Question

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

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

Answer 1

The correct option is B 40

$r=2m=200cm$
$E=5\times {10}^{-4}phot,\Theta ={60}^o$
From,
$E=\frac{Icos\Theta }{r^2},I=\frac{E{r}^2}{cos\Theta }=\frac{5\times {10}^{-4}(200{)}^2}{cos60}=40C.P$