$(i)$ If $f=0.5 \,m$ for a glass lens,what is the power of the lens?
$(ii)$ The radii of curvature of the faces of a double convex lens are $10 \,cm$ and $15 \,cm$. Its focal length is $12 \,cm$. What is the refractive index of glass?
$(iii)$ $A$ convex lens has $20 \,cm$ focal length in air. What is its focal length in water? (Refractive index of water $= 1.33$,refractive index of glass $= 1.5$)

(N/A) $(i)$ The power $P$ of a lens is given by $P = \frac{1}{f(m)}$. Substituting $f = 0.5 \,m$,we get $P = \frac{1}{0.5} = +2 \,D$.
$(ii)$ Given $f = +12 \,cm$,$R_1 = +10 \,cm$,and $R_2 = -15 \,cm$. Using the lens maker's formula $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$:
$\frac{1}{12} = (n - 1) \left( \frac{1}{10} - \frac{1}{-15} \right) = (n - 1) \left( \frac{3+2}{30} \right) = (n - 1) \left( \frac{5}{30} \right) = (n - 1) \frac{1}{6}$.
Thus,$n - 1 = \frac{6}{12} = 0.5$,which gives $n = 1.5$.
$(iii)$ In air: $\frac{1}{f_a} = (n_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{20} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{10}$.
In water: $\frac{1}{f_w} = \left( \frac{n_g}{n_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{1.5}{1.33} - 1 \right) \left( \frac{1}{10} \right) = (1.1278 - 1) \times 0.1 = 0.1278 \times 0.1 = 0.01278$.
$f_w = \frac{1}{0.01278} \approx 78.2 \,cm$.