The intensity of the light from an object varies inversely with the square of the distance. In other words, I = k over d^2
a. Solve the equation to find d in terms of I.
b. Two stars give off the same amount of light. However, from Earth their intensities differ. Let I sub 1 and I sub 2 be their intensities and let d1 and d2 be their respective distances from Earth. What is the ratio of d sub 2 to d sub 1?
c. If one star appears 9 times as intense as the other, how much closer is it to Earth?
(a) d^2 = k/I
(b) I1 d1^2 = I2 d2^2
d2^2/d1^2 = I1/I2
a. To solve the equation for d in terms of I, we can rearrange the equation:
I = k/d^2
First, we can cross multiply:
I * d^2 = k
Then, we isolate the variable d^2:
d^2 = k/I
Finally, we take the square root of both sides to solve for d:
d = √(k/I)
b. Let's use the given variables to find the ratio of d2 to d1. Since the intensities of the stars are the same, we can set up the equation as follows:
I1 = I2
k/d1^2 = k/d2^2
Cross multiplying, we get:
k * d2^2 = k * d1^2
Canceling out the k, we have:
d2^2 = d1^2
Taking the square root of both sides, we find:
d2 = d1
Therefore, the ratio of d2 to d1 is 1:1, indicating that the two stars are at the same distance from Earth.
c. Let's use the given information to determine the difference in distance between the two stars. We know that one star appears 9 times as intense as the other, so we can set up the equation:
I1 = 9 * I2
Using the inverse square law equation, we can substitute the intensities with the distances:
k/d1^2 = 9 * k/d2^2
Simplifying, we get:
d2^2 = 9 * d1^2
Taking the square root of both sides, we find:
d2 = 3 * d1
Therefore, the star that appears 9 times as intense is approximately 3 times closer to Earth than the other star.
a. To solve the equation to find d in terms of I, we can rearrange the equation as follows:
I = k / d^2
Multiply both sides by d^2 :
I * d^2 = k
Divide both sides by I :
d^2 = k / I
Take the square root of both sides:
d = √(k / I)
Therefore, d in terms of I is given by d = √(k / I).
b. Given that the intensities of two stars are I1 and I2, and their respective distances are d1 and d2, we can use the formula from part a to determine the ratio of d2 to d1.
From part a, we know that d = √(k / I). Therefore, we have:
d1 = √(k / I1)
d2 = √(k / I2)
To find the ratio of d2 to d1, we divide d2 by d1:
(d2 / d1) = (√(k / I2)) / (√(k / I1))
Simplifying the expression:
(d2 / d1) = (√(k / I2) * √(I1 / k))
The square root of k cancels out, leaving us with:
(d2 / d1) = √(I1 / I2)
Therefore, the ratio of d2 to d1 is √(I1 / I2).
c. If one star appears 9 times as intense as the other (I2 = 9 * I1), we can use the same formula from part b to calculate how much closer it is to Earth.
(d2 / d1) = √(I1 / I2)
Substituting the known values:
(d2 / d1) = √(I1 / (9 * I1))
Simplifying further:
(d2 / d1) = √(1 / 9)
(d2 / d1) = 1 / 3
Therefore, the star that appears 9 times as intense is one-third as far from Earth compared to the other star.
a. To find d in terms of I, we can rearrange the equation as follows:
I = k / d^2
Multiply both sides by d^2:
I * d^2 = k
Divide both sides by I:
d^2 = k / I
Take the square root of both sides:
d = √(k / I)
Therefore, d in terms of I is given by d = √(k / I).
b. We are given that two stars give off the same amount of light, but their intensities differ. Let I₁ and I₂ be their respective intensities, and let d₁ and d₂ be their respective distances from Earth.
Since the intensity of light from an object varies inversely with the square of the distance, we can set up the following equation:
I₁ = k / d₁^2 (Equation 1)
I₂ = k / d₂^2 (Equation 2)
Divide Equation 1 by Equation 2:
I₁ / I₂ = (k / d₁^2) / (k / d₂^2)
Simplify the expression:
I₁ / I₂ = (d₂^2 / d₁^2)
To find the ratio of d₂ to d₁, take the square root of both sides:
√(I₁ / I₂) = √(d₂^2 / d₁^2)
Simplify further:
d₂ / d₁ = √(I₁ / I₂)
Therefore, the ratio of d₂ to d₁ is given by d₂ / d₁ = √(I₁ / I₂).
c. If one star appears 9 times as intense as the other, it means that I₂ = 9 * I₁.
Using the equation from part b, d₂ / d₁ = √(I₁ / I₂), we can substitute the given values:
d₂ / d₁ = √(I₁ / (9 * I₁))
Simplify:
d₂ / d₁ = √(1 / 9)
Since the distance ratio d₂ / d₁ is always positive, we take the positive square root:
d₂ / d₁ = 1 / 3
Therefore, the star that appears 9 times as intense is one-third as far from Earth as the other star.