$$\require{cancel}$$

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

Suppose that some surface is being irradiated from a point source of radiation of intensity $$I \ \text{W sr}^{-1}$$ at a distance $$r$$. The normal flux density ("normal" meaning normal to the direction of propagation), as we have seen, is $$I / r^2$$. If the surface being irradiated is inclined so that its normal is inclined at an angle $$\theta$$ to the line joining it to the point source of radiation, the rate at which radiant energy is falling on unit area of the surface will be $$I \cos \theta / r^2$$.

In any case, the rate at which radiant energy is falling upon unit area of a surface is called the irradiance of that surface. It is denoted by the symbol $$E$$, and the units are $$\text{W m}^{-2}$$. In the simple geometry that we have described, the relation between the intensity of the source and the irradiance of the surface is

$E = (I \cos \theta) / r^2 \tag{1.10.1} \label{1.10.1}$

If we are dealing with visible radiation, the number of lumens falling per unit area on a plane surface is called the illuminance, and is expressed in lumens per square metre, or lux. Recall that a lumen is the SI unit of luminous flux, and the candela is the unit of luminous intensity, and that an isotropic point source of light radiating with a luminous intensity of $$I$$ cd (that is, $$I$$ $$\text{lm sr}^{-1}$$) emits a total luminous flux of $$4\pi \ \text{lm}$$. The relation between the illuminance of a surface and the luminous intensity of a source of light is the same as the relation between irradiance and radiant intensity, namely, equation $$\ref{1.10.1}$$, or, if the surface is being illuminated normally, equation 1.8.2. If the luminous intensity of a source of light in some direction is one candela, the irradiance of a point on a surface that is closest to the source is $$1 \text{lm m}^{-2}$$ if the distance is one metre, $$1 \text{lm cm}^{-2}$$ if the distance is one cm, and $$1 \text{lm ft}^{-2}$$ if the distance is one foot. A lumen per square metre is a lux, and a lumen per square cm is a phot. A lumen per square foot is often (usually!) given the extraordinary name of a "foot-candle". This is a most illogical misuse of language, and is mentioned here only because the term is still in frequent use in non-scientific circles. Lumen, candela and lux are, respectively, the SI units of luminous flux, luminous intensity and illuminance. Phot and "foot-candle" are non-SI units of illuminance. The exact definition of the candela will be given in section 1.12; the lumen and lux are derived from the candela. Those who are curious about other strange-sounding units encountered in the quantitative measurement of the visible portion of radiation will also find the definition of "stilb" in section 1.12.

Problem

A table is being illuminated by a light bulb fixed at a distance $$h$$ vertically above the table. The fixture is such that the socket is above the bulb, and the luminous intensity of the bulb varies as

$I (\theta ) = \frac{1}{2} I (0) (1+ \cos \theta) \tag{1.10.2} \label{1.10.2}$

where $$\theta$$ is the angle from the downward vertical from the bulb. Show that the illuminance at a point on the table at a distance $$r$$ from the sub-bulb point is

$E(r) = \frac{I(0)}{2h^2} \left[ \frac{(1+r'^2)^{\frac{1}{2}}+1}{(1+r'^2)^2}\right] \tag{1.10.3} \label{1.10.3}$

where $$r' = r/h$$, and draw a graph of this for $$r' = 0$$ to $$r' = 2$$. For what value of $$r'$$ does the irradiance fall to half of the sub-bulb irradiance?

Problem

If the table in the above problem is a circular table of radius a, show that the flux that it intercepts is

$\phi = \pi I (0) \left[\frac{3a^2+2h^2-2h(a^2+h^2)^{\frac{1}{2}}}{2(a^2+h^2)} \right] \tag{1.10.4}$

What is this if $$a = 0$$ and if $$a \rightarrow \infty$$ ? Is this what you would expect? (Compare equation 1.6.7.)