Finding the focal length of the lense

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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:
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 :)