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

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SUMMARY

The discussion centers on calculating the image distance and focal length of a concave mirror when an object is placed 22 cm in front of it, resulting in a magnification of -2. The user correctly determined the image distance (di) to be -44 cm using the magnification formula M = -di/do. The focal length (f) was calculated as 44 cm using the mirror equation 1/f = 1/do + 1/di. The confusion arose from the sign convention, where a positive magnification indicates an upright image, while a negative magnification indicates an inverted image.

PREREQUISITES
  • Understanding of concave mirrors and their properties
  • Familiarity with the mirror equation: 1/f = 1/do + 1/di
  • Knowledge of magnification formulas: M = -di/do
  • Concept of sign conventions in optics
NEXT STEPS
  • Study the derivation and applications of the mirror equation in optics
  • Learn about sign conventions in optics, particularly for mirrors and lenses
  • Explore ray diagrams for concave mirrors to visualize image formation
  • Investigate real vs. virtual images and their characteristics in optical systems
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Students studying optics, physics educators, and anyone interested in understanding the principles of image formation with concave mirrors.

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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