What is the power of the lens in diopters?

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Homework Help Overview

The problem involves determining the power of a lens in diopters, given a specific magnification factor and the distance of an object from the lens. The subject area pertains to optics and lens equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the image distance using the magnification formula and then applies the lens formula to find the power. Some participants question the units of focal length in relation to diopters, while others clarify the relationship between magnification and the signs of distances.

Discussion Status

The discussion is active, with participants providing guidance on the correct interpretation of the formulas and units involved. There is an acknowledgment of differing perspectives on the calculations, but no explicit consensus has been reached.

Contextual Notes

Participants are navigating the implications of sign conventions in optics and the conversion of focal length to appropriate units for calculating power in diopters.

KVM
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Homework Statement


It is desired to magnify reading material by a factor of 3.5× when a book is placed 9.5 cm behind a lens.
What is the power of the lens in diopters?

Homework Equations


1/f = 1/di + 1/do
M = -di/do
P=1/f

The Attempt at a Solution


I set equal the magnification to -di/do and solved for di
3.5=-di/9.5
di = -33.25 cm
Then I plugged into equation to find 1/f which = P
1/f = 1/di + 1/do
1/f = 1/-33.25 + 1/9.5
1/f = 0.075
I don't understand why this isn't right??
 
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What are the units of f in 1/f when you want 1/f to be in diopters??
 
(If 1/f is in diopters, f is in meters).

The magnification is negative; di is positive. If you do that, it should work out.
 
Go with Gene N. but also pay attention to what kuruman said. Together they have the answer!
 

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