A convex lens of refractive index 1.5 and focal length 18 cm in air, is immersed in water. The change in the focal length of the lens will be _____ cm.
(Given refractive index of water = $$\frac{4}{3}$$)

$$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{1}{f_a} = (\mu_l - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{1}{f_w} = \left(\frac{\mu_l}{\mu_w} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{f_w}{f_a} = \frac{\mu_l - 1}{\frac{\mu_l}{\mu_w} - 1}$$

$$\frac{f_w}{f_a} = \frac{\frac{3}{2} - 1}{\frac{3/2}{4/3} - 1}$$

$$\frac{f_w}{f_a} = 4$$

$$f_w = 4 \times 35 = 140\text{ cm}$$

$$\Delta f = 140 - 35$$

$$\Delta f = 105\text{ cm}$$