)
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MAGNIFIC
ATIO
NEQ
N
do
-di
MWhen,
+M = object is upright
-M = object is inverted
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Same, Reduced or Enlarged?If.
l M l > 1 then the image is enlarged
l M l =1 then the image is of thesame size of the object
l M l < 1 then the image is reduced
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Lets Do It Determine the exact location of
the image, type of the image,
orientation of the image and themagnification of the imageformed in a converging lens with
a focal length of 20mm when theobject is placed exactly at 5mmfrom the lens
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Solve: Given:
Type of Lens: Converging lens
f = +20 mmdo= 5mm
Find: di = ? Magnification = ?type of image = ? Orientation = ?
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Formula to be used:
1
f
1
do
1
di
do-diM
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transposition
1
20
1
5
1
diGET THE LCD OF 20 and 5
201- 4 1
di
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-3
20
1
diCROSSMULTIPLY
-3di = 20
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Divide both sides by -3
-3di = 20 mm
3 3
di = - 6.67 mm
Therefore,
the image is located 6.67 cm
and it is a virtual image
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Solve for
Magnification
do
-di
M
5- (-6.67)M
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M= +1.4
Then,
the image is enlarged by 1.4
and it is upright
ANSWER:
The image formed is virtual. It is6.67 mm in front of the lens whichis enlarged by 1.4 and oriented in
the upright manner
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do
di
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Lets do it together Determine the type of image,
orientation of the image,
magnification of the image and theorientation of the image formed by aconverging lens with a focal length of15mm when the object is placed at
30 mm.
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Assignment Determine the type of image,
orientation of the image,
magnification of the image and theorientation of the image formed by aconverging lens with a focal length of15mm when the object is placed at 30
mm. Will there be changes in the result if
we will use diverging lens instead of aconverging lens?
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The end..
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Targets: Identify the different
parts of the human eye
Compare the parts ofthe camera and the eye
Apply the concepts
learned in real lifesituation
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