Optics: Find & Confirm Image Positions w/ Ray Diagrams

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Discussion Overview

The discussion revolves around the use of ray diagrams and the thin lens formula to determine the image positions created by a biconvex lens with a focal length of 100 mm. Participants explore both the graphical and mathematical methods for confirming image locations for objects placed at distances of 150 mm and 75 mm from the lens.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks for confirmation of their ray diagram and inquires about the equations needed to calculate image positions.
  • Several participants suggest using the thin lens formula, specifically 1/f = 1/do + 1/di, to find image distances.
  • There is confusion regarding the application of the lens equation, particularly in relation to the object distances and the resulting image distances.
  • Participants discuss the implications of placing the object within and outside the focal length, noting that this affects the nature of the image (real or virtual).
  • One participant calculates an image distance of 250 mm for the 150 mm object and 333 mm for the 75 mm object, but there is uncertainty about the signs of these distances.
  • Another participant suggests that the biconvex lens is symmetrical, which may simplify calculations regarding the radii of curvature.
  • There is a discussion about longitudinal and transverse magnification, with participants attempting to clarify how object dimensions relate to these calculations.
  • Some participants express uncertainty about the relevance of the object's dimensions in the context of magnification calculations.

Areas of Agreement / Disagreement

Participants generally agree on the use of the thin lens formula but exhibit disagreement and confusion regarding specific calculations and interpretations of the results. The discussion remains unresolved regarding the correct application of the equations and the implications of the object dimensions on magnification.

Contextual Notes

There are limitations in the clarity of the ray diagrams and the precision of calculations presented. Some assumptions about the symmetry of the lens and the nature of the images (real vs. virtual) are not universally accepted, leading to varying interpretations of the results.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of optics, particularly those interested in understanding the practical applications of ray diagrams and the thin lens formula in determining image positions and magnifications.

  • #31
Thanks.
 
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  • #32
Hi there. Just a quick question regarding an equation. In order to find the transverse magnification of an object 10 mm high and 5 mm long, it gives the equation:

fo / xo = xi / fi = - yi / yo giving xo x xi / fo x fi

Now which part of that is supposed to be used to calculate the transverse magnification?

Am not sure why some variables are equated to other variables e.g. xo x xi = fo x fi

Thanks.
 
  • #33
questions_uk said:
Hi there. Just a quick question regarding an equation. In order to find the transverse magnification of an object 10 mm high and 5 mm long, it gives the equation:

fo / xo = xi / fi = - yi / yo giving xo x xi / fo x fi

Now which part of that is supposed to be used to calculate the transverse magnification?

Am not sure why some variables are equated to other variables e.g. xo x xi = fo x fi
The definition of transverse magnification is:
m ≡ yi/yo

It can be shown (using similar triangles) that:
m ≡ yi/yo = -si/so (Where si is the image distance; so is the object distance. Both measured from the lens.)

If you measure distances from the focal points (xi & xo are the image and object distances measured from the focal points), then you can show (using the thin lens equation) that:
xi * xo = f*f

m ≡ yi/yo = -xi/f = -f/xo

Using that last term is the easy way to calculate the transverse magnification.
 
  • #34
Thanks.
 

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