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 focal length \( f \) of a plano-convex lens in air is determined by the lensmaker's formula: \[\frac{1}{f} = \left( \frac{n - 1}{R} \right)\] Here, \( n \) represents the refractive index of the lens, and \( R \) is the radius of curvature of the lens.When the lens is submerged in a liquid with a refractive index \( n_l \), the effective refractive index of the lens relative to the liquid is given by \( n_{\text{eff}} = \frac{n}{n_l} \).Consequently, the lensmaker's formula is modified to: \[\frac{1}{f} = \frac{n_{\text{eff}} - 1}{R}\] Substituting \( n_{\text{eff}} = \frac{n}{n_l} \) yields: \[\frac{1}{f} = \frac{\left( \frac{n}{n_l} - 1 \right)}{R}\]Given the following values:- \( n = 1.5 \),- \( R = 10 \ \text{cm} = 0.10 \ \text{m} \),- \( n_l = 1.25 \).Upon substituting these values:\[\frac{1}{f} = \frac{\left( \frac{1.5}{1.25} - 1 \right)}{0.10} = \frac{\left( 1.2 - 1 \right)}{0.10} = \frac{0.2}{0.10} = 2\]Therefore, the focal length is calculated as: \[f = \frac{1}{2} = 0.5 \ \text{m} = 50 \ \text{cm}\] Final Answer: The focal length is \({50 \ \text{cm}} \).