Object magnification after Convex surface

Click For Summary

Homework Help Overview

The discussion revolves around determining the magnification of an object positioned in front of a convex surface, with specific parameters including object distance, radius of curvature, and index of refraction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of the lens maker formula and geometric considerations for calculating magnification. There is a discussion about the nature of the surface being a convex interface rather than a thin lens, prompting clarification of the appropriate equations for refraction.

Discussion Status

The conversation is ongoing, with participants providing different perspectives on the problem setup and the equations applicable to the scenario. Some guidance has been offered regarding the use of the general equation of refraction at a spherical surface.

Contextual Notes

There is a noted distinction between a thin lens and a convex interface, which may affect the approach to solving the problem. The specific parameters provided, such as distances and indices, are crucial for further discussion.

dimensionless
Messages
461
Reaction score
1
I have an object that is surrounded by air and sits a distance U=30 from a convex surface. The convex surface has a radius of R=5. The index of refraction of the refracting media is n=1.33. How do I determine the magnification of the object?
 
Physics news on Phys.org
I think, if you find out where the object falls after refraction using the lens maker formula([itex]R_2 = \infty[/itex]), you can calculate the magnification using some simple geometry
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenmak.html"
 
Last edited by a moderator:
It is not a thin lense. It is a convex interface between two materials.
 
dimensionless said:
It is not a thin lense. It is a convex interface between two materials.
My mistake then.
The general equation of refraction at a spherical surface will be

[tex]\frac{\mu_2}{v} - \frac{\mu_1}{u}=\frac{\mu_2 - \mu_1}{R} [/itex][/tex]
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
19
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K