Two lab partners, Mary and Paul are both farsighted. Mary has a near point of 6.

Question

Two lab partners, Mary and Paul are both farsighted. Mary has a near point of 6.9 cm from her eyes and Paul has a near point of 124 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.
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(b) Determine the closest object that Paul can see clearly (relative to his eyes) while wearing Mary's glasses.

Explanation / Answer

Ans (a)

The students wear their glasses 1.8 cm away.

For Paul, the near point is at 124cm away. The normal near point is 25 cm

Here u= (25-1.8) cm = 23.2 cm

and v =-(124+1.8) cm =122.2 cm

using the relation: 1/f =1/v+1/u ----(1)

we get , the focal length of Paul's glass= 28.64 cm

Now, for Mary, near point is at 6.9 cm. She now uses Paul's glasses with focal length 28.64 cm.

Hence v=( -6.9+ 1.8) cm = -5.1 cm.

Using this in equ (1):

1/(28.64) = 1/u + 1/(-5.1)

This gives

u=4.33 cm

But her glasses are at 1.8 cm.

Hence w.r.to her eyes 4.33+1.8 Ccm = 6.13 cm away

Ans (b)

Mary's near point is 6.9 cm. She wears the glass at 1.8 cm away.

hence, u= (25-1.8) cm = 23.2 cm

and v =(-6.9+1.8) cm =-5.1 cm

using the relation: 1/f =1/v+1/u from equ (1)

we get , the focal length of Mary's glass= 6.54 cm

Now, for Paul, near point is at124 cm.He now uses Mary's glasses with focal length 6.54 cm.

Hence v=( -124+ 1.8) cm = -123.2 cm.

Using this in equ (1):

1/(6.54) = 1/u + 1/(-123.2)

This gives

u=6.21 cm

But his glasses are at 1.8 cm.

Hence w.r.to her eyes 6.21+1.8 Ccm = 8.1 cm away

x

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