A double-concave lens has equal radii of curvature of 15.1 cm. An object placed 14.2 cm from the lens forms a virtual image 5.29 cm from the lens. What is the index of refraction of the lens material?
A) 1.90
B) 1.98
C) 1.82
D) 1.77

Question

Quizplus · Accepted Answer

The lens equation is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. For a double-concave lens, f is negative. We can use the thin lens equation to find the focal length: 1/f = 1/do + 1/di = 1/14.2 cm + 1/(-5.29 cm) = -0.0423 cm^-1. Therefore, f = -23.6 cm. The lens maker's equation is 1/f = (n-1)(1/R1 + 1/R2), where n is the index of refraction of the lens material, and R1 and R2 are the radii of curvature of the lens surfaces. For a double-concave lens, R1 = R2 = -15.1 cm. Substituting these values into the lens maker's equation, we get -0.0423 cm^-1 = (n-1)(1/-15.1 cm + 1/-15.1 cm). Solving for n, we get n = 1.90.

A Double-Concave Lens Has Equal Radii of Curvature of 15