Ask Question
21 August, 00:21

In order for a film camera with a lens of fixed focal length F to focus on an object located a distance x from the lens, the film must be placed a distance y behind the lens. F, y, and x are related as follows: 1/F=1/X + 1/Y. Now suppose a camera has a lens with focal length F = 65.1. Explain what happens to the focusing distance y as the object moves far away from the lens. 2. Explain what happens to the focusing distance y as the object moves closer and closer to the lens. 3. In general, why is it not possible to cross a vertical asymptote?

+2
Answers (1)
  1. 21 August, 00:27
    0
    y will increase if x is increased,

    y will decrease if x is decreased.

    Step-by-step explanation:

    1/F=1/X + 1/Y

    1/F-1/X=1/Y

    X-F/FX=1/Y

    FX/X-F=Y

    Rearranging, Y=FX/X-F

    So this is the function in terms of X, and F is just a constant

    Therefore, the function becomes Y=65X/X-65

    Explanation:

    y = focusing distance

    x = distance from lens

    Focusing distance y will increase if the object x moves far away from the lens i. e. x is increased,

    Similarly, focusing distance y will decrease if the object x moves closer and closer to the lens i. e. x is decreased.

    An asymptote is a function that mimics a curve f (x) as x approaches infinity.

    A vertical asymptote, however, cannot be crossed in a function. Remember that a function cannot have multiple y values for a given x value, hence the vertical line test for a function. If a function crossed a vertical asymptote and then went back to it, then it would have to go back over itself as it becomes arbitrarily close to the asymptote. Therefore, it is not possible to cross a vertical asymptote.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “In order for a film camera with a lens of fixed focal length F to focus on an object located a distance x from the lens, the film must be ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers