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

 

[kyle8921]

I can provide a response to this content by explaining the concepts of reflection and refraction and how they apply to this problem.

Firstly, reflection is the phenomenon where light bounces off a surface. In this case, some of the parallel rays entering the biconcave lens will be reflected off the first surface of the lens. The angle of reflection will be equal to the angle of incidence, as stated in the law of reflection.

On the other hand, refraction is the bending of light as it passes through a different medium with a different index of refraction. In this problem, the biconcave lens has an index of refraction of 2.00, which means that light passing through it will bend more than it would in air with an index of refraction of 1.00.

Now, in order to solve the problem, we need to understand how these two phenomena, reflection and refraction, work together. When light enters the lens, it will be split into two paths - some will be reflected and some will be refracted. The refracted light will continue to bend as it passes through the lens and eventually form an image on the other side. The reflected light, on the other hand, will bounce off the first surface and form a separate image.

However, since the radii of curvature of the lens are equal, the angles of incidence and reflection will also be equal. This means that the reflected rays will converge at the same point as the refracted rays. So, the image formed by the reflected light will overlap with the image formed by the refracted light, creating a single image at the same point.

In conclusion, by understanding the principles of reflection and refraction, we can see that the reflected image and the image formed by the lens will fall at the same point. This is because the rays of light are split and then converge again due to the equal radii of curvature of the lens.