Question

Find the focal length of the plano-convex lens of refractive index 1.5 and radius of curvature 10 cm when it is immersed in a liquid of refractive index 1.25.

Hint:

When the lens is immersed in a liquid, the effective refractive index is reduced. The refractive index of the lens should be divided by the refractive index of the liquid in the lensmaker's formula.

Answer 1

The formula for the focal length \( f \) of a plano-convex lens in air is given by the lensmaker's formula: \[ \frac{1}{f} = \left( \frac{n - 1}{R} \right) \] where: - \( n \) is the refractive index of the lens, - \( R \) is the radius of curvature of the lens. However, when the lens is immersed in a liquid of refractive index \( n_l \), the effective refractive index of the lens relative to the liquid is \( n_{\text{eff}} = \frac{n}{n_l} \). So, the modified lensmaker's formula becomes: \[ \frac{1}{f} = \frac{n_{\text{eff}} - 1}{R} \] Substitute \( n_{\text{eff}} = \frac{n}{n_l} \): \[ \frac{1}{f} = \frac{\left( \frac{n}{n_l} - 1 \right)}{R} \] Given: - \( n = 1.5 \), - \( R = 10 \ \text{cm} = 0.10 \ \text{m} \), - \( n_l = 1.25 \). Substituting the values: \[ \frac{1}{f} = \frac{\left( \frac{1.5}{1.25} - 1 \right)}{0.10} = \frac{\left( 1.2 - 1 \right)}{0.10} = \frac{0.2}{0.10} = 2 \] Thus, the focal length is: \[ f = \frac{1}{2} = 0.5 \ \text{m} = 50 \ \text{cm} \] Final Answer: The focal length is \({50 \ \text{cm}} \).