Question 4.8.5: Solving a Problem Involving Inverse Variation with Powers Th...
Solving a Problem Involving Inverse Variation with Powers
The intensity of light varies inversely as the square of the distance from the light source. If Rita doubles her distance from a lamp, what happens to the intensity of light at her new location?
BY THE WAY …
Light travels at a speed of 299,792,458 meters per second. The distance light travels in a year is so large that it is a useful unit of distance in astronomy.
1. One light-year is approximately 9.46 × 10^{15} m.
2. The nearest star (other than the sun) is 4.3 light-years away.
3. The Milky Way (our galaxy) is about 100,000 light-years in diameter.
4. The most distant objects that astronomers can see are about 18 billion light-years away. Thus, the light that we presently see from these objects began its journey to us about 18 billion years ago. Because this is close to the estimated age of the universe, that light is a kind of fossil record of the universe not long after its birth.
Let I be the intensity of light at a distance d from the light source. Because I is inversely proportional to the square of d, we have
I=\frac{k}{d^{2}}.
If we replace d with 2d, the intensity I_{1} at the new location is given by
\begin{array}{ll}I_{1}=\frac{k}{(2 d)^{2}} & \text { Replace } d \text { with } 2 d \text { in } I=\frac{k}{d^{2}} \\I_{1}=\frac{k}{4 d^{2}} & (2 d)^{2}=2^{2} d^{2}=4 d^{2}\end{array}
Therefore,
I_{1}=\frac{I}{4} \quad \text { Replace } \frac{k}{d^{2}} \text { with } I
This equation tells us that if Rita doubles her distance from the light source, the intensity of light at the new location will be one-fourth the intensity at the original location.