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

Hint: 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 $