the lens equation

8/9/2019 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

the lens equation

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