## EXAMPLE:

# Irradiance of a flashlight

Consider again our cheap flashlight example.

The flashlight projects a beam that is 200 mW to create a 10-cm-diameter spot on a wall 25 cm meters away. What is the irradiance at the wall?

- The geometry of the problem is:

- The diameter of the circular spot on the wall is 10 cm. Therefore, the radius r is 5 cm. The area of the spot is: A = r
^{2} = (5 cm)^{2} = 78.5 cm^{2}.
- The
**irradiance E** is: E = P/A = (0.2 W)/(78.5 cm^{2}) = 0.00254 W/cm^{2}.

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