A concave-convex lens of refractive index $$1.5$$ and the radii of curvature of its surfaces are $$30 cm$$ and $$20 cm$$, respectively. The concave surface is upwards and is filled with a liquid of refractive index $$1.3$$. The focal length of the liquid-glass combination will be

$$\frac{1}{F} = \frac{1}{f_{\text{liquid}}} + \frac{1}{f_{\text{glass}}}$$ (Net focal length)

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

$$\frac{1}{f_{\text{glass}}} = (1.5 - 1) \left( \frac{1}{-30} - \frac{1}{-20} \right)$$

$$\frac{1}{f_{\text{glass}}} = 0.5 \left( -\frac{1}{30} + \frac{1}{20} \right) = 0.5 \left( \frac{-2 + 3}{60} \right) = 0.5 \times \frac{1}{60} = \frac{1}{120}\text{ cm}^{-1}$$

$$\frac{1}{f_{\text{liquid}}} = (1.3 - 1) \left( \frac{1}{\infty} - \frac{1}{-30} \right)$$

$$\frac{1}{f_{\text{liquid}}} = 0.3 \left( 0 + \frac{1}{30} \right) = \frac{0.3}{30} = \frac{1}{100}\text{ cm}^{-1}$$

$$\frac{1}{F} = \frac{1}{100} + \frac{1}{120}$$

$$F = \frac{600}{11}\text{ cm}$$