The ciliary muscles of eye control the curvature of the lens in the eye and hence can 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 2.5cm for a grown up person.
A perosn 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 o the spectacle lens becomes object for the eye lens and whose image is formed on the retina.
The number of spectacle lens used for th eremedy of eye defect is decided by the power fo 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 use 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.

Q. Minimum focal length of eye lens of a normal peson is

Options

A. 25cm
B. 2.5cm
C. `25//9cm`
D. `25//11cm`

Answer

d.

Explanation

`(1)/(f)=(1)/(v)-(1)/(u)`
Here, `v=2.5` (distance of retina as position of image is fixed)
`u=-x`
`(1)/(f)=(1)/(2.5)+(1)/(x)`
For `f_(mi n):x ` is minimum `(1)/(f_(m i n))=(1)/(2.5)+(1)/(25)`