Question

Gbenga needs to get glasses to correct his farsightedness. His eyes currently cannot focus on objects that are within 2 ft (or 61 cm) of his eyes. This is in contrast to people with normal vision who can focus on objects as close as 25 cm in front of them. If the glasses that Gbenga will get will sit 1.6 cm in front of his eyes, what lens focal length and power would correct his vision? That is, what lens focal length and power would allow Gbenga to focus on objects that are 25 cm in front of his eyes? Will they be converging or diverging lenses?

Answer 1

Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.

Step-by-step explanation:

When an object is placed 25cm from Gbenga's eye, the glasses lens must produce an image 61cm away (Gbenga's eye near point).

An image 61cm from the eye will be (61cm - 1.6cm) from the glasses.

i.e.
`d_(i)=61cm-1.6cm=59.4cm`

and
`d_(o) = 25cm - 1.6cm = 23.4cm`

note
`d_(i)` will be negative because the image is formed on the same side as the object.

finally,
`d_(i)=-59.4cm\\d_(o)=23.4cm`

the formula for finding the focal length
`f` is given as

`f=(d_(i)d_(o) )/(d_(o)+d_(i) )`

`f=(-59.4*23.4)/(23.4-59.4) \\`

`f=(-1389.96)/(-36)`

`f=38.61cm`

The focal length is positive which indicates converging lens

power
`p=(1)/(f)`

but
`f\\` must be in metres

Therefore,
`f=38.61cm=0.3861m`

`p=(1)/(0.3861)`

`p=2.59 Diopter`