A convex lens of refractive index 1.5 and focal length f = 18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $$\alpha \times f$$. The value of$$\alpha$$ is______.
(refractive index of water = 4/3)

We need to find α where the difference in focal lengths in water and air is αf.

In air: $$f_{air} = f = 18$$ cm

In water: Using the lens maker's equation:

$$\frac{1}{f_{water}} = \frac{n_L/n_w - 1}{n_L - 1} \times \frac{1}{f_{air}}$$

$$\frac{f_{water}}{f_{air}} = \frac{n_L - 1}{n_L/n_w - 1} = \frac{1.5 - 1}{1.5/(4/3) - 1} = \frac{0.5}{1.125 - 1} = \frac{0.5}{0.125} = 4$$

$$f_{water} = 4f = 72$$ cm

Difference: $$f_{water} - f_{air} = 72 - 18 = 54 = 3f$$

So $$\alpha = 3$$.

Therefore, α = 3.