The problem states:

Light intensity varies inversely with the square of the distance from the light source. If the light intensity measures 615 lux at a distance of 28 cm, what is the light intensity at 55 cm? Round to the thousandths place.

Formula:

The inverse square law for light intensity is:

$I_1 \cdot d_1^2 = I_2 \cdot d_2^2$

Where:

• $I_1 = 615$ lux
• $d_1 = 28$ cm
• $I_2 = ?$ (light intensity at 55 cm)
• $d_2 = 55$ cm

Step 1: Rearrange for $I_2$:

$I_2 = \frac{I_1 \cdot d_1^2}{d_2^2}$

Step 2: Plug in values:

$I_2 = \frac{615 \cdot 28^2}{55^2}$

First, calculate $28^2$ and $55^2$: $28^2 = 784, \quad 55^2 = 3025$

Now substitute these: $I_2 = \frac{615 \cdot 784}{3025}$

Step 3: Simplify:

$615 \cdot 784 = 482160$

$I_2 = \frac{482160}{3025} \approx 159.392$

Final Answer:

The light intensity at 55 cm is 159.392 lux.

Would you like further clarification?

Relative Questions:

1. How does the inverse square law apply to other phenomena like sound or gravity?
2. What would the intensity be at a distance of 100 cm using the same formula?
3. How can we measure light intensity in real-world scenarios?
4. What happens to light intensity when the distance doubles?
5. How accurate is this formula for non-point light sources?

Tip:

When using inverse proportionality laws, always ensure you square the distances for light and sound problems.