Question E6F02:
What happens to photoconductive material when light shines on it?  

Choices:  
A: Resistance decreases  
B: Resistance increases  
C: Reflectivity increases  
D: Reflectivity decreases    

Correct Answer is  A: Resistance decreases  

Correct answer: A — Resistance decreases

Why A is correct:
- Photoconductive materials (e.g., CdS photoresistors/LDRs) generate electron-hole pairs when illuminated. More carriers mean higher conductivity.
- Conductivity increases with light:
  $$\sigma = q\,(n\mu_n + p\mu_p)$$
  where $q$ is charge, $n,p$ are carrier densities, and $\mu_n,\mu_p$ are mobilities. Light increases $n$ and $p$, so $\sigma$ increases.
- Resistance is inversely related to conductivity:
  $$R = \frac{\rho L}{A} = \frac{L}{A\sigma}$$
  As $\sigma$ increases with light, $R$ decreases.

Why the other choices are incorrect:
- B: Resistance increases — opposite of photoconductive behavior.
- C and D: Reflectivity changes — photoconductive devices are defined by changes in conductivity/resistance, not by their surface reflectivity.

Rules of thumb:
- Light intensity up → carriers up → conductivity up → resistance down.
- Typical CdS LDR: dark $R$ ≈ MΩ, bright light $R$ ≈ kΩ.
- Approximate relationship often follows $R \propto 1/I^\alpha$ with $\alpha \approx 0.7$–1, where $I$ is light intensity.