Question

I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limited range of accommodation. My glasses have progressive lenses. The lenses are diverging lenses, with a negative power, but the magnitude of the power varies from a maximum at the top, for distance vision, to a minimum at the bottom, for close vision. With no correction, for my left eye, my near point is 15 cm, my far point is 20 cm. With my glasses on, my near point is about 25 cm, and my far point is infinity. What is the range of powers of the lens in my glasses?

Answer 1

Answer:

The range of powers is    $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

       The far point of the left eye is $n_f = 20 cm$

       The near point of the left eye is  $n = 15cm$

       The near point with the glasses on is $n_g =25 \ cm$

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be $u = -25 \ cm$

          Image distance would be  $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

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

substituting values

              $\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

               $f = - \frac{75}{2} cm$

           converting to  meters

               $f = - \frac{75}{2} * \frac{1}{100}$

               $f = - \frac{75}{200} \ m$

   Generally the power of the lens is mathematically represented as

                $P = \frac{1}{f}$

Substituting values

                 $P = - \frac{200}{75} m$

                 $P = - 2.667 \ D$

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be $u_f = - \infty \ cm$

          Image distance would be  $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    $\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

                  $\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

                 $\frac{1}{f} = \frac{1}{-20} - 0$      

                  $f_f = \frac{20}{1} \ cm$

converting to  meters

                $f_f = - \frac{20}{1} * \frac{1}{100}$

               

Generally the power of the lens is mathematically represented as

                $P = \frac{1}{f_f}$

Substituting values

                 $P = - \frac{100}{20} m$

                 $P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

                  $- 5 \ D \le P \le - 2.667\ D$