Finding the focal length of the lense

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Homework Help Overview

The problem involves determining the focal length of a lens used in a magnifying glass that produces a five times enlarged image at a distance of 25 cm. The original poster provides initial values and an equation but indicates that their solution is incorrect.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants discuss the need to apply the formula for maximum magnification, questioning the implications of the distance to the cornea and the conditions under which maximum magnification occurs. Others provide an alternative formula and attempt to derive the focal length based on the given magnification.

Discussion Status

The discussion is exploring different interpretations of the problem, particularly regarding the conditions for maximum magnification and the derivation of relevant formulas. Some participants have offered insights and alternative approaches, but there is no explicit consensus on the correct method or solution yet.

Contextual Notes

There is mention of missing information regarding the distance to the cornea, which may affect the magnification and focal length calculations. The original poster's initial approach appears to be based on a simplified understanding of the magnification process.

Aladin
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Homework Statement


A magnifying glass gives a five times enlarged image at a distance of 25 cm from the lense.Find the focal length of the lense
d=25cm
M=5
f=?

Homework Equations


f = d/M


The Attempt at a Solution


f = 25/5
f = 5 cm

But this solution is not correct please help me and explain also.
thanks
 
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hrm interesting.
When a spectacle magnifier produces an image 25cm in front of the lens you need to use the formula for maximum magnification. The fact that the distance to the cornea has not been specified also suggests this. In other words 5x is the maximum magnification that this lens can produce, should the lens be held say 10cm from the cornea the magnification would be less. This is a special case of angular magnification where the magnification is at a maximum due to the cornea being virtually in contact with the lens and the image being produced at the reference seeing distance (25cm)
Anyways here's the formula...M(max)=1+F/4
which when rearranged gives... (M-1)*4=F
substitute into formula...(5-1)*4=F
therefore F=+16.00D
to find focal length we take the reciprocal of F...1/F = 1/16 = 0.0625m or 6.25cm

hope this is useful
 
Last edited:
M(max)=1+F/4

what is 4 in this formula?
 
Aladin said:
M(max)=1+F/4

what is 4 in this formula?

This formula as i stated previously is a special case of angular magnification when the magnification is at a maximum. It is derived from;
M = qL / 1-dL', where q is the least distance of distinct vision, L and L' are the vergences of light entering and leaving the lens respectively and d is the distance to the cornea.
The proof for maximum magnification is as follows;
Angular mag will be at a maximum when the image is formed at least distance of distinct vision ( l'=-0.25m) and the eye is placed close to the lens so that d=0, substituting into the equation with d=0, L = L' - F = -4 - F,
where L'=1/l' =1/-0.25 = -4D then,
M = qL / 1-dL'
M = (-0.25)(-4-F) / 1-(0xL')
M = 1 + 0.25F
M(max) = 1 + F/4

Hope this clarifies things :)
 

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