The intensity

The light intensity on a screen is inversely proportional to the square of the distance between the screen and the light source.

If a screen is illuminated by a light source 20 m away, the intensity is one-fifth of what is required. Where should the light be placed?

Correct answer:

x₂ =  8.9443 m

Step-by-step explanation:

$$\begin{gathered} I(x)=1/x^2 \\ {x}_1=20\text{m} \\ {I}_1=1/5=\frac{1}{5}=0.2 \\ {I}_2=1 \\ I\cdot x^2=K \\ K=I_1\cdot x_1^2=I_2\cdot x_2^2=const \\ K=I_1\cdot x_1^2=0.2\cdot 20^2=80 \\ {x}_2=\sqrt{K/I_2}=\sqrt{80/1}=4\sqrt{5}\doteq 8.9443\text{m} \\ \text{ Verifying Solution: } \\ {I}_{11}=K/x_1^2=80/20^2=\frac{1}{5}=0.2 \\ {I}_{22}=K/x_2^2=80/8.9443^2=1 \end{gathered}$$

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