HELP Finding the focal length of a lens, I'm totally stuck

[Kalibasa]

Homework Statement

A slide projector needs to create a 81 cm high image of a 2.0 cm tall slide. The screen is 360 cm from the slide. Assume that it is a thin lens. What focal length does the lens need? How far should you place the lens from the slide?

Homework Equations

1/f = 1/s + 1/s' (f is the focal length, s is the object distance, s' is the image distance)
M= h'/h = -s'/s (M= magnification)

The Attempt at a Solution

This problem is driving me crazy! From the wording of the question, it sounds like s' + s= 360 cm. I also know that M= h'/h= 40.5cm, and therefore -s'/s = 40.5cm. Almost everything I've tried has involved these equations.

I tried plugging 360-s' in for s, so -s'/(360-s')=40.5. But when I solved this I got 369.1 cm for the image distance, and I don't see how that can be right given that this is more than the image distance and the object distance added together. It should be less than 360 cm.

All the other substitutions I've tried have gotten me nowhere, as I keep running into the same wall- even if I get an equation down to only s' or s, I still know neither value and can't solve it. :(

 

[mgb_phys]

First draw a diagram.

Second, you are given the total distance, object->lens->image ie 360 = s + s'
Most of the optics equations use these distance separately so you have an extra step.

 

[Kalibasa]

I already know that s' + s= 360, but that hasn't helped. I know it's a vital piece of information, but I don't know how to use it- everything I've tried hasn't worked

 

[mgb_phys]

You have 1/f = 1/s + 1/s' and you know s+s' = 360cm
M= h'/h = s'/s = 81cm/2cm = 40.5

So just from simultaneous equations you should be able to get s and s' (without any optics considerations)

 

[Kalibasa]

Simultaneous equations? What do you mean? Haha I haven't taken algebra in ten years and maybe it's showing... :uhh:

 

[mgb_phys]

s + s' = 360cm
s'/s = 40.5 so, s'=40.5 s

Substituting in the first equation
s + 40.5 s = 360cm so, s ( 1+40.5) = 360cm

 

[Kalibasa]

Oh! That's exactly what I did except that I did -40.5 + 1 = -39.5 instead of 40.5+ 1= 41.5. Maybe I got too caught up in the signs from the equation...

Ok thank you so much! Haha that took a lot of posts

 

[mgb_phys]

. Maybe I got too caught up in the signs from the equation...

That's why you always draw a diagram first !

It also helps to have a guess at the asnwer. You know that the distance from the slide to the lens in a projector is only going to be 10cm or so, and if you have a magnification of about 40 this distance is going to be about 1/40 of the total distance.
So you are looking for an answer around 10cm