Question

Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye. Part A If you hold the book 28.0 cmcm from your eye, what is your nearpoint distance

Answer 1

Answer:

The near point is  $n =44.8 \ cm$

Explanation:

From the question we are told that

   The power is  $P = 1.50$

   The  distance from the eye is  $k = 1.8 \ cm$

    The distance of the book from the eye is $z = -28 \ cm$

Generally the focal length of the glasses is  

       $f = \frac{1}{P}$

=>   $f = \frac{1}{1.50 }$

=>   $f = 0.667 \ m$

=>   $f = 66.7 \ cm$

The object distance is evaluated as

     $u = z + k$

=>   $u = -28 + 1.8$

=>  $u = -26.2 \ cm$

The image distance is evaluated from lens formula as

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

=>   $\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}$

=>   $v=- \frac{1}{0.0232}$

=>    $v=- 43 \ cm$

The  near point is evaluated as

      $n = -v + k$

=>    $n =-(-43) + 1.8$

=>    $n =44.8 \ cm$