The luminous intensity of a small plane source of light along the forward normal is 160 candela. Assume the source to be perfectly diffused, find the luminous flux emitted into a cone of solid angle 0.02 sr around a line making an angle of `60^@` with the forward normal.

Explanation

The situation is shown in ure. By Lambert's cosine law, intensity in the direction SB is
`I=I_0cos60^@`
Where e`I_0=160` candela is the intensity along the forward normal
` Thus I=(160 candela)(1/2)`
=80 candela.
The luminous flux emitted in the cone shown in the ure is
` /_\f=I/_\omega`
`=(80 candela)(0.02sr)`
`=1.6 lumen.