Can someone explain to me how i can figure out the sign of the focal length with lens

1. Mar 15, 2006

mr_coffee

Hello everyone, i'm confused on this issue. How can I tell if the focal length is going to be postive or negative when 2 lens are combined like the ones below:
Here is the directions:
You grind the lenses shown in Figure 34-49 from flat glass disks (n = 1.5) using a machine that can grind a radius of curvature of either 15 cm or 30 cm. In a lens where either radius is appropriate, you select the 15 cm radius. Then you hold each lens in sunshine to form an image of the Sun.

What is the focal length f for each lens?

for some reason i'm getting this wrong, all i'm doing is taking the center of radius and dividing it by 2 to get the focal length. But I'm not getting that right either. I don't see where i'm messing up. THe professor just said, take the given cirvature of raidus which is 15, divide it by 2 and thats ur answer, then you have to decide if its positve or negative. I tried both postive and negative both wrong. Any ideas?

In my notes i have it saying:
Focal length of a convex lens is +
Focal Length of a concave lens is -
so if i had a bi convex, is that like a convex and a concave put together, there for, a end result of a negative focal length becuase a postive times a negative is -?

Here is my work:

http://img146.imageshack.us/img146/5817/lastscan5bq.jpg [Broken]

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2. Mar 16, 2006

hellraiser

For lenses the f = R/2 is not valid. It is for mirrors.
The correct one is
1/f = (u-1) (1/r1 - 1/r2)

(Use the New Cartesian Convention. Dunno which one you use. But we are supposed to use this :) )

so 1/f = (1.5-1) (1/15+1/15)
so f = 15cm.

And for combination of lenses placed together you may use the power of a lens thing.
P = 1/f f is in m
so combined power is P = 1/f1 + 1/f2
use the sign of focal length with sign.

and as far as i remember bi convex is a lens with both sides convex i.e both sides outward. Then there is a plano concave, plano convex, bi convex. Although I never mess with them, I always get confused.

PS If you don't use the New Cartesian Convention better wait for someone who can help with the Real Virtual thing.

3. Mar 16, 2006

mr_coffee

Thanks hell raiser, the question wants

So if i'm finding the combined power, is that still the focal length?
Like for instance if i had a (4) bi-concave, which looks like #4 in the drawing.
I would have
f = (1-1.5)(1/15+1/15)
f = 15cm;

) would be concave
( would be convex
and a bi-concave looks like those 2 put together )(
The rule i have is, focal length of a convex is +
The focal length of a concave is a -
so would i do this?
combined power is P = 1/f1 + 1/f2
P = -1/15 + 1/15 = 0?
That makes no sense, do u see where I messed up on understanding what you said? Thanks for the reply!

4. Mar 16, 2006

Staff: Mentor

Looks to me that you are confusing combining surfaces with combining lenses. (They are not the same! The power of a lens surface depends on what side of the lens it's on.) You need to use the Lens Maker's formula to find the focal length of these lenses (as hellraiser described). Here's a useful page describing that equation and its sign convention: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenmak.html#c1

Here's a quick tip: You can quickly tell if a lens is converging (+ focal length) or diverging (- focal length) by comparing the thickness of the middle to the thickness of the edges: if the middle is thicker, it's converging.

5. Mar 17, 2006