Question

At noon on a sunny day, the intensity of sunlight is 1000 W/m2. A student uses a circular lens 6 cmin diameter to focus light from the sun onto a 3-mm-diameter spot. Part A: How much light power does the lens capture? Part B: What is the intensity in the focused spot?

Answer 1

A)

Light power,

P = I*A

where I = intensity = 1000 W/m2

A = area = pi*d^2/4 = pi*(0.06)^2/4

So, P = 1000*(pi*0.06^2/4) = 2.83 W <------answer

B)

Intensity , I in the focussed spot,

I = P/A = 2.83/(pi*0.003^2/4) = 4*10^5 W/m2