Question

the intensity i of light varies inversely as the square of the distance d from the source. if the intensity of illumination on a screen 49 ft from a light is 3.8 foot - candles, find the intensity on a screen 70 ft from the light.

a. 7.76 foot - candles
b. 5.43 foot - candles
c. 2.66 foot - candles
d. 1.862 foot - candles

Explanation:

Step1: Write the inverse - square formula

Since the intensity $I$ of light varies inversely as the square of the distance $D$ from the source, we have the formula $I=\frac{k}{D^{2}}$, where $k$ is a constant.
We know that when $D = 49$ ft and $I=3.8$ foot - candles. Substitute these values into the formula: $3.8=\frac{k}{49^{2}}$.

Step2: Solve for the constant $k$

Multiply both sides of the equation $3.8=\frac{k}{49^{2}}$ by $49^{2}$ to find $k$.
$k = 3.8\times49^{2}=3.8\times2401 = 9123.8$.

Step3: Find the intensity at a new distance

Now we want to find the intensity $I$ when $D = 70$ ft. Substitute $k = 9123.8$ and $D = 70$ into the formula $I=\frac{k}{D^{2}}$.
$I=\frac{9123.8}{70^{2}}=\frac{9123.8}{4900}=1.862$ foot - candles.

Answer:

D. 1.862 foot - candles