Gbenga needs to get glasses to correct his farsightedness. His eyes currently cannot focus on objects that are within 2 ft (or 61 cm) of his

Question

Gbenga needs to get glasses to correct his farsightedness. His eyes currently cannot focus on objects that are within 2 ft (or 61 cm) of his eyes. This is in contrast to people with normal vision who can focus on objects as close as 25 cm in front of them. If the glasses that Gbenga will get will sit 1.6 cm in front of his eyes, what lens focal length and power would correct his vision? That is, what lens focal length and power would allow Gbenga to focus on objects that are 25 cm in front of his eyes? Will they be converging or diverging lenses?

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Hưng Khoa 5 years 2021-07-20T19:07:59+00:00 1 Answers 28 views 0

Answers ( )

  1. Amity
    0
    2021-07-20T19:09:30+00:00
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    Answer:

    Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.

    Explanation:

    When an object is placed 25cm from Gbenga’s eye, the glasses lens must produce an image 61cm away (Gbenga’s eye near point).

    An image 61cm from the eye will be (61cm – 1.6cm) from the glasses.

    i.e. d_{i}=61cm-1.6cm=59.4cm

    and d_{o} = 25cm - 1.6cm = 23.4cm

    note d_{i} will be negative because the image is formed on the same side as the object.

    finally, d_{i}=-59.4cm\\d_{o}=23.4cm

    the formula for finding the focal length f is given as

    f=\frac{d_{i}d_{o} }{d_{o}+d_{i} }

    f=\frac{-59.4*23.4}{23.4-59.4} \\

    f=\frac{-1389.96}{-36}

    f=38.61cm

    The focal length is positive which indicates converging lens

    power p=\frac{1}{f}

    but f\\ must be in metres

    Therefore, f=38.61cm=0.3861m

    p=\frac{1}{0.3861}

    p=2.59 Diopter

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