Solving Inverted Image w/Concave Mirror: 22 cm, 44 cm, & Focal Length

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Homework Help Overview

The problem involves determining the image distance and focal length of a concave mirror given that an inverted image is magnified by 2 when the object is placed 22 cm in front of the mirror.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the magnification formula and the implications of the negative sign in relation to image orientation. There is exploration of the conventions used in optics regarding image characteristics.

Discussion Status

Participants are actively clarifying the conventions related to magnification and image orientation. Some guidance has been provided regarding the interpretation of the magnification value as negative for inverted images, but no consensus has been reached on the application of the formulas.

Contextual Notes

There is an ongoing discussion about the conventions used in optics, particularly regarding the signs in formulas and their implications for image characteristics. The original poster expresses confusion about the relationship between the magnification value and the nature of the image.

noobie!
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Homework Statement



An inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror. Determine the image distance and the focal length of the mirror.

Homework Equations



don't have,sorry

The Attempt at a Solution


firstly,i use M= -di/do then i got di=-44cm ,then substitute in into the equation 1/f=1/do + 1/di ..1/f= 1/22 + (- 1/44) =44cm..did i get my steps correct?then i refer to answers,i noticed that their solution is 1/f= 1/22cm + 1/44cm without the negative..why could it be so?so,are my steps correct?the solution also stated image is real if its being describe upright..according to what I've read,it should be when a real image will describe as inverted image while virtual is to be described as upright and located behind the mirror,its so contradicting..so could you please help me to clear my doubts..thanks for your kind help!:smile:
 
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LowlyPion said:
http://en.wikipedia.org/wiki/Concave_mirror#Analysis

I think by the statement of the problem your magnification is -2 to begin with.

Also you will be dealing with this case:
http://en.wikipedia.org/wiki/File:Concavemirror_raydiagram_2FE.svg

oh..did you actually mean that convention merely a choice for me..so,if i place a either a negative sign or i don't will also be correct?..then the problem the magnification should be -2 instead of 2?if yes then i understand completely...?
 
noobie! said:
oh..did you actually mean that convention merely a choice for me..so,if i place a either a negative sign or i don't will also be correct?..then the problem the magnification should be -2 instead of 2?if yes then i understand completely...?

The sign is apparently a convention that indicates the case that you have.

What they gave you was that the magnification is 2. But inverted ... so it's -2.

The formula is m = - di/do = -2 ... so the signs cancel looks like to me.

2 = di/do
 
LowlyPion said:
The sign is apparently a convention that indicates the case that you have.

What they gave you was that the magnification is 2. But inverted ... so it's -2.

The formula is m = - di/do = -2 ... so the signs cancel looks like to me.

2 = di/do

ah..i got your point..understood..thanks for your kind help..thanks!:smile:
 

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