Plano convex lenses and focal length

[physics user1]

Does the orientation of a plan convex lens affect it's focal lenght?
If I put il like in the first example in the photo and makeep the calculation using the lens maker equation I get f positive
What happens if I turn the curved face on the right?
Using the lens maker equation now the radius is negative so the focal length should be -f...

Is that Right?

 

[Drakkith]

Does the orientation of a plan convex lens affect it's focal lenght?

It does not.

If I put il like in the first example in the photo and makeep the calculation using the lens maker equation I get f positive
What happens if I turn the curved face on the right?
Using the lens maker equation now the radius is negative so the focal length should be -f...

That not correct. The lens maker equation has two terms in it that matter here: $\frac{1}{R1}-\frac{1}{R2}$
If you flip your lens around, the sign of the radii of curvatures change, but then surface 1 becomes surface 2 and vice versa. If R1 is initially 100 and R2 is ∞, then $\frac{1}{100}-\frac{1}{∞} = 0.01 - 0 = 0.01$

Flip the lens around and you get: $\frac{1}{∞}-\frac{1}{-100} = 0 - (-0.01) = 0.01$

 

[Ibix]

Clearly not. Imagine a symmetric biconvex lens - that would have zero power if you were correct.

You need to be careful with sign conventions in optics. If you are switching signs on your radius of curvature when you switch sides then the relevant formula uses $(1/r_1-1/r_2)$ so you get the same overall sign. Sometimes you see the convention that all convex curves are positive, but that would give $(1/r_1+1/r_2)$.

Edit: It does affect aberration. Best to put the curved face on the side where the rays are nearer parallel.

 

[physics user1]

Thank you guys I got it now

 

[physics user1]

But, how do I choose surface one and 2?

 

[Ibix]

Surface 1 has the lower x coordinate of you're using the $(1/r_1-1/r_2)$ form. It doesn't matter if you're using the other form.

Keep in mind a simple case, such as a biconvex lens. You know that f is positive. That let's you make a workable choice for convention: both positive and use $(1/r_1+1/r_2)$, or surface 1 positive and 2 negative and use $(1/r_1-1/r_2)$. Then you can flip signs as appropriate for the surfaces of the lens you actually have.

 

[physics user1]

Correct? The surface 1 is then the firSt thwt the light hits

 

[Ibix]

It's all just convention. Usually light rays enter the system from the left, moving to the right, which is the +x direction. And curvatures are defined as positive if the inside of the curve is on the +x side of the surface. And $r_1$ would be the radius of the first surface. In that case, yes you are right.

But I advise you to keep a simple case (e.g. a symmetric biconvex lens) in mind and figure out the signs as you go. With the symmetric biconvex lens you know that the two $1/r$ terms must end up with positive signs somehow (either because the radii have opposite signs and the two terms do too or because the two terms have the same sign and so do the radii) because otherwise the power is zero.

Then you won't be fooled by people picking different conventions.