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
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.
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
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)
Nice work
Well, let's shed some light on this problem!
So, according to the question, the illumination of the object is directly proportional to the candle-power of the lamp and inversely proportional to the square of the distance between the lamp and the object.
Let's say the original distance between the lamp and the object is "d" (in cm) and the illumination is "I". We can represent this relationship mathematically as:
I ∝ (C/d^2)
Where C is the candle-power of the lamp.
Now, let's calculate the original illumination (I1) when the candle-power (C1) is 8 and the distance (d) is 150 cm. We'll also calculate the new illumination (I2) when the candle-power (C2) is 5 and the distance (d2) is unknown.
Using the given information for the original setup:
I1 = (C1/d^2) = (8/150^2)
Now, we want to find the new distance (d2) that will give us the same illumination as before. So, we set up the equation:
I2 = (C2/d2^2) = I1
Substituting the values we know:
(5/d2^2) = (8/150^2)
Now, let's solve for d2:
5 * 150^2 = d2^2 * 8
112,500 = 8d2^2
d2^2 = 112,500 / 8
d2 ≈ 188.98
So, the new distance (d2) should be approximately 188.98 cm.
But wait, we're not done yet! Since the light bulb was lowered, we need to find out the difference in height between the two setups.
The original height (h1) was 150 cm, and the new height (h2) is unknown. Let's call it "x."
We can set up another equation to find the difference in height:
h1 - h2 = 150 - x
Since we know that the new distance (d2) is equal to the new height (h2), we can substitute it into the equation:
150 - d2 = 150 - x
From this, we can solve for x:
150 - 188.98 = 150 - x
-38.98 = -x
x ≈ 38.98
So, the new light should be lowered approximately 38.98 cm to give the object the same illumination as before.
I hope that brightened up your understanding of the problem!
To solve this problem, we can use the formula for direct and inverse variation. Let's break down the problem and find a solution step by step:
1. Given information:
- Initial lamp candle power: 8 candle power
- Final lamp candle power: 5 candle power
- Initial lamp height: 150 cm
2. Let's represent the variables:
- Initial illumination: I1
- Final illumination: I2
- Initial distance between lamp and object: d1
- Final distance between lamp and object: d2
3. According to the problem, the illumination varies directly as candle power and inversely as the square of the distance. Therefore, we can set up the following equation:
I1 = k * (8 / d1^2)
I2 = k * (5 / d2^2)
Here, k is a constant that remains the same for both equations.
4. Since we want to find the new distance, let's start by finding the value of the constant k. We can do this by equating the initial and final illuminations and solving for k:
I1 = I2
k * (8 / d1^2) = k * (5 / d2^2)
Cancelling out the constant k from both sides:
(8 / d1^2) = (5 / d2^2)
5. Now, let's plug in the given values:
- d1 = 150 cm
Substituting the values and solving for d2:
(8 / 150^2) = (5 / d2^2)
8 / 150^2 = 5 / d2^2
Cross-multiplying and then isolating d2^2:
8 * d2^2 = 5 * 150^2
d2^2 = (5 * 150^2) / 8
d2^2 = 703.125
6. Finally, find the square root of both sides to get the final distance:
d2 = √703.125
d2 ≈ 26.51 cm
So, the new light needs to be lowered to approximately 26.51 cm to provide the same illumination as before.