Question

In this example we will determine what type of lens should be used to correct the vision of a hyperopic eye. Assume that the near point of the hyperopic eye is 100 cm in front of the eye. What lens should be used to enable the eye to see clearly an object that is 25 cm in front of the eye?A 60-year-old man has a near point of 300 cm . What lens should he have to see clearly an object that is 25 cm in front of the lens?

Answer 1

Complete Question

  The diagram for this question is shown on the first uploaded image  

Answer:

a

 The lens to be used is a positive converging lens with focal length of  f = 33.33 cm

b

  The lens to be used is a positive converging lens with focal length of  f = 27.3 cm

Explanation:

a

  From the  question and the diagram  we are told that

          The image  distance is   $v = - 100cm$ in front of the eye

           The object distance is  $u = -25cm$

According to the Lens formula the focal length of the lens to be used to correct the hyperopic eye is      

            $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

           $\frac{1}{-100} - \frac{1}{(-25)} = \frac{1}{f}$

           $\frac{1}{f} = \frac{1}{25} - \frac{1}{100}$

           $\frac{1}{f} = \frac{75}{25 * 100}$

           $f = \frac{100}{3}$

           $f = 33.33cm$

Since the focal length is  positive it means that the lens to use is  a positive

converging lens

b

From the question we are told

                The image  distance is   $v = -300cm$ in front of the eye

           The object distance is  $u = -25cm$

According to the Lens formula the focal length of the lens to be used to correct the hyperopic eye is      

            $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

           $\frac{1}{-300} - \frac{1}{(-25)} = \frac{1}{f}$

           $\frac{1}{f} = \frac{1}{25} - \frac{1}{300}$

           $\frac{1}{f} = \frac{275}{25 * 300}$

           $f = 27.3$

 Since the focal length is  positive it means that the lens to use is  a converging lens