How Do You Calculate the Focal Length of a Plane Convex Lens?

[Lemmy]

Homework Statement

A plane convex lens is made of glass (index 1.42) with one flat surface and the other having a radius of 24 cm. What is the focal length of the lens? Anwer in united of cm.

Homework Equations

Lensmaker Equation
1/f=(n-1)*((1/r1)-(1/r2))

The Attempt at a Solution

f=1/[(n-1)*(1/r1)-(1/r2))]

f=1/[(1.42-1)*((1/-24cm)-(1/0))]
f=-57.142857

Is this right?? If the surface is flat on the lens, radius is given as 0 then correct? please check my work.

 

[Doc Al]

If the surface is flat on the lens, radius is given as 0 then correct?

Think about it. As the radius of a sphere gets smaller, does the surface get flatter or more curved?

 

[Lemmy]

Think about it. As the radius of a sphere gets smaller, does the surface get flatter or more curved?

Oh wait hrmm, when the probelm refers to the surface, is it talking about the tangent line?
or would the image look like this (| which i thought the lens looked like, but you must be saying it looks like this instead of ()/ with the / being the surface.

OHHhhh hrmm, lol CONVEX meaning Radius is POSITIVE!

so i know that it's not 1/-24, but 1/24 but is r2 still 1/0?

 

[Doc Al]

Oh wait hrmm, when the probelm refers to the surface, is it talking about the tangent line?
or would the image look like this (| which i thought the lens looked like, but you must be saying it looks like this instead of ()/ with the / being the surface.

Yes, the lens looks like (| or |).

OHHhhh hrmm, lol CONVEX meaning Radius is POSITIVE!

Whether the radius is positive or negative depends on its orientation and your sign convention. (See the two possibilities above.) But that won't change the focal length.

so i know that it's not 1/-24, but 1/24

OK, if the shape is (|.

but is r2 still 1/0?

Not sure what you mean here. What's the "radius" of a flat surface? (See my hint earlier.)

 

[Lemmy]

Not sure what you mean here. What's the "radius" of a flat surface? (See my hint earlier.)

for the "but is r2 still 1/0?"

what i am asking , is the radius of curvature for the second lens surface given as 0 since it is flat? Because what flat surface has a radius? is what i am proposing.

In the lens maker equation light is traveling through the convex lenx first then, through the flat surface which is 1/r2 in the equation. so what i am asking is am i correct by saying 0 is the radius of curvature for the flat surface?

Light ray -------> (|

 

[Doc Al]

what i am asking , is the radius of curvature for the second lens surface given as 0 since it is flat? Because what flat surface has a radius? is what i am proposing.

I understand. But I want you to figure it out for yourself!

Try this: Compare a giant ball (large radius) with a tiny ball (small radius). Which surface is flatter? (What I'm getting at is: Would you represent a flat surface as a spherical surface with large radius or small radius?)

 

[Lemmy]

I understand. But I want you to figure it out for yourself!

Try this: Compare a giant ball (large radius) with a tiny ball (small radius). Which surface is flatter? (What I'm getting at is: Would you represent a flat surface as a spherical surface with large radius or small radius?)

Well okay, Giant ball = Earth, surface looks flat. Small ball tiny ball, looks curved. Lol but i still don't understand my 1/0 lol

But i would represent a flat surface as a spherical surface with a large radius. So isn't 1/0 flat?

 

[cepheid]

I think you've got the answer, so I don't know what your issue is. Think logically:

- surface 2 is flat
- flat means zero curvature (the surface is not curved)
- the definition of curvature is 1/(radius of curvature) = 1/r2
- therefore, 0 = 1/r2
- r2 = ?

 

[Doc Al]

But i would represent a flat surface as a spherical surface with a large radius.

Exactly. So r = infinity would be the radius for a flat surface. (That's different from what you were saying earlier.)

 

[Lemmy]

Exactly. So r = infinity would be the radius for a flat surface. (That's different from what you were saying earlier.)

AHhhh i see, thank you :)