Precalculus question 1/x+1/y=1/F?
For a 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, where F, x, and y are related by
1/x+1/y=1/F
suppose the camera has a 55-mm lens (F=55)
a. Express y as a function of x
b. What happens to the focusing distance y as the object moves far away from the lens?
c. What happens to the focusing distance y as the object moves close to the lens?
Thanks!! :)
More Questions From This User See All
Answers & Comments
Verified answer
Ah...
1/x + 1/y = F
Multiply both sides by xy to get...
y + x = Fxy
Rearrange the terms to make one side having y variables.
y - Fxy = -x
y(1 - Fx) = -x
y = -x/(1 - Fx)
So this is the function in terms of x [Don't forget that F is just a constant].
Now...
b. What happens to the focusing distance y as the object moves far away from the lens?
c. What happens to the focusing distance y as the object moves close to the lens?
Let F = 1. Then, we have y = -x/(1 - x) where...
y = focusing distance
x = distance from lens
If you increase the x distance from lens, then y decreases.
If you decrease the x distance from the lens, then y increases.
Check the relationship between x and y. http://www.wolframalpha.com/input/?i=-x%2F%281+-+x...
Good luck!