Question

The near point (smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people.
near point(cm) far point(cm)
Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]

Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required. To rank items as equivalent, overlap them.

Answer 1

Answer:

Edouard > Berenica  > Chdwick > Francesca =  Avishka = Danya

Explanation:

The near and far vision points for a normal person are

     Close view 25 cm

     Far vision  infinite

Let's use the constructor equation to raise the object to these points

Close vision

           1 / f = 1 / o + 1 / i

where f is the focal length, or the distance to the object and i the distance to the image, in this case i must be 25cm for everyone, as this distance is on the same side as the object is negative, according to the optical sign convention

   

Avishka

 The distance to the object is 40 cm

      1 / f = 1 / 0.40 - 1 / 0.25 = -1.5

       

Focal power P is the inverse of the focal length in meters

      P = 1 / f = -1.5 D

The vision for the far point is normal and therefore does not require correction   P=0

Berenice

Distance to the object that can see o = 0.30 m

     1 / f = P = 1 / o + 1 / i

     P = 1 / 0.30 -1 / 0.25

     P = -0.67 D

Far vision correction

    o = 500cm = 5m

    1 / f = 1 / o + 1 / i

    P = 1/3 + 1 / inf

    P = 0.33 D

Chadwick

Close point of view is normal, so does not require correction

Far vision point

       1 / f = 1/5 + 1 / inf

       P = 0.2 D

Danya

Has normal vision does not require glasses

        P = 0

Edouard

Close view point

     o = 80cm = 0.80m

     1 / f = 1 / o + 1 / i

     1 / f = 1 / 0.80 - 1 / .25

     P = -2.75 D

Far vision point

     o = 200cm = 2m

      P = 1 / f = ½ + 1 / onf

      P = 0.5 D

Francesca

Close vision point

      o = 50 cm = 0.50 m

       1 / f = 1 / 0.5 - 1 / .25

       1 / f = -2 D

Normal far vision point

We write in summary the focal power for each person and each point

Person    Near point    Far point

Avishka       -1.5D               0

Berenice     -0.67D            0.33D

Chadwick     0                    0.2D

Danya           0                    0

Edouard       -2.75D            0.5D

Francesca     -2D                 0

In order of the lenses required from highest to lowest we have , por teh near ponint myopia

 Edouard> Francesca> Avishka> Berenice> Chanwick> Danya

Para el point of the far vision  (hyperopic)

Edouard > Berenica  > Chdwick > Francesca =  Avishka = Danya