Question

A point source emitting 628 lumen of luminous flux uniformly in all directions is placed at the origin. Calculate the illuminance on a small area placed at (1.0 m, 0, 0) in such a way that the normal to the area makes an angle of 37° with the X-axis.

Answer 1

Given,
Luminous flux = 628 lumen
Angle made by the normal with the x axis ($\theta$) = 37^ₒ
Distance of point, r = 1 m
Since the radiant flux is distributed uniformly in all directions, the solid angle will be 4$\mathrm{\pi }$.
∴ Luminous intensity, l =$\frac{\mathrm{Luminous}\mathrm{Flux}}{\mathrm{Solid}\mathrm{angle}}$
$=\frac{628}{4\mathrm{\pi }}=50\mathrm{candela}$

$$\begin{gathered} I\mathrm{lluminance}\left(E\right)\mathrm{is}\mathrm{given}\mathrm{by}, \\ E=l\cos \frac{\theta }{r^2} \end{gathered}$$
On substituting the respective values we get,
$$\begin{gathered} E=50\times \frac{\cos 37°}{1^2} \\ =50\times \frac{(4/5)}{1}=40\mathrm{lux} \end{gathered}$$
So, the illuminance on the area is 40 lux.