As we know,
\[ \frac{1}{f} = \left[\frac{\mu_L}{\mu_m} - 1\right] \left[\frac{1}{R_1} - \frac{1}{R_2}\right] \]
For air, \(\mu_m = 1\):
\[ \frac{1}{12} = [1.6 - 1]\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \] \[ \frac{1}{12} = \frac{6}{10}\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \]
\[ \left[\frac{1}{R_1} - \frac{1}{R_2}\right] = \frac{10}{72} \]
For water:
\[ \frac{1}{f} = \left[\frac{1.6}{1.28} - 1\right]\left[\frac{10}{72}\right] \] \[ \frac{1}{f} = \frac{32}{128} \times \frac{10}{72} \] \[ \frac{1}{f} = \frac{1}{4} \times \frac{10}{72} \] \[ f = 28.8 \, \text{cm} \] \[ f = 288 \, \text{mm} \] \[ \boxed{f = 28.8\,\text{cm or } 288\,\text{mm}} \]
Light from a point source in air falls on a spherical glass surface (refractive index, \( \mu = 1.5 \) and radius of curvature \( R = 50 \) cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is 1cm.

Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: