COMEDK Ray Optics — practice questions

Lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. Sign convention: $u = -20$ cm (object on the left), $v = +60$ cm (real image on the right). We have $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{-20}$. That is $\dfrac{1}{f} = \dfrac{1}{60} + \dfrac{3}{60}$. That is $\dfrac{1}{f} = \dfrac{4}{60}$. So $\dfrac{1}{f} = \dfrac{1}{15}$. So $f = 15$ cm. **Trap.** Forgetting the sign on $u$ gives $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{20} = -\dfrac{1}{30}$, leading to a negative $f$ — that's how option B (12) or A (10) creeps in.

**For thin lenses in contact**, powers add: $P = P_1 + P_2$. $P_1 = \dfrac{1}{f_1\text{ (m)}} = \dfrac{1}{0.10} = 10$ D. $P_2 = \dfrac{1}{0.20} = 5$ D. $P_{\text{total}} = 10 + 5 = 15$ D. Equivalently, $\dfrac{1}{F} = \dfrac{1}{10} + \dfrac{1}{20} = \dfrac{3}{20}$ giving $F = \dfrac{20}{3}$ cm $\approx 6.67$ cm so $P = \dfrac{100}{6.67} = 15$ D.

**Critical angle formula** (denser → rarer): $\sin C = \dfrac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} = \dfrac{1}{\mu_{\text{denser}}} = \dfrac{1}{1.5} \approx 0.667$. $C = \sin^{-1}(0.667) \approx 41.8° \approx 42°$. **Why option C (60°) is tempting.** Some recall $\sin^{-1}(0.866) = 60°$ — but that's $\mu = 1/0.866 \approx 1.15$, not $1.5$. **A useful sanity check:** as $\mu$ rises, the critical angle drops (denser glass holds light in more aggressively). Diamond's $C \approx 24°$ — that's why it sparkles.