Page 391 - Schaum's Outline of Theory and Problems of Applied Physics

376 LIGHT [CHAP. 30

The bulb emits the total luminous flux

F = 4π I = (4π)(150 cd) = 1885 lm

Because the radius R of the illuminated circle is small compared with the radius r of a sphere whose center is the
2
spotlight, we can neglect the difference between the circle’s plane area of R and its area measured on the surface of
the sphere. Hence the solid angle of the spotlight beam is
A π R 2 (π)(0.8m) 2
= = = = 0.0032 sr
r 2 r 2 (25 m) 2

The luminous intensity I of the spotlight looking into its beam is therefore
F 1885 lm
5
I = = = 5.89 × 10 cd
0.0032 sr
The source brightness has been increased by a factor of I /I = 3927 by the optical system of the spotlight.

SOLVED PROBLEM 30.7
An illumination of about 800 lx is recommended for reading. How far away from a book should a 75-W
lamp of intensity 90 cd be located if the angle between the light rays and the plane of the opened book is
60 ?
◦
2
◦
The angle between the light rays and the normal to the book is θ = 30 . Since E = (I cos θ)/R ,

I cos θ (90 cd)(cos 30 )
◦
R = = 2 = 0.31 m = 31 cm
E 800 lm/m
SOLVED PROBLEM 30.8
A 60-W light bulb whose luminious efficiency is 14 lm/W is suspended 2 m over a table. (a) What is the
illumination on the table directly under the bulb? (b) How high over the table should the bulb be in order
to double that illumination?

(a) The luminous flux emitted by the bulb is
F = (60 W)(14 lm/W) = 840 lm

Since F = 4π I, the intensity of the bulb is

F 840 lm
I = = = 66.8cd
4π 4π
The illumination at a distance of R = 2mis

I 66.8cd 2
E = = = 16.7 lm/m = 16.7lx
R 2 (2m) 2
2
(b) Since E = I/R ,
I 2 R 2
E 2 1
= 2
E 1 I 1 R
2
Here I 1 = I 2 and E 2 /E 1 = 2, so

1
E 1
R 2 = R 1 = (2m) = 1.41 m
E 2 2