Problem Statement
The photopic luminous efficiency function peaks at $\lambda = 555$ nm with $V(555) = 1.0$ and has $V(450) = 0.038$, $V(650) = 0.107$. A lamp emits equal radiant flux at all three wavelengths. Find the relative luminous fluxes.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Luminous flux $\Phi = 683\int P(\lambda)V(\lambda)d\lambda$. For equal radiant power $P$ at each wavelength:
Step 2 — Apply the relevant physical law or equation: $$\Phi_{450}:\Phi_{555}:\Phi_{650} = V(450):V(555):V(650) = 0.038:1.0:0.107$$
$$\boxed{= 0.038:1.000:0.107 \text{ (violet:green:red)}}$$
Step 3 — Solve algebraically for the unknown: The eye is 26× more sensitive to green than to red, and 26× more sensitive to red than to violet.
Worked Calculation
$$\Phi_{450}:\Phi_{555}:\Phi_{650} = V(450):V(555):V(650) = 0.038:1.0:0.107$$
$$\boxed{= 0.038:1.000:0.107 \text{ (violet:green:red)}}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
Luminous flux $\Phi = 683\int P(\lambda)V(\lambda)d\lambda$. For equal radiant power $P$ at each wavelength:
$$\Phi_{450}:\Phi_{555}:\Phi_{650} = V(450):V(555):V(650) = 0.038:1.0:0.107$$
$$\boxed{= 0.038:1.000:0.107 \text{ (violet:green:red)}}$$
The eye is 26× more sensitive to green than to red, and 26× more sensitive to red than to violet.
Answer
$$\boxed{= 0.038:1.000:0.107 \text{ (violet:green:red)}}$$
Physical Interpretation
The Carnot efficiency sets an absolute upper bound imposed by thermodynamics — no real engine, however well engineered, can exceed it. Higher hot-reservoir temperature or lower cold-reservoir temperature both increase efficiency.
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