Question

Review Constants In this example, we will apply the thin-lens equation to a lens that has two concave surfaces. The two surfaces of the lens shown in (Figure 1) have radii of curvature with absolute values of 20 cm and 5.0 cm. The index of refraction is 1.52. What is the focal length f of the lens? SOLUTION SET UP AND SOLVE The center of curvature of the first surface is on the incoming side of the light, so R, is negative: R = -20 cm. The center of curvature of the second surface is on the outgoing side of the light, so R is positive: R2 = 5.0 cm. Then, } = (n-1)(-+) = (1.52 – 1) (-20 cm 5 cm), f = -7.7 cm REFLECT This lens is a diverging lens--a negative lens--with a negative focal length. Part A - Practice Problem: A double-concave lens with index of refraction n =1.40 has two surfaces with radii of curvature with absolute values 15.0 cm and 12.0 cm What is the focal length of the lens? Express your answer with the appropriate units. }= -13.2 cm Submit Previous Answers Request Answer * Incorrect; Try Again Provide Feedback Next > Figure < 1 of 1

Answer 1

SET UP AND SOLVE: The center of curvature of the first surface is on the incoming side of the the light, so R, is negative R, = - 1 sem The centre of the second surface is on the outgoing side of the light. So R2 is positive R2 = 12cm f = (n-1) (1 h - 12) = (1.40-1) 75 - 1 1 2 f=-16.67cm Reflects: This lens is a negative lens... with a negative focal length.