Question

Two lab partners, Mary and Paul are both farsighted. Mary has a near point of 6.7 cm from her eyes and Paul has a near point of 130 cm from his eyes. Both students wear glasses that correct their vision to a normal near point of 25.0 cm from their eyes, and both wear glasses 1.80 cm from their eyes. In the process of wrapping up their lab work and leaving for their next class, they get their glasses exchanged (Mary leaves with Paul's glasses and Paul leaves with Mary's glasses). When they get to their next class, find the following.

(a) Determine the closest object that Mary can see clearly (relative to her eyes) while wearing Paul's glasses. m

(b) Determine the closest object that Paul can see clearly (relative to his eyes) while wearing Mary's glasses. m

Answer 1

a] Mary has a near point of 6.7 cm. This means that Mary will not be able to see clearly the objects which are closer than 6.7 cm (being farsighted).

Normal near point of a person = 25 cm

This normal near point is achieved because of the glasses. This means that if an object is kept at a distance of 25 cm, the glasses should create an image at a distance of 6.7 cm (near point) so that Mary sees the object clearly.

Object distance = u = 25 - 1.8 = 23.2 cm

Image distance = v = - (6.7 - 1.8) = - 4.9 cm

the negative sign comes because the image is on the same side as the object (It is not between the lens and the eyes)

use lens formula,

1/f = 1/v + 1/u

=> f = - 6.212 cm

this is the focal length of the lens required for Mary.

Similarly, the focal length of lens required for Paul is, f' = +28.326 cm

If Mary wears Paul's glasses, the image produced by them should be at a distance of v = - (6.7 - 1.8) cm, so that the object is seen clearly.

1/28.326 = 1/(-4.9) + 1/u'

=> u' = 4.177 cm

so, the closest object that Mary can see clearly (relative to her eyes) while wearing Paul's glasses is u'' = 4.177 + 1.8 = 5.977cm = 0.05977m.

Similarly, the closest object for Paul will be: u''' = 6.528 + 1.8 = 8.328 cm.