Question

An electric bulb illuminates a plane surface. The intensity of illumination on the surface at a point 2m away from the bulb is $5 \times 10^{- 4}$ phot (lumen/cm²). The line joining the bulb to the point makes an angle of 60^o with the normal to the surface. The intensity of the bulb in candela is

• A. $40\sqrt{3}$
• B. 40
• C. 20
• D. $40 \times 10^{- 4}$

Answer 1

Answer: (B)

40

Answer 2

$I = \frac{L\cos\theta}{r^{2}}$ $\Rightarrow \mspace{6mu}\mspace{6mu} L = \frac{I \times r^{2}}{\cos\theta}$

$= \frac{5 \times 10^{- 4} \times 10^{4} \times 2^{2}}{\cos 60{^\circ}}\mspace{6mu} = \mspace{6mu} 40\mspace{6mu} Candela$