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8/7/2019 IMAGE FORMATIONS IN LENSES AND THIN LENS EQUATIONS
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IMAGE FORMATIONS IN LENSESAND THIN LENS EQUATIONS
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The image formed by a converging
lens can be made using only threeprincipal rays.
Ray 1 is the ray which travels parallel to the
axis and after going through the lens it passesthrough the focal point.
Ray 2 passes through the center of the lens.
Ray 3 goes through the focal point and thentravels parallel to the axis after passingthrough the lens.
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The image formed by a diverging
lens can be made using only threeprincipal rays:
Ray 1 is the ray which travels parallel to the
axis and after going through the lens it passesthrough the focal point.
Ray 2 passes through the center of the lens.
Ray 3 goes through the focal point and thentravels parallel to the axis after passingthrough the lens.
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Thin LENS Equation
You can find the distance of the image from the
lens by the following equation:
1/q + 1/p = 1/f,
where q is the distance from the lens to theimage, p is the distance from the object to
the lens, and fis the focal distance of the lens
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Thin LENS Equation
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Thin LENS Equation
Example:
A converging lens of focal length 10.0 cm
forms images of objects placed 30.0 cm.Construct a ray diagram, find the imagedistance and describe the image.
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Thin LENS Equation
Solution:
Construct a ray diagram as shown below.
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Thin LENS Equation
Solution:
Use the thin lens equation to find the image
distance.
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Thin LENS Equation
Solution:
Substituting values,
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Thin LENS Equation
Solution:
Solving for the magnification
The image is reduced by one half and the negative
sign for M tells us that the image is inverted.
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Thin LENS Equation
Example:
A diverging lens of focal length 10.0 cm forms
images of objects placed 30.0 cm. Constructa ray diagram, find the image distance anddescribe the image.
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Thin LENS Equation
Solution:
Construct a ray diagram as shown below.
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Thin LENS Equation
Solution:
Use the thin lens equation to find the image
distance.
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Thin LENS Equation
Solution:
Substituting values,
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Thin LENS Equation
Solution:
Solving for the magnification
The image is reduced by one fourth and the positive
sign for M tells us that the image is upright.
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EXERCISE (1 whole by pair)
1. Find graphically the image location for an objectat each of the following distances from aconverging lens that has a focal length of 20cm.
Check your results by calculating the imageposition and the lateral magnification,respectively.
A. do = 50 cm
B. do = 20 cm C. do = 15 cm D. do = - 40 cm
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EXERCISE (1 whole by pair)
2.)You are given a thin diverging lens.You findthat a beam of parallel rays spreads out afterpassing through the lens, as though all therays came from a point 20.0 cm from thecenter of the lens.You want to use this lens toform an erect virtual image that is 1/3 theheight of the object. (a) Where should the
object be placed? (b) Draw the principal raydiagram.