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Elena L [17]
3 years ago
7

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.
Physics
1 answer:
kiruha [24]3 years ago
6 0

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

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