Question

The ciliary muscles of the eye control the curvature of the lens in the eye and hence 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 the 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 from the retina to the eye lens. It is about 2.5cm for grown up persons.
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 the 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 the eye defects uses spectacles (eye glasses). The function of the lens of the spectacles is to form the image of the objects within the range in which a person can see clearly. The image of the spectacle lens becomes the object for the eye lens and whose image is formed on the retina.
The number of the spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of the spectacle lens 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 the lens will be +3.
For all calculations required, you can use the lens formula and the lens makers formula. Assume that the eye lens is an equiconvex lens. Neglect the distance between the eye lens and the spectacle lens.
A farsighted man cannot see objects clearly unless they are at least at 100cm from his eyes The number of spectacle lens that will make his range of clear vision equal to an average grown up person will be

Answer 1

It is given in the question that the object has to be at least be placed at a distance of 100m from the eye of the so that the man can see the object clearly. Hence he has to use correction glasses(spectacles) so that the image is formed on his retina of the eye. The eye defect he is suffering from is called hypermetropia where in the image is formed behind the retina. Hence he has to use converging lenses (convex lens) to correct his vision. For the man to see the object, the lens used should be such that the image at the least distance of distinct vision should be formed at 100cm from the eye so that the image formed by the lens can be viewed as the object. Therefore using the lens formula we can calculate the focal length and determine the lens number.

Complete step by step answer:

In the above diagram fig 1,we can see that the image is formed behind the retina, therefore we need a converging lens to bring back the image to the retina.

In the above figure we can see that the object kept at 25 cm should be at least formed at 100 cm, so that the man is able to see the object. Hence the focal length of the lens, using the lens formula is,
$\dfrac{1}{f}=\dfrac{1}{V}-\dfrac{1}{U}$ where f is the focal length of the lens, U is the object distance and V is the image distance from the centre of the lens. IN the above lens diagram the image distance and the object distance are negative, Therefore the focal length of the lens is,
$\begin{aligned} & \dfrac{1}{f}=\dfrac{1}{V}-\dfrac{1}{U} \\\ & \dfrac{1}{f}=\dfrac{1}{-100}-\dfrac{1}{-25} \\\ & \dfrac{1}{f}=\dfrac{25-100}{-100\times 25}=\dfrac{-75}{-100\times 25}=\dfrac{3}{100} \\\ & f=\dfrac{100}{3}cm=\dfrac{1}{3}m \\\ \end{aligned}$
Hence the power of the lens P is equal to,
$\begin{aligned} & P=\dfrac{1}{f} \\\ & P=\dfrac{1}{\left( 1/3 \right)m}=+3D \\\ \end{aligned}$
Hence the number of the lens is equal to +3.

Note:
In an optical system the distances are always measured from the optical centre of the lens. If the distance traversed from the optical centre to the object or the image is in the direction of the incident light, the distance is taken as positive. If the distance traversed from the optical centre to the object or the image is in the direction opposite to the incident light, the distance is taken is negative.