Question

What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43 cm away? Your answer should be a number with two decimal places. Hint: you need to first determine the focal distance of the lens.

Answer 1

Answer:

The magnification is $m = 0.3674$

Explanation:

From the question we are told that

  The  power of the lens is  $P = -4.00 D(dioptre)$

Generally  $1 dioptre = 1 \ meter$

  The object distance is $u = -43 \ cm$ the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

 Generally the focal length is mathematically represented as

          $f = \frac{1}{P}$  

   =>$f = \frac{1}{4.00 }$  

  =>  $f = 0.25 \ m$

converting to  cm  

 =>   $f = 0.25 \ m = 0.25 * 100 = 25 \ cm$

Generally from lens equation  we have that  

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

=>  $\frac{1}{25} +\frac{1}{v} -\frac{1}{-43}$

=>   $v = -15.8 \ cm$

Generally the magnification is mathematically represented as

      $m = \frac{v}{u}$

=>    $m = \frac{- 15.8}{-43}$

=>    $m = 0.3674$