Worksheet 4: Radiometry and Photometry

Part 1

The bulb uses \(25\,\mathrm{W}\) electrically, but only \(20\%\) becomes light, so the radiant power is \(5\,\mathrm{W}\).

\[ E_{\text{photon}} = \frac{hc}{\lambda} \]

For \(\lambda = 500\,\mathrm{nm}\), with \(h = 6.626\cdot10^{-34}\,\mathrm{J\,s}\) and \(c = 3.00\cdot10^{8}\,\mathrm{m/s}\), one photon carries

\[ E_{\text{photon}} = \frac{(6.626\cdot10^{-34})(3.00\cdot10^{8})}{500\cdot10^{-9}} \approx 3.97 \cdot 10^{-19}\,\mathrm{J}. \]
\[ \frac{\text{photons}}{s} = \frac{5}{3.97 \cdot 10^{-19}} \]

About \(1.26 \cdot 10^{19}\) photons/s.

Part 2

Under the ideal assumption (efficiency \(\varepsilon = 100\%\)) all of the electrical power \(P = VC\) becomes radiant power, where \(V\) is the voltage and \(C\) the current:

\[ \Phi = \varepsilon\,VC = 2.4 \cdot 0.7 = 1.68\,\mathrm{W} \]

Since the bulb emits equally in all directions:

\[ I = \frac{\Phi}{4\pi} = \frac{1.68}{4\pi} = 0.134\,\mathrm{W/sr} \]

The bulb radius is \(0.005\,\mathrm{m}\), so its surface area is:

\[ A = 4\pi r^2 = 3.14 \cdot 10^{-4}\,\mathrm{m^2} \]
\[ M = \frac{\Phi}{A} = \frac{1.68}{3.14 \cdot 10^{-4}} = 5.35 \cdot 10^3\,\mathrm{W/m^2} \]

In 5 minutes, the time is \(300\,\mathrm{s}\):

\[ Q = \Phi t = 1.68 \cdot 300 = 504\,\mathrm{J} \]

Radiant flux: \(1.68\,\mathrm{W}\). Radiant intensity: \(0.134\,\mathrm{W/sr}\). Radiant exitance: \(5.35 \cdot 10^3\,\mathrm{W/m^2}\). Energy in 5 min: \(504\,\mathrm{J}\).

Part 3

At distance \(r = 1\,\mathrm{m}\), the irradiance from an isotropic point source is:

\[ E = \frac{I}{r^2} = \frac{0.134}{1^2} \]

The pupil size is not needed for the irradiance itself; it only sets the total power collected. With pupil diameter \(6\,\mathrm{mm}\) (radius \(3\,\mathrm{mm}\)):

\[ A_{\text{pupil}} = \pi(0.003)^2 = 2.83\cdot10^{-5}\,\mathrm{m^2}, \qquad \Phi_{\text{eye}} = E\,A_{\text{pupil}} = 0.134 \cdot 2.83\cdot10^{-5} \approx 3.78\cdot10^{-6}\,\mathrm{W}. \]

Irradiance at the eye: \(0.134\,\mathrm{W/m^2}\). The \(6\,\mathrm{mm}\) pupil receives about \(3.78 \cdot 10^{-6}\,\mathrm{W}\) total.

Part 4

Only 20% of the 200 W bulb becomes red light:

\[ \Phi = 200 \cdot 0.20 = 40\,\mathrm{W} \]
\[ I = \frac{\Phi}{4\pi} = 3.18\,\mathrm{W/sr} \]

The table is 2 m below the bulb and faces the bulb directly, so:

\[ E = \frac{I}{r^2} = \frac{3.18}{2^2} = 0.796\,\mathrm{W/m^2} \]

For photometry at \(650\,\mathrm{nm}\), use \(V(\lambda) = 0.1\):

\[ E_v = E \cdot 685 \cdot V(\lambda) = 0.796 \cdot 685 \cdot 0.1 \approx 54.5\,\mathrm{lux} \]

Irradiance: \(0.796\,\mathrm{W/m^2}\). Illuminance: \(54.5\,\mathrm{lux}\).

Part 5

Both sides of the screen are equally illuminated, so the two inverse-square illuminances match:

\[ \frac{I_s}{0.35^2} = \frac{I_x}{0.65^2} \]
\[ I_x = 40 \left(\frac{0.65}{0.35}\right)^2 \]

The unknown luminous intensity is \(138\,\mathrm{cd}\).

Part 6

For a diffuse emitter, radiance and radiant exitance are related by:

\[ M = \pi L \]

With \(L = 5000\,\mathrm{W/(sr\,m^2)}\):

\[ M = \pi \cdot 5000 = 1.57 \cdot 10^4\,\mathrm{W/m^2} \]

The source area is \(0.10 \cdot 0.10 = 0.01\,\mathrm{m^2}\), so the emitted power is:

\[ \Phi = MA = 1.57 \cdot 10^4 \cdot 0.01 \]

Radiosity/radiant exitance: \(1.57 \cdot 10^4\,\mathrm{W/m^2}\). Total emitted power: \(157\,\mathrm{W}\).

Part 7

Here the source is not diffuse because radiance depends on angle: \(L(\theta) = 6000\cos\theta\).

Radiant exitance integrates outgoing radiance over the hemisphere:

\[ M = \int_{\Omega} L(\theta)\cos\theta\,d\omega \]
\[ M = \int_0^{2\pi}\int_0^{\pi/2} 6000\cos^2\theta\sin\theta\,d\theta\,d\phi \]

The \(\phi\)-integral contributes \(2\pi\), and \(\int_0^{\pi/2}\cos^2\theta\sin\theta\,d\theta = \left[-\tfrac{1}{3}\cos^3\theta\right]_0^{\pi/2} = \tfrac{1}{3}\):

\[ M = 6000 \cdot 2\pi \cdot \frac{1}{3} = 4000\pi \approx 1.26\cdot10^{4}\,\mathrm{W/m^2} \]

The area is again \(0.01\,\mathrm{m^2}\).

Radiant exitance: \(1.26 \cdot 10^4\,\mathrm{W/m^2}\). Total power: \(126\,\mathrm{W}\).

Part 8 Optional

Radiance from the Sun is the same on Earth and Mars in empty space. What changes is the solid angle: Mars is farther away, so the Sun appears smaller and contributes less irradiance.

The Sun is diffuse, so:

\[ L = \frac{\Phi}{\pi A_{\text{sun}}} = \frac{3.91 \cdot 10^{26}}{\pi \cdot 6.07 \cdot 10^{18}} \]

Solar radiance at both planets: \(2.05 \cdot 10^7\,\mathrm{W/(m^2\,sr)}\).

The Sun is seen as a disk of projected area \(\pi R^2\), a quarter of its total surface area \(A_{\text{sun}} = 4\pi R^2\), so the solid angle subtended at distance \(d\) is \(\Omega = \pi R^2/d^2 = A_{\text{sun}}/(4d^2)\):

Solid angle from Earth: \(6.74 \cdot 10^{-5}\,\mathrm{sr}\). Solid angle from Mars: \(2.92 \cdot 10^{-5}\,\mathrm{sr}\).

The irradiance on a surface facing the Sun is \(E = L\,\Omega = \Phi/(4\pi d^2)\) (the two forms agree); over a \(1\,\mathrm{m^2}\) surface the received power equals \(E\):

Received by a \(1\,\mathrm{m^2}\) surface: Earth \(1.38 \cdot 10^3\,\mathrm{W}\), Mars \(5.99 \cdot 10^2\,\mathrm{W}\).

Part 9 Optional

For a uniform sky over the full hemisphere:

\[ E = \int_{\Omega} L\cos\theta\,d\omega = \pi L \]

Full sky irradiance: \(3.14 \cdot 10^3\,\mathrm{W/m^2}\).

For a visible cone around the zenith with half-angle \(\theta_{\max} = 30^\circ\), only the upper limit of the \(\theta\)-integral changes:

\[ E = L\int_0^{2\pi}\!\!\int_0^{\theta_{\max}}\cos\theta\sin\theta\,d\theta\,d\phi = \pi L \sin^2\theta_{\max} = \pi L \sin^2(30^\circ) = \tfrac{1}{4}\pi L \]

Cone-only irradiance: \(785\,\mathrm{W/m^2}\).