A lamp illuminates a screen of area \(2.0m^{2}\), which is \(4 m\) away. The irradiance at the screen is \(0.02Wm^{-2}\).
a) Calculate the power of the incident beam.
b) If the screen is moved to a distance of \(1.5 m\) from the lamp, what would the new irradiance be?
Solution
a) To calculate the power use the formula \( I = \frac{P}{A}\)
This can be rearranged into \(P=IA\)
\(=0.02 \times 2=0.04W\)
b) To find the new irradiance:
\(I_{1}d_{1}\,^{2}=I_{2}d_{2}\,^{2}\)
\(0.02\times 4^{2}=I_{2}\times 1.5^{2}\)
\(I_{2}=\frac{0.02\times 4^{2}}{1.5^{2}}\)
\(=0.142Wm^{-2}\)
Question
A 60 W lamp is placed in the centre of a spherical chamber with a radius of 3m. Calculate the irradiance of the lamp on the inner surface of the chamber.
(You need to use the relationship - surface area of a sphere \(= 4\pi {r^2}\)).