A double-concave lens has equal radii of curvature of magnitudes 15.1 cm. An object placed 14.2 cm from the lens forms a virtual image 5.00 cm from the lens. What is the index of refraction
Of the lens material?
A) 1.98
B) 1.84
C) 2.06
D) 1.90

Question

Quizplus · Accepted Answer

The correct answer is A.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/di1/f = 1/14.2 cm + 1/(-5.00 cm)f = -7.10 cmThe radius of curvature is given by R = 2f, so R = -14.2 cm. The index of refraction is given by n = 1 + (R/f), so n = 1 + (-14.2 cm/-7.10 cm) = 1.98.

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