Find the change in focal length

[lha08]

Homework Statement

A nearsighted person has eyeglasses with a -1D prescription. If the near point is at 23 cm without eyeglasses, where is it when they are put on?

Homework Equations

(1/f)=(1/do)+(1/di)

The Attempt at a Solution

I got the answer which is 0.298701298 meters by -1= (1/do)-(1/0.23)...but the problem is that i don't know why it's like that...Any ideas would be extremely helpful! Thanks!

 

[mgb_phys]

You have that equation wrong,
f = the focal length of the lens ( -0.23m)
di is the image distance (which you don't know - but can assume is the same )
do is the object distance

Then you need the formula for a combination of two lenses (ie the eye + glasses) to find the change in focal length.

 

[nlightner]

As a scientist, it is important to understand the reasoning behind equations and solutions. In this case, the equation (1/f)=(1/do)+(1/di) relates the focal length (f) of a lens to the object distance (do) and image distance (di). In this scenario, the nearsighted person has a prescription of -1D, which means their lens has a power of -1 diopters. This power is inversely proportional to the focal length, meaning that a higher power results in a shorter focal length and vice versa.

The near point, or the closest point at which the person can see clearly without glasses, is given as 23 cm. This is the object distance (do) in the equation. When the glasses are put on, the person's vision is corrected and the image distance (di) becomes the focal length (f). Therefore, we can substitute the values for do and di into the equation as follows:

(1/f)=(1/0.23)+(1/di)

Since the prescription is -1D, we can also substitute -1 for the power (P) in the equation P=1/f. This gives us:

-1=1/0.23+1/di

Solving for di, we get:

di=0.298701298 meters

This is the location of the near point when the person is wearing their glasses. In summary, the change in focal length is due to the power of the lens, which is determined by the prescription. The higher the power, the shorter the focal length and the closer the near point will be.