Near & Far Point of Eye: Power of Lens & Max Distance

[prat]

the near point and far point of an eye are 35cm and 300cm respectively.what is power of lens so that the person can see objects at 25cm .with this lens what is the maximum distance he can see.

f=xd/x-d p=1/f in m

i could only solve the first part of the problem.p=1.14d f=35*25/35-25=87.5cm=0.875m=1/f=1/0.875=1.14d

Homework Statement

Homework Equations

The Attempt at a Solution

 

[berkeman]

the near point and far point of an eye are 35cm and 300cm respectively.what is power of convex lens for seeing 25cm .with this lens what is the maximum distance he can see.

f=xd/x-d p=1/f in m

i could only solve the first part of the problem.p=1.14d

Welcome to the PF!

Your post is a bit confusing. Could you please post the full text of the question, and show your work on the initial solution?

 

[andrevdh]

the near point and far point of an eye are 35cm and 300cm respectively.what is power of lens so that the person can see objects at 25cm .with this lens what is the maximum distance he can see.

f=xd/x-d p=1/f in m

i could only solve the first part of the problem.p=1.14d f=35*25/35-25=87.5cm=0.875m=1/f=1/0.875=1.14d

The convex lens need to form a virtual image of the object in front of the eye within the range 35 -> 300 cm in order for the eye to see it. Your calculation gives a virtual image forming at 35 cm for an object at 25 cm in front of it.

The furthest distance this virtual image can be from the eye, with the lens, is 300 cm. This is the furthest (negative, since it needs to be virtual) image distance that the eye will be able to focus on. What will the accompanied object distance then be for this particular image distance?