Question

A person whose near-point distance is 45.5 cm wears a pair of glasses that are 2.1 cm from her eyes. With the aid of these glasses, she can now focus on objects 25 cm away from her eyes.

Required:
a. Find the focal length of her glasses.
b. Find the refractive power of her glasses.

Answer 1

Answer:

a

   $f = 0.4848 \ m$

b

  $p = 2.063 D$

Explanation:

From the question we are told that

   The  near point distance is  $k = 45.5 \ cm$

    The  distance of the glasses from the eye is  $y = 2.1 \ cm$

    The  distance of an object she can focus with the glass is  $i= 25 \ cm$

Generally the image distance is mathematically evaluated  as

     $v = -(45.5 - 2.1)$

     $v = -43.4 \ cm$

Generally the object distance is mathematically represented as

     $u = (25 -2.1)$

     $u = 22.9 \ cm$

The  negative sign tells us that the image was formed behind the eye  

Generally the lens formula  is mathematically represented as  

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

=>   $\frac{1}{f} = \frac{1}{22.9} + \frac{1}{ - 43.4}$

=>  $f = 48.48 \ cm$

 converting to meters  

   $f = 0.4848 \ m$

Thus the refractive power is mathematically represented as

    $p = \frac{1}{f}$

=>  $p = \frac{1}{0.4848 }$

=>   $p = 2.063 D$