Optics(Converging lens and Diverging lens)

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Homework Help Overview

The discussion revolves around the application of optics principles involving converging and diverging lenses. Participants are examining the relationships between object distances, image distances, and magnification using relevant equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate image distances and magnifications for a system of two lenses but expresses uncertainty regarding the object distance for the diverging lens and the significance of the object's height.
  • Some participants note that the image from the converging lens serves as a virtual object for the diverging lens, prompting questions about the spatial relationship between the two lenses.
  • There is a discussion about using the lens formula for the diverging lens, with participants questioning the sign convention for the focal length.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem setup and the implications of the object height. Some guidance has been provided regarding the virtual object concept and the use of the lens formula, but no consensus has been reached on the overall approach.

Contextual Notes

Participants are navigating potential constraints related to the object distance for the diverging lens and the implications of the object's height in the context of magnification calculations.

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Homework Statement


http://www.imageurlhost.com/images/r0its2lyef7v3s0ppms2.png

Homework Equations


Mirror Equation
(1/q) + (1/p) = (1/f) = (2/r)
Magnification
M = (q/p)

for two lenses
M1 & M2
Total magnification = M1*M2

The Attempt at a Solution


we have p1 = 40cm and f1 = 30cm
then the image distance of lens 1 will be q1 = 120

then we have diverging lens 2
with focal length f2 = 20cm

then I am stuck i don't know what to do with the object distance of the diverging lens
i also don't know what to do with the fact that the object is 2cm high
 
Last edited:
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The image of the converging lens acts as a VIRTUAL object for the diverging lens.
 
grzz said:
The image of the converging lens acts as a VIRTUAL object for the diverging lens.

is it between the diverging and converging lens?

this is how i do to find the image distance of the divergng lens
between C lens and D lens (110cm)
but there is an image distance q1 from the C lens (120cm)
the object distance from the D lens p2 might be (110cm - 120cm)(1/q2) + (1/(110-120)) = (1/20)
q2 = 6.67
then i can find the magnification of the Divirging lens
M2 = q2/p2
total
M = M1M2

whats the use of that 2cm thing above the object?
 
If the formula (1/u) + (1/v) = 1/f is to be used then the focal length of the diverging lens is taken as -20.
 

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