02562 - Rendering - Worksheet 4

Part 1

To calculate the number of photons emitted per second by the light, we can simply divide the effective power output of the light by the energy of each photon, giving us the following formula:

\frac{\epsilon P}{\frac{hc}{\lambda}} = \frac{20\% \cdot 25\ \text{W}}{\frac{6.626 \cdot 10^{-34}\ \text{Js/photons} \cdot 2.9979 \cdot 10^8\ \text{m/s}}{500\ \text{nm}}} = 1.26 \cdot 10^{19}\ \frac{\text{photons}}{\text{s}}

Part 2

The values calculated using their respective formulas:

• Radiant flux: \Phi = VC = 2.4\ \text{V} \cdot 0.7\ \text{A} = 1.7\ \text{W}
• Radiant intensity: I = \frac{\text{d}\Phi}{\text{d}\omega} = \frac{1.7\ \text{W}}{4\ \pi\ \text{sr}} = 0.13\ \frac{\text{W}}{\text{sr}}
• Radiant exitance: M = \frac{\text{d}\Phi}{\text{d}A} = \frac{1.7\ \text{W}}{4 \cdot \pi \cdot (0.005\ \text{m})^2} = 5.3 \cdot 10^3\ \frac{\text{W}}{\text{m}^2}
• Emitted energy in 5 minutes: E = \Phi \cdot \text{d}t = 1.7\ \text{W} \cdot 300\ \text{s} = 5.0 \cdot 10^2\ \text{J}

Part 3

Since we do not know the size of the eye, the best assumption is that the pupil is a circle of area A = \pi \cdot (0.003\ \text{m})^2 = 2.8 \cdot 10^{-5}\ \text{m}^2. The angle between the surface normal and the central ray of the solid angle we assume to be 0 since the eye is pointed directly at the light. From this we can find the proportion of the area of the light bulb that focuses on the eye:

\text{d}\omega = \frac{A \cdot \cos \theta}{r^2} = \frac{2.8 \cdot 10^{-5}\ \text{m}^2 \cdot \cos 0}{(1\ \text{m})^2}\ \text{sr} = 2.8 \cdot 10^{-5}\ \text{sr}

E = \frac{I \cdot \text{d}\omega}{\text{d}A} = \frac{0.13\ \text{W}/\text{sr} \cdot 2.8 \cdot 10^{-5}\ \text{sr}}{2.8 \cdot 10^{-5}\ \text{m}^2} = 0.13\ \frac{\text{W}}{\text{m}^2}

Part 4

With the size and shape of the table being unknown, we assume that the entire surface of the table is 2 meters away from the light bulb. Again we assume \theta to be 0. Using this information, we can calculate the lightbulb's flux and intensity which we can use to calculate the irradiance:

\Phi = \epsilon P = 20\% \cdot 200\ \text{W} = 40\ \text{W}

I = \frac{\text{d}\Phi}{\Omega} = \frac{40\ \text{W}}{4\pi\ \text{sr}} = 3.18\ \frac{\text{W}}{\text{sr}}

E_{\text{radiometric}} = \frac{I \cdot \cos \theta\ \text{sr}}{r^2} = \frac{3.18\ \text{W}/\text{sr} \cdot \cos 0}{(2\ \text{m})^2} = 0.80\ \frac{\text{W}}{\text{m}^2}

Then we can calculate the illuminance given the new formula and given luminous efficiency:

E_{\text{photometric}} = E_{\text{radiometric}} \cdot 685 \cdot V(\lambda) = 0.80\ \frac{\text{W}}{\text{m}^2} \cdot 685 \cdot 0.1\ \frac{\text{lm}}{\text{w}} = 54.5\ \text{lux}

Part 5

We assume the two lights' wavelengths are equal. We can then set up an equality of the two equations for each of the lights' intensities:

E_s = E_x \Leftrightarrow I_s \frac{cos \theta}{r_s^2} = I_x \frac{cos \theta}{r_x^2}

Here we assume \theta to be equal as well, and since we know all variables aside from I_x, we can solve for that:

I_s \frac{cos \theta}{r_s^2} = I_x \frac{cos \theta}{r_x^2} \Leftrightarrow I_x = I_s \frac{r_x^2}{r_s^2} = 40\ \text{cd} \cdot \frac{(0.65\ \text{m})^2}{(0.35\ \text{m})^2} = 138\ \text{cd}

Part 6

To calculate the radiosity of the diffuse light, we can use the following formula and calculation:

B = L\pi = 5000\ \frac{\text{W}}{\text{sr}\ \text{m}^2} \cdot \pi = 15708\ \frac{\text{W}}{\text{m}^2}

To calculate how much energy is emitted by the light, also known as the flux, we can use the following formula:

B = \frac{\Phi}{A} \Leftrightarrow \Phi = B \cdot A = 15708\ \frac{\text{W}}{\text{m}^2} \cdot (0.01\ \text{m})^2 = 157.08\ \text{W}

Part 7

Radiant exitance is defined by the following the integral:

M = \int_{2\pi} L \cos \theta\ d\omega

M = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} 6000\ \frac{\text{W}}{\text{m}^2\ \text{sr}} \cos^2 \theta \sin \theta\ d\theta\ d\phi = 4000 \pi \frac{\text{W}}{\text{m}^2} = 12566\ \frac{\text{W}}{\text{m}^2}

M = \frac{\Phi}{A} \Leftrightarrow \Phi = M \cdot A = 12566\ \frac{\text{W}}{\text{m}^2} \cdot (0.01\ \text{m})^2 = 125.66\ \text{W}