A farsighted person has a near point of 55 cm rather than the normal 25 cm . Part A What strength lens, in diopters, should be prescribed to correct this vision problem

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Answer:

We have do = 25cm and di = -55cm

So 1/f = 1/do + 1/di = 1/25 + 1/(-55) = 2.214x10^-2

So f = 1/2.214x10^-2 = 45.2cm

So power = 1/0.452m = +2.21 Diopters (converging lens)

The strength lens to be prescribed to correct this vision problem is 2.18 diopters.

From the given information;

Using the Lens formula, we have:

[tex]\mathbf{\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} }[/tex]

where

  • f = focal length
  • v = image distance
  • u = object distance

For a farsighted person with a near point (v) = 55 cm rather than the normal object distance of u = 25 cm.

Then, The focal length can be determined as follow:

[tex]\mathbf{\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} }[/tex]

since the image must be on the same side of the lens, then the image distance must be negative.

[tex]\mathbf{\dfrac{1}{f} = \dfrac{1}{25} + \dfrac{1}{-55} }[/tex]

[tex]\mathbf{\dfrac{1}{f} =0.0218 cm }[/tex]

[tex]\mathbf{f = \dfrac{1}{0.0218 \ cm } }[/tex]

f = 45.8 cm

f = 0.458m

Recall that:

The dioptic power is inverse of the focal length strength;

[tex]\mathbf{D = \dfrac{1}{f}}[/tex]

[tex]\mathbf{D = \dfrac{1}{0.458 }}[/tex]

D = 2.18 diopters.

Learn more about the lens here:

https://brainly.com/question/3225974?referrer=searchResults