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Physics Bootcamp

Section 43.5 Photometry

Radiometry gives us the objective measure of energy and power of light sources. However, human eye does not respond to all colors of visible light equally. Consequenctly, we perceive intensities differently than an objective instrument. For instance, experiments have shown that, during the day (i.e., photopic vision), the eye is most sensitive to the yellow-green (at around wavelength of \(555\,\text{nm}\)) light than to the red and blue lights of the spectrum. Therefore, yellow-green light will appears brighter to our eyes than either the red or the blue light of the same intensity. At night (i.e., scotopic vision) the sensitivity blue shifts with peak around \(505\,\text{nm}\) wavelength.
To take the difference between actual physical power emitted by a light source and as perceived by a human eye, it is necessary to weigh the actual power by the human eye response function, called spectral luminance efficiency or the \(V(\lambda)\) function, whose value is between 0 and 1 and varies with wavelength of the light. These weighted quenties are called photometric quantities to distinguish them from physical radiometric quantities.
If we kept the photometric units same as the radiometric units we will have the following relation between photometric power emitted \(F_{\text{photo},\lambda}\text{,}\) called luminous flux, and the radiometric cousin \(P_{\text{radio},\lambda}\) (the power) at some wavelength \(\lambda\text{.}\)
\begin{equation} F_{\text{photo},\lambda}(\text{Units: W}) = V(\lambda)\times P_{\text{radio},\lambda}(\text{Units: W}).\tag{43.13} \end{equation}
But we express photometric quantities in lumin (\(lm\)) and radiometric quantities in Watts (\(W\)). The conversion factor is chosen to be
\begin{equation*} 1\,\text{W} = 683\,\text{lm}. \end{equation*}
Thus, we include the conversion factor when writing Eq. (43.13).
\begin{equation} F_{\text{photo},\lambda}(\text{Units: lm}) = V(\lambda)\times P_{\text{radio},\lambda}(\text{Units: W}) \times \frac{683\,\text{lm}}{1\,\text{W}}.\tag{43.14} \end{equation}
Luminous flux gives total power in wavelength \(\lambda\) over the entire solid angle of \(4\,\pi\,\text{str}\text{.}\) The luminous flux per unit solid angle is called lumninous intensity. Let us denote it by \(I_{\text{photo},\lambda}\text{.}\)
\begin{equation*} I_{\text{photo},\lambda} = \frac{F_{\text{photo},\lambda}}{4\,\pi}\,\frac{\text{lm}}{\text{str}}. \end{equation*}
The unit \(\text{lm.str}^{-1}\) is called a candela (\(\text{Cd}\)), the base SI unit of luminous intensity. Or, multiply \(\text{lm.str}^{-1}\) by \(4\,\pi\ \text{str}\) and you get
\begin{equation*} 1.0\,\text{Cd} = 4\,\pi\,\text{lm}. \end{equation*}
A typical wax candle has luminous intensity of about \(1\,\text{Cd}\text{.}\)
Table 43.14 shows the units of radiometry and photometry side by side. Various sources of light are often labelled with their luminance, which is the photometric term for the total power radiated per unit are per unit solid angle. Table 43.15 gives approximate luminance of several sources of light.
Table 43.14. Radiometric and Photometric Quantities and Units
Radiometry Photometry
Definition Name Unit Name Unit Unit
Conversion
Power Power Watt Luminous flux Lumin \(1\text{ W}\)
(\(\text{W}\)) (lm) \(= 683\text{ lm}\)
Power per unit Radiant Luminous Candela \(1\text{ W/sr}\)
unit solid intensity W/sr intensity (\(\text{Cd}\)) \(= 683\text{ Cd}\)
angle
Power per Irradiance/ Illuminance \(\text{lm/m}^2\) \(1\text{ W/m}^2\)
unit area Emittance/ \(\text{W/m}^2\) Illuminance (lux(\(\text{lx}\)) ) \(= 683\text{ lx}\)
Intensity
Power per Radiance \(\text{W/(m}^2\text{sr)}\) Luminance \(\text{Cd/m}^2\) \(1 \text{ W/m}^2 \text{sr}\)
unit area \((\text{nit})\) \(= 683\text{ nit}\)
per unit
solid angle
Table 43.15. Luminance of Various Sources (Ref: National Physical Laboratory, UK.)
Light Source Luminance (\(\text{nits}\))
Atomic fission bomb \(2 \times 10^{12}\)
(\(0.1\text{ msec}\) after firing)
Lightning flash \(8 \times 10^{10}\)
Sunlight on the surface of Earth \(1.6 \times 10^9\)
(at meridian)
Photoflash lamps \(2.5 \times 10^8\)
\(60\text{ W}\) bulb \(1.2 \times 10^5\)
Sperm candle flame \(1 \times 10^4\)
Clear blue sky \(4 \times 10^3\)
Moonlight on Earth \(3 \times 10^3\)
Starlit sky \(5 \times 10^{-4}\)

Exercises Exercises

1. Luminance of Moon.

What will be the luminance of moon if observed from half the distance to the moon? Luminance of Moonlight on Earth \(= 3000\text{ nits}\text{.}\)
Hint.
Luminosity varies as inverse square of distance from an isotropic source. Assume moon as an isotropic source.
Answer.
\(1.2 \times 10^4\text{ nits}\text{.}\)
Solution.
The luminosity will vary in space from the source (moon here) as the intensity varies. Let us assume moon to be an isotropic source of secondary light. Therefore, the luminosity at half the distance will be 4 times as much.
\begin{equation*} L_{x/2} = \left(\dfrac{x}{x/2} \right)^2 \times L_x = 4\times 3000 = 12,000\:\text{nit}. \end{equation*}

2. Luminosity of a Laptop Monitor at the Eye.

The luminance of a laptop monitor is \(150\text{ nits}\) at the eye \(25\text{ cm}\) from the screen. How much energy in Joules enters the eye in one hour if the cross-section of the pupil is assumed to be a circle of radius \(5\text{ mm}\text{?}\)
Hint.
Answer.
\(1.55 \times 10^{-5}\text{ J}\text{.}\)
Solution.
The solid angle \(\Delta\Omega\) subtended at the cross-section of the pupil by a point source 25 cm from the eye is given by \(4\pi\) times the ratio of the area of the eye to the area of the spherical surface from the source.
Figure 43.16. The solid angle and area element at the eye for Exercise 43.5.2.
\begin{equation*} \Delta\Omega = \dfrac{ \pi (0.5\:\text{cm})^2}{(25\:\text{cm})^2} = 2.5\times 10^{-4}\:\text{sr}, \end{equation*}
and the luminosity given in nits can be expressed in SI units to give
\begin{equation*} 150\:\text{nit} = \dfrac{150}{683}\:\dfrac{\text{W}}{\text{m}^2\text{sr}}. \end{equation*}
Therefore,
\begin{align*} \text{Energy} \amp = \dfrac{150}{683}\:\dfrac{\text{W}}{\text{m}^2\text{sr}}\times 2.5\times 10^{-4}\:\text{sr} \times \pi(0.005\:\text{m})^2 \\ \amp = 1.55\times 10^{-5}\:\text{J}. \end{align*}

3. Illuminance of Sunlight from Irradiance.

The irradiance of the sunlight on the Earth is approximately \(1.4\text{ kW/m}^2\text{.}\) Determine the illuminance.
Hint.
Use definition of illuminance.
Answer.
\(9.6 \times 10^5\text{ lx}\text{.}\)
Solution.
Using the definition of illuminance we get
\begin{equation*} \text{Illuminance} = 1.5\times 10^3\:\frac{\text{W}}{\text{m}^2}\times \dfrac{683\:\text{lux}}{1\:\text{W/m}^2} = 9.56\times 10^5\:\text{lux}. \end{equation*}

4. Radiant Intensity and Luminous Intensity of the Sun from Irradiance.

The irradiance of the sunlight on the Earth is approximately \(1.4\text{ kW/m}^2\text{.}\) The average distance between the Earth and the Sun is \(1.5\times10^8\text{ km}\text{.}\) Determine the radiant intensity and luminous intensity of the Sun.
Hint.
Use definitions of radiant intensity and luminous intensity.
Answer.
\(3.15 \times 10^{22}\text{ kW/sr}\text{,}\) \(2.15 \times 10^{28}\text{ cd}\text{.}\)
Solution.
Using the definitions we get
\begin{align*} \text{Radiant intensity} \amp = \dfrac{\text{Total power}}{4\pi\:\text{str}} = \dfrac{1.4\:\text{kW/m}^2\times 4\pi\times R_{\text{ES}}^2}{4\pi},\\ \amp = 1.4\:\text{kW/m}^2\times(1.5\times 10^{11}\:\text{m})^2 \\ \amp = 3.15\times 10^{22}\:\text{kW/str}. \end{align*}
\begin{align*} \text{Luminous } \amp \text{intensity} = \text{Radiant intensity}\times 683\times 10^{3}\:\text{Cd.str/kW}, \\ \amp = 3.15\times 10^{22}\:\text{kW/str}\times 683\times 10^{3}\:\text{Cd.str/kW} \\ \amp = 2.15\times 10^{28}\:\text{Cd}. \end{align*}