Optics: Thin Lens and Image Formation

[crzcrcketer89]

Homework Statement

A student is working in an optics lab and has a light bulb which is 100 cm from a screen. She needs to make a real image a real image of the bulb on the screen and she has a converging lens with a focal length of +15 cm. Find two places she can put the lens and give object and image distances to both setups.

Homework Equations

1/p + 1/q = 1/f

The Attempt at a Solution

f = +15 cm, p = 100 cm, so
1/.15 - 1/1.00 = 1/q
q = 17.6 cm

and

f= + 10 cm, p = 66.66 cm, so
1/.10 - 1/.6666 = 1/q
q = 11.7 cm

Are these two distances correct?

 

[Carid]

1/p + 1/q = 1/f

and

p + q = 100cm

Two equations, two unknowns.

Might need this

http://www.math.com/students/calculators/source/quadratic.htm

 

[flicker128]

Use substitution taking the above two equations. When you solve one variable (p or q), you can use them vice versa to find your other value, since you can do either p=100-q or q=100-p.

Put these substitutions into the thin lens equation and voila. I think that's how you do it.

 

[flicker128]

crzcrcketer89, you can't just use random values, and 100 cm is the distance between the LIGHT BULB and the SCREEN. The LENS goes in between, so p can't equal 100.