The following video will explain focal length, lens distance, and image distance for a lens.

In the video, the variables were listed as o (object distance), i (image distance), and f (focal length), and the equation relating these three quantities was one over 0 plus one over i equals 1 over f. 1 o + 1 i = 1 f .

To avoid confusion with o and the number 0, we are going to use the variables:

D_o = Object Distance
D_i = Image Distance
f = Focal length

The equation changes to 1 D_o + 1 D_i = 1 f .

The Hubble Space Telescope has a focal length of 57.6 meters.

Source: Hubble Space Telescope Illustration, NASA Marshall Space Flight Center Collection

1. Write an equation for D_o, the object distance, in terms of the image distance, D_i, and the focal length, 57.6 meters.
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Our original equation was 1 D_o + 1 D_i = 1 f. If we substitute the given focal length of 57.6 meters in for the variable f, we have:
1 D_o + 1 D_i = 1 57.6.
We are asked to write an equation for D_o in terms of D_i and the value of f. So we need to solve this equation for D_o. If we multiply the entire equation through by a common denominator, we'd have:

Now we bring all the terms with D_o to one side of the equation and factor.

And finally, divide both sides of the equation by(D_i - 57.6)to solve for D_o.


2. A graph of the equation D_o = fifty-seven point six times D sub i 57.6D_i D_i − 57.6 is shown below.

What is the domain of this function? Is it the same or different than the relevant domain of the problem situation?


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• The original equation was D_o = fifty-seven point six times D sub i 57.6D_i D_i − 57.6 . The domain of this equation is all real number, except D_i ≠ 57 (this value would make the denominator of the function 0, and hence make the function undefined).
• The domain of the problem situation is different from the domain of the entire function equation because the object distance or image distance cannot be negative, only positive values of D_othat yield positive D_i values are relevant. Therefore, the domain of the problem situation is D_o > 57.6.

3. What is the location of the vertical asymptote, and what meaning does it have in the context of this problem?
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• The equation of the vertical asymptote is D_o = 57.6.
• Values of D_o close to 57.6 produce image distances near infinity.

4. What is the equation of the horizontal asymptote, and what meaning does it have in the context of the problem situation?
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• The equation of the horizontal asymptote is D_i = 57.6.
• As the object distance, D_o , increases and gets very large, the image distance approaches a value of 57.6.