A convex lens in a human eye adjusts its focal length based on the object position. If it can form a clear image on retina for objects as close as 25 cm and as far as infinite distance, find the range offocal length of lens in the human eye?

Q. A human eye can form sharp images of different objects lying at different distances by changing the focal length of the eye lens on the retina. This is done to achieve

Q. The ability of an eye lens to adjust its focal length to form sharp images on the retina is known as its power of accommodation.

Q. The ciliary muscles of eye control the curvature of the lens in the eye and hence and alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means, radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about $25cm$ for a grown up person.
A person can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around $25cm$.
A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image of the spectacle lens becomes object for the eye lens and whose image is formed on the retina.
The number of spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is $+3D$ (converging lens of focal length $100/3cm$), then number of lens will be $+3$.
For all the calculations required, you can used the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens.
A near sighted man can clearly see objects only upto a distance of $100cm$ and not beyond this. The number of the spectacle lenses necessary for the remedy of this defect will be

Q. Assertion :A normal human eye can clearly see all the objects at the different distance. Reason: The human eye has the capacity to suitably adjust the focal length of its lens to a certain extent.