The intensity produced by a long cylindrical light source at a sm… - Vidyadip

Question

The intensity produced by a long cylindrical light source at a small distance r from the source is proportional to:

$\frac{1}{\text{r}^2}$
$\frac{1}{\text{r}^3}$
$\frac{1}{\text{r}}$
None of these.

✓

Answer

$\frac{1}{\text{r}}$

Explanation:

Let us consider two coaxial cylindrical surfaces at distances r and r' from the axis.
Let areas dA and dA' subtend the solid angle $\text{d}\omega$ at the central axis.
The height of the area element will be same, i.e. equal to dy.
Let the breath of dA be dx and that of dA' be dx'.
Now from the arcs,
${\text{dx}}=\text{rd}\theta$
$\text{dx} '=\text{r}'\text{d}\theta$
Now,
$\text{dA}'=\text{dxdy}=\text{rd}\theta\text{dy}$
$\text{dA}'=\text{dx}'\text{dy}=\text{r}'\text{d}\theta\text{dy}$
$\frac{\text{dA}}{\text{dA}'}=\frac{\text{r}}{\text{r}'}$
$\Rightarrow\frac{\text{dA}}{\text{r}}=\frac{\text{dA}^,}{\text{r}^,}=\text{d}\omega$
The luminous flux going through the solid angle dω will be:
$\text{dF} = \text{I}\text{d}\omega$
Now,
$\text{dF}=\text{I}\frac{\text{dA}}{\text{r}}$
If the surfaces are inclined at an angle $\alpha,$
$\text{dF}=\text{I}\frac{\text{dA}\cos{\alpha}}{\text{r}}$
Now, illuminance is defined as
$\text{E}=\frac{{\text{dF}}}{\text{dA}}=\text{I}\frac{\text{dA}\cos\alpha}{\text{r}}$
$\Rightarrow \text{E}\propto\frac{1}{\text{r}}$

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