Calculate Focal Length to Reduce Viewing Distance to 25cm

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SUMMARY

The discussion focuses on calculating the focal length of a converging lens required to enable a patient to see objects at a distance of 25cm, given that they can currently only see objects at 1m. The Lensmaker's equation is utilized, with the correct approach being to use the modified formula 1/u - 1/v = 1/f. Substituting u=1m and v=0.25m yields a focal length of 33cm, indicating that the image formed is virtual.

PREREQUISITES
  • Understanding of the Lensmaker's equation
  • Knowledge of virtual and real images in optics
  • Familiarity with the concepts of object distance (u) and image distance (v)
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation and applications of the Lensmaker's equation
  • Explore the characteristics of virtual images in optical systems
  • Learn about different types of lenses and their focal lengths
  • Investigate practical applications of optics in vision correction
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Students studying optics, optical engineers, and healthcare professionals involved in vision correction techniques.

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Homework Statement


A patient is unable to see objects closer than 1m from the eye. Calculate the focal length of the converging lens which is required to reduce this to 25cm.


Homework Equations



Lensmaker's equation: 1/u + 1/v = 1/f

u= distance from object to lens
v= distance of image to lens
f = focal length


The Attempt at a Solution



I tried using the above equation by substituting in u=1m, v=0.25m. But that doesn't help. The solution should be 33cm.
 
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Here the image is virtual.
So the formula should be
1/u - 1/v = 1/f.
 

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