- #1
RicXeoR
- 2
- 0
hello all,
i've been working on this whole day, still can't figure out, any help would be appreciated! thanks a bunch!
Problem as follows:
Bill is farsighted and has a near point located 140 cm from his eyes. Anne is also farsighted, but her near point is 75.6 cm from her eyes. Both have glasses that correct their vision to a normal near point (25.0 cm) from their eyes, and both wear glasses 2.0 cm from the eyes. Relative to the eyes, what is the closest object that can be seen clearly (a) by Anne when she wars Bill’s glasses and (b) by Bill when he wears Anne’s glasses.
This is what I have interpreted: I have S(near point) as 140-2 and 75.6-2, d of object as 25 cm, and d of image as -138 and -73.6 because it's on the same side as the object. With these numbers i plug into the thin lens equation to get 1/f or power of the lens. Then I switched the powers around, used d of image as 140 and 75.6, and solved for each d of object.
Those answers turn out as wrong. What I'm doing wrong here, any help is much appreciated...Thanks!
i've been working on this whole day, still can't figure out, any help would be appreciated! thanks a bunch!
Problem as follows:
Bill is farsighted and has a near point located 140 cm from his eyes. Anne is also farsighted, but her near point is 75.6 cm from her eyes. Both have glasses that correct their vision to a normal near point (25.0 cm) from their eyes, and both wear glasses 2.0 cm from the eyes. Relative to the eyes, what is the closest object that can be seen clearly (a) by Anne when she wars Bill’s glasses and (b) by Bill when he wears Anne’s glasses.
This is what I have interpreted: I have S(near point) as 140-2 and 75.6-2, d of object as 25 cm, and d of image as -138 and -73.6 because it's on the same side as the object. With these numbers i plug into the thin lens equation to get 1/f or power of the lens. Then I switched the powers around, used d of image as 140 and 75.6, and solved for each d of object.
Those answers turn out as wrong. What I'm doing wrong here, any help is much appreciated...Thanks!