A convex lens $\left(n_{g}=1.5\right)$ of focal length 10 cm in air when immersed in a liquid behaves as a diverging lens of focal length 45 cm. The refractive index of the liquid:

1.68

The correct answer is Option (4) → 1.68

$\frac{1}{f_{air}}=(n_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f_{liquid}}=(n_g-n_{liquid})\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$⇒\frac{1}{f_{liquid}}=\frac{n_g-n_{liquid}}{n_g-1}.\frac{1}{f_{air}}$

$\frac{1}{-45}=\frac{1.5-n_{liquid}}{0.5}.\frac{1}{10}$

$-\frac{1}{45}=\frac{1.5-n_{liquid}}{10}$

$n_{liquid}-1.5=\frac{1}{9}=0.111$

$n_{liquid}≃1.6$