How Is the Radius of Curvature 8.68 mm Calculated for a Lens?

[roam]

Homework Statement

Parallel light in air enters a transparent medium of refractive index 1.33 and is focused 35 mm behind the surface. Calculate the radius of curvature of the surface of the medium

Homework Equations

$f = \frac{R}{2}$

$\frac{1}{f}=(n-1) \left( \frac{1}{R_1}-\frac{1}{R_2} \right)$

The Attempt at a Solution

The correct answer must be 8.68 mm, I can't see how they got this answer.

We know that the focus is 35 mm, so if we use the equation

$R=2f=2 \times 35 = 70 \ mm$

But this is not correct and it doesn't take into account the refractive index.

So, I also tried using the lens-maker's equation:

$\frac{1}{35} = (1.33-1) \left( \frac{1}{R}- 0 \right) \implies R = 17.9$

I wasn't sure what to use for the second radius so I used 0, and I didn't get the correct answer. So how can I get 8.68 mm?

 

[Mindscrape]

I don't think you understood the problem correctly. You have light in one medium entering another medium through a parabolic surface. Not sure the level of your class, but since it's posted in intro phys, I'll just give you the formula.

$$F=n_ 2 R/(n_2-n_1)$$