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IMAGE FORMATIONS IN LENSES AND THIN LENS EQUATIONS

Apr 08, 2018

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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.