Question

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 the lens increases. That means, the 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 the eye lens. It is about $2.5cm$ 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, the minimum distance of the object should be around $25cm$.

A person suffering from eye defects uses spectacles (eye glass). The function of a 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 an object for the eye lens and whose image is formed on the retina.

The number of spectacle lenses used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle lenses is equal to the numerical value of the power of the lens with sign. For example, if the power of the lens required is $+ 3D$ (converging lens of focal length $100/3cm$), then the number of lenses will be $+ 3$.For all the calculations required, you can use the lens formula and lensmaker's formula.

Assume that the eye lens is an equiconvex lens. Neglect the distance between the eye lens and the spectacle lens.A farsighted 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

A. $+ 1$
B. $- 1$
C. $+ 3$
D. $- 3$

Answer 1

We know that Myopia or nearsightedness happens when the eye loses its ability to focus on far-off objects as the lenses do not possess a long focal length. It is also known as Myopia Objects that are near are clearly visible to persons who has this problem. Here we will use the lens formula for the given condition and then find the power of the spectacle lenses necessary for the remedy of this defect.

Formulas used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$,
where, $f$is the focal length, $u$ is the object distance and $v$ is the image distance
$P = \dfrac{1}{f}$,
where $P$ is the power and $f$ is the focal length

Complete step by step answer:
Here, we are given that a near sighted man can clearly see objects only upto a distance of 100cm and not beyond this.Thus the number of the spectacle lens should be such that the image of an object at infinite distance can be obtained at a distance of $100cm$.Therefore, the object distance will be infinite and the image distance will be $100cm = 1m$.Applying the formula $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$, we get
$\dfrac{1}{f} = \dfrac{1}{{ - 1}} - \dfrac{1}{{ - \infty }} = - 1 + 0 = - 1$
We know that $P = \dfrac{1}{f} = - 1$
The number of the spectacle lenses necessary for the remedy of this defect will be $- 1$.

Hence, option B is the right answer.

Note: Here, we have determined the power by using lens formula. However, when we know the focal length of the human eye and the lens required, we can determine the power of lens by using the formula $P = \dfrac{1}{f} = \dfrac{1}{{{f_1}}} - \dfrac{1}{{{f_2}}}$, where ${f_1}$ and ${f_2}$ are the focal length of lens required and the focal length of a human eye and $f$ is the effective focal length.