# steve reiny proportion help maths reiny!!!

The illumination of a small object by a lamp varies
directly as the candle-power of the lamp and inversely
as the square of the distance between the lamp and
the object.If a light-bulb of 8 candle power,fixed
150cm above a table,is replaced by a 5 candle power
bulb,how far must the new light be lowered to give the object thesame illumination as before

plz show step

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1. i = k cp /d^2

k*8/150^5 = k*5/d^2

8 d^2 = 5*150^2

d^2 = (5/8)150^2
d = 150 sqrt(5/8)
d = 150(.79)
d = 118.5cm
lower it 150-118.5 = 31.5 cm

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2. If
y is the illumination
c is the candlepower
d is the distance, then

y = kc/d^2

so, if y is constant,

c/d^2 = y/k, which is constant

So, you want d such that

8/150^2 = 5/d^2
d^2 = 5*150^2/8
d = 118.56 cm

or, using a proportion, you use
c2/c1 = d2^2/d1^2
d2/d1 = √(c2/c1)

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3. Let illumination be I and C.P be the candle-power and distance be d. I=k(C.P /dxd .Let I=1
dxd/C.P=k
Where d=150
C.P=8
150x150/8 =8
K=2812.5
I=2812.5(C.P/dxd)
Where C.P=5
I=2812.5(5/dxd)
2812.5x5/dxd=1
d=118.59m(by finding the square root )
d=150m_118.59m
d=31.41m.

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4. Nice work

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