EXAMPLE:

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 = r² = (5 cm)² = 78.5 cm².
• The irradiance E is: E = P/A = (0.2 W)/(78.5 cm²) = 0.00254 W/cm².