Question

The illuminance of a surface distance 10 m from a light source is 10 lux. The luminous intensity of the source for normal incidence will be
(A) ${10^1}Cd$
(B) ${10^2}Cd$
(C) ${10^3}Cd$
(D) None of these

Answer 1

Illuminance is defined as the luminous flux per unit surface area. While luminous flux is the luminous intensity through a unit solid angle. Assume the source light diverges in all directions.

Formula used: In this solution we will be using the following formulae:
$\Rightarrow {E_v} = \dfrac{{{\phi _v}}}{{4\pi {R^2}}}$ , where ${E_v}$ is the illuminance, ${\phi _v}$ is the luminous flux, and $R$ is the distance of the surface from the source.
$\Rightarrow {l_v} = \dfrac{{{\phi _v}}}{\theta }$ where ${l_v}$ is the luminous intensity and $\theta$ is the solid angle measured in Steradians.

Complete step by step solution:
The illuminance of a surface is the luminous flux incident on that surface. It is given as luminous flux per unit area.
For a source, diverging in all directions, we have that the total surface area is the surface area of a sphere, hence $A = 4\pi {R^2}$ . Then the illuminance becomes
$\Rightarrow {E_v} = \dfrac{{{\phi _v}}}{{4\pi {R^2}}}$
Calculating luminous flux by inserting all known values, we have
$\Rightarrow {\phi _v} = {E_v} \times 4\pi {R^2} = 10 \times 4\pi \left( {{{10}^2}} \right) = 4\pi \times {10^3}lm$ , $lm$ is called the lumen.
Now, luminous intensity is given by the luminous flux per unit solid angle. Hence
$\Rightarrow {l_v} = \dfrac{{{\phi _v}}}{\theta }$ where ${l_v}$ is the luminous intensity and $\theta$ is the solid angle measured in Steradians.
For a complete sphere, the solid angle subtended is $4\pi$ i.e. $\theta = 4\pi$ (analogous to a circle subtending a plane angle of $2\pi$ ).
Hence,
$\Rightarrow {l_v} = \dfrac{{{\phi _v}}}{\theta } = \dfrac{{4\pi \times {{10}^3}}}{{4\pi }} = {10^3}cd$
Hence the correct option is C.

Note:
Alternatively, without firstly calculating the flux, we substitute the expression for luminous flux as a function of luminous intensity into ${E_v} = \dfrac{{{\phi _v}}}{{4\pi {R^2}}}$ as in:
From
$\Rightarrow {l_v} = \dfrac{{{\phi _v}}}{\theta } \Rightarrow {\phi _v} = {l_v}\theta$
Hence, ${E_v} = \dfrac{{{\phi _v}}}{{4\pi {R^2}}}$ becomes
$\Rightarrow {E_v} = \dfrac{{{l_v}\theta }}{{4\pi {R^2}}}$ . Making ${l_v}$ subject we have
$\Rightarrow {l_v} = \dfrac{{{E_v} \times 4\pi {R^2}}}{\theta }$
Inserting values, we get
$\Rightarrow {l_v} = \dfrac{{10 \times 4\pi \times {{10}^2}}}{{4\pi }} = {10^3}cd$