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Answer
FRP3
t = 8640
...
do = 0
...
Plens = -5
...
do = 1 m , the distance of the board from his eyes

Explanation
FRP3
Given
m = 500g, mass of carbon
N(t) = 2690 decays/min , decay rate
T1/2 = 5730
λ = ln2/T1/2 , decay constant

Required
t = ? age of the sample in years

Solution
;activity for 500 g
No = 500x(255Bq/kg)/1000g
No = 127
...
83 decays/s
N(t) = Noe-λ t ; ln both sides
t = ln(N(t)/No) /-λ
= ln(N(t)/No) /-(ln2/T1/2)
=-5730( ln ( 44
...
5)) / ln2
= 5989
...
62 years , the age of the sample in years

FRP4
a
...
00 D
Ptotal = 50
...
0 = -4
...
0 D
Now, solve for her far point , do :
Peyes = 1/do + 1/di
54
...
02m
1/do = 54
...
0
1/do = 4
...
0
do = 0
...

Given
do = 20cm or 0
...
0 D, total power of the eyes
di = 2 cm , the normal distance between eyes lens and retina

Required
Plens = ? power of contact lens
Solution
First, solve for Peyes ;
Peyes = 1/do + 1/di
= 1/0
...
02
Peyes = 55
...
0 = Plens + 55
...
0 - 55
...
00 D , power of ther contact lens when her far point is changed to 20 cm
c
...
02 m , the normal distance between eyes lens and retina
Peyes = 51
...
0 = 1/do + 1/0
...
0-50
...
0
do = 1 m , the distance of the board from his eyes



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