A convex spherical refracting surface

Question

I'm trying to solve the following problem:

A convex spherical refracting surface with a radius of 0.6 m separates a medium whose index of refraction is 1.5 from another whose index of refraction is 2.0. An object is placed in the first medium 1.0 m from the surface. Determine the focal distances of the object and image, the position of the image, and the magnification of the image.

Here is my attempt:

The focal length of a lens or a refracting surface can be determined using the lens formula:

$1/f = (n_2 - n_1)(1 / R_1 - 1 / R_2)$

Since we have a single convex spherical refracting surface, we can assume the other surface has an infinite radius of curvature, so the lens formula simplifies to:

$1/f = (n_2 - n_1) \cdot (1 / R_1)$

Substituting the given values:

$1/f = (2.0 - 1.5) \cdot (1 / 0.6)$

Solving for $f$:

$1/f = 0.5 \cdot (5 / 3)$

Therefore, $f = 1.2 \, \text{m}$

But I should obtain that it is 1.8m. Can someone help me?

You can't apply the lens formula. That one assumes that the light moves from $n_1$ to $n_2$ and back to $n_1$. — Andrei, Dec 06 '23 at 20:44
@Andrei Could you please explain how I can approche this problem then? — user1259404, Dec 06 '23 at 21:13

score 0 · Answer 1 · answered Dec 07 '23 at 04:32

Any introductory optics book contain a chapter about refraction from spherical surfaces. For example, you can take a look at this. The formula you want to use is $$\frac{n_1}{d_1}+\frac{n_2}{d_2}=\frac{n_2-n_1}R$$ Notice that you have different focal lengths on the two sides of the surface. If the object is at infinity, $d_1=\infty$, so the image is in the focal point $f_2$: $$f_2=R\frac{n_2}{n_2-n_1}=2.4m$$ If you want the image to be at infinity, $d_2=\infty$, you put the object at $f_1$: $$f_1=R\frac{n_1}{n_2-n_1}=1.8m$$

A convex spherical refracting surface

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