Given that, Refractive index of the glass lens, $\mu_g = 1.5$ Refractive index of water, $\mu_w = 1.33$ Focal length of the lens in air, $f_a = 8 \text{ cm}$ According to the lens maker's formula, the focal length of a lens in a medium is given by: $$ \frac{1}{f} = (_{\text{medium}}\mu_{\text{lens}} - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) $$ For the lens in air: $$ \frac{1}{f_a} = (_a\mu_g - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \quad \dots (i) $$ For the lens in water: $$ \frac{1}{f_w} = (_w\mu_g - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \quad \dots (ii) $$ Dividing equation (i) by equation (ii), we get: $$ \begin{aligned} \frac{f_w}{f_a} &= \frac{_a\mu_g - 1}{_w\mu_g - 1} \\ &= \frac{\mu_g - 1}{\frac{\mu_g}{\mu_w} - 1} \end{aligned} $$ Substituting the values: $$ \begin{aligned} \frac{f_w}{8} &= \frac{1.5 - 1}{\frac{1.5}{1.33} - 1} \\ \frac{f_w}{8} &= \frac{0.5}{1.1278 - 1} \\ \frac{f_w}{8} &= \frac{0.5}{0.1278} \\ \frac{f_w}{8} &= 3.912 \\ f_w &= 8 \times 3.912 \\ f_w &\approx 31.3 \text{ cm} \end{aligned} $$ Note: If we use the approximation $\mu_w = \frac{4}{3}$, the calculation becomes: $$ \begin{aligned} \frac{f_w}{f_a} &= \frac{1.5 - 1}{\frac{1.5}{4/3} - 1} = \frac{0.5}{1.125 - 1} = \frac{0.5}{0.125} = 4 \\ f_w &= 4 \times f_a = 4 \times 8 = 32 \text{ cm} \end{aligned} $$ Thus, the focal length of the lens in water is $32 \text{ cm}$.