What is the power of the lens in diopters?

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SUMMARY

The power of a lens in diopters can be calculated using the formula P = 1/f, where f is the focal length in meters. In this discussion, the goal is to achieve a magnification of 3.5× with an object distance (do) of 9.5 cm. The derived image distance (di) is -33.25 cm, leading to the calculation of the focal length. The correct approach involves ensuring that the signs of di and do are properly accounted for, as negative magnification indicates an inverted image.

PREREQUISITES
  • Understanding of lens formulas: 1/f = 1/di + 1/do
  • Knowledge of magnification concepts: M = -di/do
  • Familiarity with unit conversions: meters to centimeters
  • Basic principles of optics and lens power calculations
NEXT STEPS
  • Study the relationship between focal length and lens power in various lens types.
  • Learn about the implications of negative magnification in optical systems.
  • Explore practical applications of diopter measurements in optical devices.
  • Investigate common mistakes in lens calculations and how to avoid them.
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding lens power and magnification calculations.

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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