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 \(=4 / 3\) )

Explanation:
According to the Lens Maker's Formula, the focal length of a lens is given by:
\( \frac{1}{f} = (\mu_{rel} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)

1. When the lens is in the air:
\( \frac{1}{f_{air}} = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Substituting \( \mu_g = 1.5 \) and \( f_{air} = f \):
\( \frac{1}{f} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \) --- (Equation 1)

2. When the lens is immersed in water:
The relative refractive index becomes \( \frac{\mu_g}{\mu_w} \).
\( \frac{1}{f_{water}} = \left(\frac{\mu_g}{\mu_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Substituting \( \mu_g = 1.5 \) and \( \mu_w = \frac{4}{3} \):
\( \frac{1}{f_{water}} = \left(\frac{1.5}{4/3} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{4.5}{4} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( \frac{1}{f_{water}} = \left(1.125 - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.125 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \) --- (Equation 2)

3. Taking the ratio of Equation 1 and Equation 2:
\( \frac{f_{water}}{f} = \frac{0.5}{0.125} = 4 \)
Therefore, \( f_{water} = 4f \)

4. Finding the difference:
Difference in focal lengths = \( f_{water} - f_{air} \)
Difference = \( 4f - f = 3f \)

Given that the difference is \( \alpha \times f \), we can clearly see that \( \alpha = 3 \).

Hence, Option C is the correct choice.