Question

A person with myopia (near-sightedness) has a far point of 45.0cm while their near point is 15.0cm. Upon wearing glasses, they have corrected their sight and can view distant objects clearly with relaxed eyes. With the glasses on, what is the closest object this person can focus now

Answer 1

Answer:

p = 22.5 cm

Explanation:

For this exercise we must use the equation of the constructor

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

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

Let's start with the far point, the object is very far away (p = ∞) and the image must be formed at the far point of view of the person q = 45.0 cm

since the image is on the same side as the object according to the sign convention the distance is negative

         $\frac{1}{f} = \frac{1}{\infty } + \frac{1}{-45}$

          f = -45.0 cm

now let's use the near point (q = 15.0 cm) at what distance the object should be

          $\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$

          $\frac{1}{p} = \frac{1}{-45} - \frac{1}{-15}$1 / p = 1 / -45 - 1 / -15

         $\frac{1}{p} = - \frac{1}{45} + \frac{1}{15}$ 1 / p = -1/45 + 1/15

         $\frac{1}{p}$ = 0.0444

          p = 22.5 cm

this is the closest distance you can see an object clearly