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

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 is 40 lux

Explanation:

Step 1:

Given data in the question  :

Luminous  emitting  flux = 628 lumen

Angle rendered with the x axis by default (θ) $=37^{\circ}$

Gap of point, r = 1 m

Due to the uniform distribution of the radiant flux in all directions, the solid angle will be 4π.

Step 2:

∴ Luminous intensity, l = $\frac{Luminous Flux}{Solid angle}$

  $=\frac{628}{4 \pi}=\frac{628}{4 \times 3.14}$

 = 50 candela

Illuminance (E) is indicated by,

$E=l \frac{\cos \theta}{r^{2}}$

Step 3:

Upon replacing the respective values we obtain,

$E=50 \times \frac{\cos 37^{\circ}}{1^{2}}$

 $=50 \times \frac{\frac{4}{5}}{1}$

$=50 \times \frac{4}{5}$

$=40 \mathrm{lux}$

So, the area illuminance is 40 lux.