Question

The intensity produced by a long cylindrical light source at a small distance r from the source is proportional to
A) $\dfrac{1}{{{r^2}}}$
B) $\dfrac{1}{{{r^3}}}$
C) $\dfrac{1}{r}$
D) None of these

Answer 1

Hint
As illumination is defined as the flux per unit area then first find the value of flux through solid angle dω by using formula $dF = Id\omega $. Now, for $dω$ we will use $angle = arc/radius$. After substituting the values and on simplification we will get illumination or intensity.

Complete Step by step solution
Step 1-
Let's find two cylindrical coaxial surfaces at distances r and r ' from the axis. Let dA and dA’ subtend at the central axis the solid angle dω. The height of the area dimension is going to be the same, that is, equal to dy.
Let the breath of dA be dx and that of dA' be dx' Of the arcs now,
$dx = rd\theta $
And $dx' = r'd\theta $
Now, as area is the product of dx and dy, we get
$dA = dxdy = rd\theta dy$
And $dA' = dx'dy = r'd\theta dy$
Divide the above equations, we get
$ \Rightarrow \dfrac{{dA}}{{dA'}} = \dfrac{r}{{r'}}$
$ \Rightarrow \dfrac{{dA}}{r} = \dfrac{{dA'}}{{r'}} = d\omega $ ………………. (1)
Step 2-
The luminous flux of the strong angle dω below will be:
$ \Rightarrow dF = Id\omega $
Now, put the value of dω from equation (1), we get
$ \Rightarrow dF = I\dfrac{{dA}}{r}$
Step 3-
If the surfaces are α-angle oriented, then
$ \Rightarrow dF = \dfrac{{IdA\cos \alpha }}{r}$
Now illumination is known as
$ \Rightarrow E = \dfrac{{dF}}{{dA}} = \dfrac{{I\cos \alpha }}{r}$
As from the above equation E is inversely proportional to r.
Then, (C) option is correct.

Note
There is another method, we know that area is inversely proportional to intensity and area of the cylinder is $2πrh$, $r$ is radius of cylinder and $h$ is the height of cylinder then we get intensity is inversely proportional to the radius of cylinder. Hence, again the (C) option is correct.