I have one question about optics because I start interested in it. If an object is placed a distance p from a thin glass lens (index of refraction n), we can see its image on a screen that is placed a distance d from the lens. Do you know any formula which can describe this situacion? I found it in Czech internet, but I did not find some good. However, the most important is this part. Light does not spread in the liquid as well as in the air. So if I take this system to liquid, something will change. We have different conditions... The distance d isn't same if we want see image on a screen... Will be larger or smaller? And why? How can I calculate it? Do you know any internet page on this topic, or video with the attempt? Thanks very much and sorry for my English.
Hello Numeriprimi, I believe the equation you are looking for is [itex]\frac{1}{f} = \frac{1}{p} + \frac{1}{d}[/itex] where [itex]f[/itex] is the focal distance of the lens, and [itex]p[/itex] and [itex]d[/itex] are the object's distance from the lens and the image's distance from the lens, respectively. As you could expect, [itex]f[/itex] will depend on the the refractive index of the medium, [itex]n_{m}[/itex], and on the refractive index of the glass (of which the lens is made of), [itex]n_{g}[/itex]. The equation that dictates this dependence is [itex]\frac{1}{f} = ( \frac{n_{g}}{n_{m}} - 1 )( \frac{1}{R_{1}} + \frac{1}{R_{2}} )[/itex] being [itex]R_{1}[/itex] and [itex]R_{2}[/itex] the radius of curvature of the two surfaces of the lens. Let's now suppose that he have a lens with a certain shape, and so [itex]R_{1}[/itex] and [itex]R_{2}[/itex] are fixed. As you can see from the above expression, depending on the ratio between the refactive indices, [itex]\frac{n_{g}}{n_{m}}[/itex], the focal distance can be positive or negative, and this will ultimately determine the behavior of the lens as being convergent or divergent. This is why the same lens can have different properties when placed inside diferent media. If you want to make some calculations on your own, be careful with the sign conventions! For example, for concave surfaces, the radius of curvature is negative; while for convex surfaces, the radius of curvature is positive. A quick guide to the sign convention of the other variables involved is: http://www.tutorvista.com/content/physics/physics-ii/light-refraction/convention-lenses.php I hope this helps! Best, Zag
Thanks very much :-) Are there any other conditions that affect image on a screen? If the liquid is too thick, it can be?
You are welcome! I am not sure if it would be possible to relate the viscosity of a fluid with its refractive index in a general way. I believe it will really depend on the material of which your fluid is made of since viscosity and refractive index seem to emerge from different phenomena. But I am not sure about that, so I will leave your question to be answered by someone who knows more than me about this topic! ;) Best, Zag