1. For convex lens R1 = +ve and R2 = -ve
\(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}+\frac{1}{R_2}\Big)\)
2. Yes.
P = \(\frac{1}{f}\) = \(\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}+\frac{1}{R_2}\Big)\)
From the above equation it is clear that, Pα \(\frac{n_2}{n_1}\)
lense in air, \(\frac{n_{glass}}{n_{air}}\) = \(\frac{1.5}{1}\) = 1.5
lense in water, \(\frac{n_{glass}}{n_{water}}\) = \(\frac{1.5}{1.33}\) = 1.12
In water,\(\frac{n_2}{n_1}\) is less. Hence power decreases.
3. P = +ve, converging
P = -ve, diverging
P = 0, Plane glass