Page 390 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 30] LIGHT 375

Fig. 30-3

1 2 1
means reducing the illumination to ( ) = its former value. For light perpendicularly incident on a surface,
2 4
2
θ = 0 and cos θ = 1, so in this situation the illumination is E = I/R .
SOLVED PROBLEM 30.3
A 10-W fluorescent lamp has a luminous intensity of 35 cd. Find (a) the luminous flux it emits and (b) its
luminous efficiency.

(a) F = 4π I = (4π)(35 cd) = 440 lm
F 440 lm
(b) Luminous efficiency = = = 44 lm/W
P 10 W

SOLVED PROBLEM 30.4
A spotlight concentrates all the light from a 100-cd bulb in a circle 1.3 m in radius on a wall. If the
spotlight beam is perpendicular to the wall, find the illumination it produces.

The luminous flux emitted by the bulb is

F = 4π I = (4π) (100 cd) = 400π lm
2
2
2
The area of a circle 1.3 m in radius is A = πr = (π)(1.3 m) = 1.69π m . Hence the illumination is
F 400π lm 2
E = = = 237 lm/m = 237 lx
A 1.69π m 2
SOLVED PROBLEM 30.5
What area on the surface of a sphere of radius 60 cm is cut by a solid angle of 0.2 sr from its center?
2
Since = A/r , here
2
2
A = r = (0.2sr)(60 cm) = 720 cm 2
We note that the steradian, which is a dimensionless ratio, does not appear in the result.

SOLVED PROBLEM 30.6
A spotlight concentrates the light from a 150-cd bulb into a circle 0.8 m in radius a distance 25 m away.
Find the luminous intensity of the source looking into the beam. This is the luminous intensity of an
isotropic source that provides the same luminous flux on the illuminated circle.