Solve Concave Mirror Problems: Find Image Distance & Focal Length

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Homework Help Overview

The discussion revolves around a concave mirror problem involving the calculation of image distance and focal length based on the characteristics of virtual images. The original poster presents a scenario where a virtual image is three times the height of the object, with the object positioned 21 cm in front of the mirror.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationships between object distance, image distance, and magnification, questioning the application of formulas for virtual images. There is an exploration of potential errors in the original calculations and a request for clarification on the implications of the image being virtual versus real.

Discussion Status

The discussion is ongoing, with participants providing guidance on re-evaluating the equations used and the relationships between the distances. Some participants express confusion regarding the magnification factor and its impact on the calculations, while others suggest revisiting the original equations for clarity.

Contextual Notes

Participants are navigating the complexities of mirror equations and the specific conditions that apply to virtual images, indicating a need for further exploration of these concepts. There is also mention of the original poster's struggle with the problem's parameters, particularly the magnification aspect.

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Homework Statement


1)A concave mirror produces a virtual image that is three times as tall as the object.
a) If the object is 21 {\rm cm} in front of the mirror, what is the image distance?
b) What is the focal length of this mirror?

Homework Equations


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


The Attempt at a Solution


hi/ho = 3 thus do/di = 3 a) di = -do /3 = -7.0cm (was wrong)
thus i could not do the second part. Where did I go wrong? And if it was a real image instead of a virtual image, what what you do different? Thanks
 
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Hello matt72lsu,

matt72lsu said:

Homework Statement


1)A concave mirror produces a virtual image that is three times as tall as the object.
a) If the object is 21 {\rm cm} in front of the mirror, what is the image distance?
b) What is the focal length of this mirror?

Homework Equations


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

hi/ho and do/do are not really equations. If you put these in full equation form, ensuring that they apply to virtual images in concave mirrors, it might help.

The Attempt at a Solution


hi/ho = 3 thus do/di = 3 a) di = -do /3 = -7.0cm (was wrong)

I believe you've got your do and di switched around.

thus i could not do the second part. Where did I go wrong? And if it was a real image instead of a virtual image, what what you do different? Thanks

For one thing, knowing whether the image is real or virtual gives you information about where the object is relative to the mirror's focal length (is the object between the focal length and the mirror or is it on the far side of the focal length?).
 
i'm sorry, i just have no clue what to do. the whole 3 times larger part is throwing me off
 
matt72lsu said:
i'm sorry, i just have no clue what to do. the whole 3 times larger part is throwing me off

If the image is 3 times as large, than it's 3 times as far away. :wink:

Start by going back to your original equations.

Earlier you calculated
hi/ho = 3 thus do/di = 3

But that should be di/do = 3.
 
i got this one already
 

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