Assume that I have got a bowl half-filled with mercury. In this bowl, I place a convex lens which make one side of the lens fully reflecting. What will happen to the lens, will it behave like a concave mirror and what will be the change in the sign of the focal length??
[Physics] What happens when one side of a convex lens is made fully reflecting
lensesopticsreflection
Related Solutions
The focal length tell us how much the light rays will be bent. Think of the light rays as a paper cone, just like the one you get when you buy a snow cone. The mouth of the cone is the size of the lens, and the point of the cone is the focal point. The length of the cone is the focal length. Now picture a cone where the point is very close to the mouth. It would be a very steep short cone. The light rays would come in at very steep angles, and after they crossed the focal point they would spread out quickly. In fact, they would make another cone leaving the focal point that matches the cone that entered the focal point. The rays would continue on spreading out wider and wider. Now picture a cone that is very long, say three feet, a novelty snow cone, notice how these rays come into focus at a much smaller angle, and after they pass the focal point they will form another long cone on there way out. Well, there you have it, the focal length tells us how long the snow cone is. I hope your not offended by my simple examples and language in this answer. It's how I think of focal length. With a little imagination you can see how different cone sizes would be applicable to different applications.
The lens equation and the ray tracing does agree. Let's denote the distance from the object and image distances to the first convex lens as $o_1$ and $i_1$ respectively. Let's denote the object and image distances to the second concave lens as $o_2$ and $i_2$. The focal lengths are $f_1=2 \text{ cm}$ and $f_2=-2 \text{ cm}$.
The lens equation reads $\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$.
For the first lens, $\frac{1}{3\text{ cm}}+\frac{1}{i_1}=\frac{1}{2\text{ cm}}$.
Solving for the image distance yields $i_1=6\text{ cm}$.
The image of the first lens becomes the object for the second lens, and since the two lenses are $4\text{ cm}$ apart, we have $o_2 = 4\text{ cm}-i_1 = -2\text{ cm}$.
The lens equation for the second lens reads as $\frac{1}{-2\text{ cm}}+\frac{1}{i_2}=\frac{1}{-2\text{ cm}}$.
Thus $1/i_2 = 0$, and $i_2=\infty$. There is neither a real image (converging rays in front of the second lens) nor a virtual image that is formed (diverging rays). Rather the rays are parallel.
Well, we can see this agrees with the ray tracing diagram (made with OpticalRayTracer software ver. 9.6):
Best Answer
It will behave as a concave mirror with a convex lens in front of it, or just as a convex mirror if the other side is reflecting (with a lens behind it that doesn't really do anything because it's blocked). You would have to do ray tracing to see more precise behavior for the former case.