Optics: Spherical Interface -- Real and Virtual Images

AI Thread Summary
In discussions about spherical interfaces between different media, the nature of the image—real or virtual—depends on the object's distance from the interface and the focal length of the system. When the object is closer than the focal length, the image formed is virtual. Snell's Law is applicable, but the curvature of the interface complicates direct calculations. Resources like the provided link can help clarify the relationship between object distance and image distance. Understanding these principles is essential for solving problems related to refraction at spherical surfaces.
Chetlin
Messages
35
Reaction score
0

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.

eeEtiUM.png

Homework Equations



I'm sure there is a simple one somewhere but I can't find it in my notes or book :(--
Thanks!
 
Physics news on Phys.org
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!
 
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top