Question

A convex lens (\( n = 1.52 \)) has a focal length of 15.0 cm in air. Find its focal length when it is immersed in a liquid of refractive index 1.65. What will be the nature of the lens?

Hint:

Immersing a lens in a liquid with a higher refractive index decreases the focal length and may change the nature of the lens.

Answer 1

Let $f_a$ be the focal length of the convex lens in air and $f_l$ be the focal length of the convex lens in liquid. Let $n_g$ be the refractive index of the lens material, $n_a$ be the refractive index of air, and $n_l$ be the refractive index of the liquid. Let $R_1$ and $R_2$ be the radii of curvature of the lens surfaces. The lens maker's formula for a lens in air is given by: $$\frac{1}{f_a} = \left(\frac{n_g}{n_a} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ The lens maker's formula for a lens in liquid is given by: $$\frac{1}{f_l} = \left(\frac{n_g}{n_l} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ We are given that $n_g = 1.52$, $n_a = 1$, and $n_l = 1.65$. Dividing the second equation by the first equation, we get: $$\frac{f_l}{f_a} = \frac{\frac{n_g}{n_a} - 1}{\frac{n_g}{n_l} - 1} = \frac{\frac{1.52}{1} - 1}{\frac{1.52}{1.65} - 1} = \frac{1.52 - 1}{\frac{1.52 - 1.65}{1.65}} = \frac{0.52}{\frac{-0.13}{1.65}} = \frac{0.52 \times 1.65}{-0.13} = \frac{0.858}{-0.13} = -6.6$$ Thus, $f_l = -6.6 f_a$. Given that $f_a = 15$ cm, we have: $f_l = -6.6 \times 15 = -99$ cm. Since $f_l$ is negative, the convex lens in the liquid behaves as a diverging lens, or a concave lens. Final Answer: The final answer is $\boxed{-99cm}$