Creating an Image 2x Size of Object w/ Convex Lens: Half Focal Length Away

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Homework Help Overview

The discussion revolves around a problem involving a converging (convex) lens that creates an image twice the size of the object. Participants are examining the placement of the object relative to the focal length of the lens.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are analyzing the relationship between object distance, image distance, and focal length using the lens formula. There is confusion regarding the interpretation of the focal length in relation to the object distance.

Discussion Status

Some participants are exploring different interpretations of the relationship between the focal length and object distance. There is an ongoing clarification regarding the mathematical expressions and their implications, with no explicit consensus reached.

Contextual Notes

Participants are grappling with the definitions and relationships in lens optics, particularly how to interpret the focal length in the context of the problem. The discussion reflects a mix of understanding and uncertainty regarding the setup and calculations involved.

Gear2d
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Homework Statement


A converging lens (convex lens) creates an image that is 2 times the size of the object. The object is placed:


Homework Equations



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

The Attempt at a Solution



I thought the answer was that the object is place two focal lengths in front of the lens. But the answer says its half a focal length in front of the lens.

I mean: -di/do = hi/ho = 2
-di = 2do
1/f = 1/di + 1/do
= -1/2do + 1/do = 1/2do
f = 2do

Isn't f=2do saying two focal lengths?
 
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Gear2d said:

Homework Statement


A converging lens (convex lens) creates an image that is 2 times the size of the object. The object is placed:


Homework Equations



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

The Attempt at a Solution



I thought the answer was that the object is place two focal lengths in front of the lens. But the answer says its half a focal length in front of the lens.

I mean: -di/do = hi/ho = 2
-di = 2do
1/f = 1/di + 1/do
= -1/2do + 1/do = 1/2do
f = 2do

Isn't f=2do saying two focal lengths?

I'm not sure what you mean. If [itex]f=10[/itex] for example, then [itex]d_o=5[/itex], so [itex]d_o[/itex] is half the focal length.
 
So simple, thanks. I guess I just missed an easy answer by thinking too much (or thinking to less).

Thanks again
 
Gear2d said:
Isn't f=2do saying two focal lengths?

No.

f=2do is saying do=f/2, or half a focal length.

(do=2f would be saying two focal lengths.)
 

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