Calculating Distance and Size of Image on Blackboard Reading

[thompson.1674]

Homework Statement

A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 61.20 diopters, and the lens-to-retina distance is 1.643 cm.
(a) How far (in meters) is the blackboard from her eyes?
(b) If the writing on the blackboard is 6.00 cm high, what is the size of the image on her retina (including the proper algebraic sign)? (in cm)

Homework Equations

p=1/f
1/f=(1/do)+(1/di)
m=hi/ho=-di/do

The Attempt at a Solution

(a) 61.20=1/f
f=.0163 m
1/.0163=(1/.01643m)+(1/do)
do=2.06 m
(b) m=-1.643cm/206cm= -.00798 cm x 6cm= -.0479cm.

I am not really sure what I did wrong on these two problems?? Any help would be appreciated

 

[LowlyPion]

Homework Statement

A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 61.20 diopters, and the lens-to-retina distance is 1.643 cm.
(a) How far (in meters) is the blackboard from her eyes?
(b) If the writing on the blackboard is 6.00 cm high, what is the size of the image on her retina (including the proper algebraic sign)? (in cm)

Homework Equations

p=1/f
1/f=(1/do)+(1/di)
m=hi/ho=-di/do

The Attempt at a Solution

(a) 61.20=1/f
f=.0163 m
1/.0163=(1/.01643m)+(1/do)
do=2.06 m
(b) m=-1.643cm/206cm= -.00798 cm x 6cm= -.0479cm.

I am not really sure what I did wrong on these two problems?? Any help would be appreciated

For one thing part a) looks like the math is a little off.
1/f = 61.20
1/.01643 = 60.8643
difference = .3357 That suggests do = 2.98 m

 

[thompson.1674]

k cool thanks i got it now