Diamond Ring in Front of a Convex Lens?

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Homework Help Overview

The problem involves a diamond ring placed in front of a convex lens, with specific dimensions and distances provided. Participants are tasked with determining the position and size of the image, as well as the magnification of the lens.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster identifies the object distance and height but expresses uncertainty regarding the focal length of the lens. Some participants suggest using the lens maker's formula to find the focal length, while others propose a simplified assumption based on the radius of curvature.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the type of optical device involved. There is a shift in understanding as one participant reveals that the teacher intended to refer to a mirror instead of a lens, which appears to clarify the problem for some.

Contextual Notes

There is a lack of information regarding the index of refraction of the lens material, which affects the calculation of the focal length. The initial setup is also complicated by the potential miscommunication about whether to consider a lens or a mirror.

itsgood819
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Homework Statement



A 1.5 cm high diamond ring is placed 20 cm in front of a convex lens whose radius of curvature is 30 cm.

a) What is the position and the size of the image?
b) What magnification does this lens have?

Homework Equations



1/f= 1/di + 1/do
m= -di/do or hi/ho

The Attempt at a Solution



the 20 cm is do, and the ho is 1.5 cm. Since it's a lens, I don't know the focal length. I tried using the 1/f= 1/di + 1/do, but I don't any variable besides the do.

How would I solve this problem? Help please, and thank you! Also, I'm new to this forum so my posting might look a bit weird.
 
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Is it a double convex lens with both sides having the same radius of curvature? If so, you can use the lens maker's formula to figure out the focal length.
 
Because, the index of refraction of the lens material is not given, I would assume that f=R/2.
 
thank you for your suggestions! it turns out that my teacher actually meant to put mirror instead of lens.
 
itsgood819 said:
it turns out that my teacher actually meant to put mirror instead of lens.
Aha! Now the problem makes sense. :wink:
 

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