Concept:An LDR is made of a semiconductor (such as Cadmium Sulfide, CdS). When illuminated with photons where \( hf \ge E_g \), it absorbs the energy and generates electron-hole pairs, increasing carrier concentration and reducing resistance.
Formula:$$R_{\text{LDR}} \propto \frac{1}{\Phi^\gamma} \quad (\text{where } \Phi \text{ is light intensity and } 0.7 < \gamma < 0.9)$$
Solution:- In darkness, carrier concentration is low, resulting in a very high 'dark resistance' (typically \( 1\text{–}10\text{ M}\Omega \)).
- Under illumination, photo-generated carriers increase electrical conductivity (\( \sigma = q n \mu_n + q p \mu_p \)), lowering the resistance to a few hundred ohms (\( 100\text{–}1000\ \Omega \)).
Why other options are incorrect:- Option A: Resistance decreases with illumination; it does not increase.
- Option C: LDRs are specifically designed to have light-dependent resistance.
- Option D: Passive photoconductors cannot have negative electrical resistance.
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