AP Intermediate 2nd Year Physics — practice by topic

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.

Sustained interference needs sources with a constant phase relationship over time, which is what coherence means. Two ordinary lamps emit light with phases that vary randomly at $10^{-8}$ s scale so their interference pattern washes out almost instantly. The two slits in Young's experiment achieve coherence because both are illuminated by the same primary source, inheriting its phase. Equal intensity (B) improves fringe contrast but is not required. Same lamp (C) is one way to ensure coherence but not the only way (e.g. laser pairs). Polarisation alignment (D) matters for contrast, not for the existence of fringes.