Solving the Biconcave Lens Problem with Reflection and Refraction

[NutriGrainKiller]

Here's the problem:

Parallel rays along the central axis enter a biconcave lens, both of whose radii of curvature are equal. Some of the light is reflected from the first surface, and the remainder passes through the lens. Show that, if the index of refraction of the lens (which is surrounded by air) is 2.00, the reflected image will fall at the same point as the image formed by the lens.

Here's what I understand:

I know what a biconcave lens looks like, and how it behaves (for the most part). Of course I know that the index of refraction is ~1.00. And since the radii are equal that makes the equation somewhat easier. The thick lens equation is:

(Sorry I am not yet familiar with LaTex, so just *imagine* that (1/R1)-(1/R2) isn't there).

What I don't understand:

Where is the image diverging from? "Parallel rays from the central axis convirge into a reflected and transmitted image"..does this just mean where the rays convirge?

Any guidance would be appreciated. Thanks as always!

 

[OlderDan]

Here's the problem:



Here's what I understand:

I know what a biconcave lens looks like, and how it behaves (for the most part). Of course I know that the index of refraction is ~1.00. And since the radii are equal that makes the equation somewhat easier. The thick lens equation is:

(Sorry I am not yet familiar with LaTex, so just *imagine* that (1/R1)-(1/R2) isn't there).

What I don't understand:

Where is the image diverging from? "Parallel rays from the central axis convirge into a reflected and transmitted image"..does this just mean where the rays convirge?

Any guidance would be appreciated. Thanks as always!

So, can we reduce the equation to

1/f = d/2R^2

For parallel incoming rays the "object" is at infinity and the reflected image is a point. For spherical reflectors this point is half a radius from the surface. I'm not sure I follow the sign convention for the formula, and it's not immediately clear to me how the "transmitted image" will appear. A biconcave lens would form a virtual image, and with thin lenses that would mean a negative focal length. I know that not all sign conventions are the same, so maybe all is OK. I can see how the "transmitted image" could mean the virtual image of the diverging lens. It's not obvious that d has no effect on the position of the virtual image, but maybe that is the point of the problem