Solve for di: Negative Effect on Object Distance?

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Homework Help Overview

The discussion revolves around solving for the image distance (di) in a scenario involving object distance (do), object height (ho), and image height (hi). The context is related to optics and image formation, particularly focusing on the implications of negative values in the formula used.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to apply the formula -di/do = hi/ho, particularly the role of the negative sign in determining the image distance. Some participants question the implications of a negative image height and whether it indicates an inverted image.

Discussion Status

Participants are exploring the relationship between the variables and the significance of negative values in the context of image formation. There is an ongoing clarification regarding the formula and its components, with some guidance provided on manipulating the equation.

Contextual Notes

The original poster mentions using made-up values for the variables and expresses confusion specifically about the negative aspect of the formula. There is a lack of a complete problem statement, which may affect the clarity of the discussion.

Grekory
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When I'm given do, ho, and hi. How would i solve for di
Normally i would do -di/do = hi/ho and cross-multiply but I'm not sure how the negative comes into effect.

do = object distance
di = image distance
ho = object height
hi = image height



If...

ho = 3cm
hi = 4cm
do = 5cm

What would be di?

Thanks
 
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Please provide the complete statement of the problem you are trying to solve. (I presume that a negative image height means that the image is inverted.)
 
It's just made up values and i want to know how i would solve for di. The negative in the formula confuses me
 
Start with:
-di/do = hi/ho

You can solve for di by cross-multiplying if you like, or just multiply both sides by -do. As I said before, a negative value for di just means that the image is inverted. (Real images are inverted.)
 

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