What is the focal length of the lens? given magnification and distance

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SUMMARY

The focal length of a lens required to achieve a magnification of +5.00 with an image distance of 25 cm is calculated to be +6.25 cm. The student utilized the magnification formula, m = -d(image)/d(object), to derive the object distance, which is 0.5 cm. The focal length was then determined using the lens formula, 1/f = 1/d(object) + 1/d(image), leading to a focal length of approximately 0.04167 m. The discussion highlights the importance of recognizing that a virtual image implies a negative image distance in the calculations.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/d(object) + 1/d(image)
  • Knowledge of magnification concepts in optics
  • Familiarity with virtual and real images in lens systems
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the properties of converging lenses and their applications
  • Learn about virtual images and their implications in optical systems
  • Explore advanced lens formulas and their derivations
  • Investigate practical applications of magnification in biology and microscopy
USEFUL FOR

This discussion is beneficial for biology students, optics enthusiasts, and anyone interested in understanding the principles of lens magnification and focal length calculations.

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Homework Statement


A biology student wishes to examine a bug at a magnificaiton of +5.00 by looking through a single lens. the student wants the image of the bug to be a distance of 25cm from the lens. what is the focal length of the lens?

Answer:+6.25cm

Homework Equations


m=-d(image)/d(object)
1/f=[itex]\frac{1}{d(object)}[/itex]+[itex]\frac{1}{d(image)}[/itex]


The Attempt at a Solution


first i solve for distance of object, d(object), by using the first equation:
d(object) = (25x10-2)/5

then i calculated focal length by using the second equation
f = ((1/0.5) + (1/0.25))-1 = 0.041667

i don't know what I am doing wrong
 
Physics news on Phys.org
This is a simple magnifying glass using a converging lens.
The image is VIRTUAL...what does this mean about the image distance in the equation?
 

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