A converging lens with two convex surfaces has a front surface with radius of curvature of 13.1 cm;the back surface has radius of curvature of 16.5 cm,and it is made from material with an index of refraction of 2.10.What is the focal length of the lens?
A)6.64 cm
B)57.8 cm
C)−57.8 cm
D)3.48 cm
E)30.3 cm

Question

Quizplus · Accepted Answer

The focal length $f$ of a lens is given by the Lensmaker's formula:$$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$where $n$ is the index of refraction of the lens material, $R_1$ is the radius of curvature of the first surface (positive for convex surfaces), and $R_2$ is the radius of curvature of the second surface (negative for convex surfaces when following the sign convention that treats rays as moving from left to right).Given:- $n = 2.10$- $R_1 = 13.1 \, \text{cm}$ (convex surface, so positive)- $R_2 = -16.5 \, \text{cm}$ (convex surface, but since it's the second surface and we're assuming light travels from left to right, it's considered negative)Plugging these values into the Lensmaker's formula:$$\frac{1}{f} = (2.10 - 1) \left( \frac{1}{13.1} - \frac{1}{-16.5} \right)$$$$\frac{1}{f} = 1.10 \left( \frac{1}{13.1} + \frac{1}{16.5} \right)$$$$\frac{1}{f} = 1.10 \left( \frac{16.5 + 13.1}{13.1 \times 16.5} \right)$$$$\frac{1}{f} = 1.10 \left( \frac{29.6}{216.15} \right)$$$$\frac{1}{f} \approx 1.10 \times 0.1368 \approx 0.1505$$$$f \approx \frac{1}{0.1505} \approx 6.64 \, \text{cm}$$Therefore, the correct answer is A) 6.64 cm.

A Converging Lens with Two Convex Surfaces Has a Front