Question

One of the inevitable consequences of aging is a decrease in the flexibility of the lens. This leads to the farsighted condition called presbyopia (elder's eye). Almost every aging human will experience it to some extent. However, for the myopic, or nearsighted, the person it is possible that far vision will be limited by a subpar far point and near vision will be hampered by an expanding near point. One solution is to wear bifocal lenses that are diverging in the upper half to correct the nearsightedness and converging in the lower half to correct the farsightedness.
Suppose one such individual asks for your help. The patient complains that she can’t see far enough to safely drive (her far point ????far=108 cm ) and she can’t read the font of her smartphone without holding it beyond arm’s length (her near point ????near=82.0 cm ). Prescribe the bifocals that will correct the visual issues for your patient.
a. What is the focal length ????ns of the part of the bifocal lens that corrects the patient's nearsightedness?
b. What is the focal length ????fs of the part of the bifocal lens that corrects the patient's farsightedness?

Answer 1

Answer:

a)   f = 19.16 cm , b)    f = 108 cm

Explanation:

For this exercise we use the constructor equation

       1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

a) we seek correction for near vision,

we place the object in the point of near vision of a normal person p = 25 cm, the image must be formed in the point of view of the person with presbyopia q = 82 cm

    1 / f = 1/25 + 1/82

    1 / f = 0.052195

    f = 19.16 cm

b) we repeat the same operation for the distant point of view, the object is placed at infinity p = ∞, the image is formed at q = 108 cm

   

      1 / f = 1 /∞ + 1/108 = 0 + 1/108

       f = 108 cm

Answer 2

a)   The focal length for nearsightedness will be f = 19.16 cm

b)   The focal length for farsightedness will be f = 108 cm

,

what is focal length?

The focal length of the  lens is  defined as  when an  object is  placed in  front  of the  lens, then  the  light  coming  from  the  object  is  reflected towards the focus of the lens and the distance of the focus point is called as the focal length of the lens.

The equation of the lens formula is given as:

$\dfrac{1}{f} =\dfrac{1}{p}+\dfrac{1}{q}$      

where f is the focal length, p the distance to the object and q the distance to the image

a) Now for the correction for near vision,

we place the object in the point of near vision of a normal person p = 25 cm, the image must be formed in the point of view of the person with presbyopia q = 82 cm

$\dfrac{1}{f}=\dfrac{1}{25}+\dfrac{1}{82}=0.052195$  

$\dfrac{1}{f}=19.16\ cm$  

b) we repeat the same operation for the distant point of view, the object is placed at infinity p = ∞, the image is formed at q = 108 cm

$\dfrac{1}{f}=\dfrac{1}{oo}+\dfrac{1}{108}= 0+\dfrac{1}{108}$    

$f=108 \ cm$      

Hence  the focal length for nearsightedness will be f = 19.16 cm the focal length for farsightedness will be f = 108 cm

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