A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the scre

Question

A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the screen. In terms of f, what is the smallest value d can have for an image to be in focus on the screen?

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Thu Cúc 6 months 2021-08-09T05:20:28+00:00 1 Answers 18 views 0

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    2021-08-09T05:22:08+00:00

    Answer:

    2f

    Explanation:

    The formula for the object – image relationship of thin lens is given as;

    1/s + 1/s’ = 1/f

    Where;

    s is object distance from lens

    s’ is the image distance from the lens

    f is the focal length of the lens

    Total distance of the object and image from the lens is given as;

    d = s + s’

    We earlier said that; 1/s + 1/s’ = 1/f

    Making s’ the subject, we have;

    s’ = sf/(s – f)

    Since d = s + s’

    Thus;

    d = s + (sf/(s – f))

    Expanding this, we have;

    d = s²/(s – f)

    The derivative of this with respect to d gives;

    d(d(s))/ds = (2s/(s – f)) – s²/(s – f)²

    Equating to zero, we have;

    (2s/(s – f)) – s²/(s – f)² = 0

    (2s/(s – f)) = s²/(s – f)²

    Thus;

    2s = s²/(s – f)

    s² = 2s(s – f)

    s² = 2s² – 2sf

    2s² – s² = 2sf

    s² = 2sf

    s = 2f

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