Optics: Spherical Interface -- Real and Virtual Images

[Chetlin]

Homework Statement

If you have a spherical interface between two different "media" (like air and water), and an object is placed in the one with the lower index of refraction, with the interface being convex toward the object, how can you tell if the image will be real or virtual?

Here's a picture of the two different possibilities. The white area has a lower index of refraction than the blue area, so the rays will bend toward the normal when they strike it. But how far they bend determines if the image will be real or virtual. I know about Snell's Law, but I think it would be difficult to apply here since the interface is not flat. Also, I am pretty sure whether the image is real or virtual depends on the distance the object is from the interface, but I don't know how to find that distance.

Homework Equations

I'm sure there is a simple one somewhere but I can't find it in my notes or book :(--
Thanks!

 

[ehild]

Read http://www.math.ubc.ca/~cass/courses/m309-01a/chu/MirrorsLenses/refraction-curved.htm

ehild

 

[Chetlin]

Thanks for the link! So it all just comes from Snell's Law? That page doesn't have any example of a case where the interface is convex but the image is virtual, however this must happen at least sometimes, since the thin-lens formula is derived from the first interface giving a virtual image.

I just found something in my notes talking about the focal length, and if the object's distance from the interface is less than the focal length, then the image will be virtual. I'll look up how to find what this focal length is. Thanks!

 

[ehild]

The focus is the image point when the object is at infinity. If the object is at the other focus, the refracted rays are parallel with the principal axis, so the image distance is infinity. You find the relation between object distance and image distance at that URL. I am sure, you also find these topics in your lecture notes.
Or browse "refraction at spherical surfaces". You find lot of places.

ehild

 

[HalfLostAndSearching]

I can provide some insights into this problem. First, let's define what is meant by a real and virtual image. A real image is one that can be projected onto a screen, whereas a virtual image cannot be projected and is only seen through the eye or a camera.

In this scenario, the object is placed in a medium with a lower index of refraction, which means that the light rays will bend towards the normal when they hit the interface. The amount of bending, or refraction, depends on the angle at which the light rays hit the interface and the index of refraction of the two media. This can be calculated using Snell's law.

Now, to determine if the image will be real or virtual, we need to consider the position of the object in relation to the interface. If the object is located beyond the focal point of the interface, the image formed will be real. This is because the light rays will converge and can be projected onto a screen.

On the other hand, if the object is located between the interface and the focal point, the image formed will be virtual. This is because the light rays will diverge and cannot be projected onto a screen. Instead, the image will be seen through the eye or a camera.

To determine the distance at which the object must be placed to form a real or virtual image, we can use the thin lens equation:

1/f = 1/do + 1/di

Where f is the focal length of the interface, do is the distance of the object from the interface, and di is the distance of the image from the interface.

Solving this equation for different values of do will give us the corresponding values of di. If di is positive, the image is real, and if di is negative, the image is virtual.

In conclusion, the position of the object in relation to the interface and the focal length of the interface can determine if the image formed will be real or virtual. Snell's law can also be used to calculate the amount of refraction at the interface, which is necessary in determining the position of the image.