Question

The intensity produced by a long cylindrical light source at a small distance r from the source is proportional to
(a) $\frac{1}{r^2}$

(b) $\frac{1}{r^3}$

(c) $\frac{1}{r}$

(d) none of these

Answer 1

Correct option (c).

Let us consider two coaxial cylindrical surfaces at distances r and r' from the axis. Let areas dA and dA' subtend the solid angle d​ω 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'.
$$\begin{gathered} \mathrm{Now}\mathrm{from}\mathrm{the}\mathrm{arc}s, \\ \mathrm{dx}=\mathrm{rd\theta } \\ \mathrm{dx}\text{'}=\mathrm{r}\text{'}\mathrm{d\theta } \\ \mathrm{Now}, \\ \mathrm{dA}=\mathrm{dxdy}=\mathrm{rd\theta dy} \\ \mathrm{dA}\text{'}=\mathrm{dx}\text{'}\mathrm{dy}=\mathrm{r}\text{'}\mathrm{d\theta dy} \\ \frac{\mathrm{dA}}{\mathrm{dA}\text{'}}=\frac{\mathrm{r}}{\mathrm{r}\text{'}} \\ \Rightarrow \frac{\mathrm{dA}}{\mathrm{r}}=\frac{\mathrm{dA}\text{'}}{\mathrm{r}\text{'}}=\mathrm{d\omega } \end{gathered}$$

The luminous flux going through the solid angle d​ω will be:
dF = I​dω

$$\begin{gathered} \mathrm{Now}, \\ \mathrm{dF}=\mathrm{I}\frac{\mathrm{dA}}{\mathrm{r}} \\ \mathrm{If}\mathrm{the}\mathrm{surfaces}\mathrm{are}\mathrm{inclined}\mathrm{at}\mathrm{an}\mathrm{angle}\mathrm{\alpha }, \\ \mathrm{dF}=\mathrm{I}\frac{\mathrm{dAcos\alpha }}{\mathrm{r}} \\ \mathrm{Now},\mathrm{illuminance}\mathrm{is}\mathrm{defined}\mathrm{as} \\ \mathrm{E}=\frac{\mathrm{dF}}{\mathrm{dA}}=\mathrm{I}\frac{\mathrm{dAcos\alpha }}{\mathrm{r}} \\ \Rightarrow \mathrm{E}\propto \frac{1}{\mathrm{r}} \end{gathered}$$