Question

The thick lens shown at right is made from glass with index of refraction 1.50. The thick lens has thickness of 6.0 cm and radii of curvature of 8.00 cm and 10.00 cm. (a) Find the focal length of this thick lens in air and the position of its focal points (F and F) and its principal points (H, and H,) with respect to the front and back of the lens respectively. (b) Draw a sketch showing the locations of the focal points and the principle planes. Indicate all distance on the sketch. 1.

Answer 1

Index of refraction = 1.5 = (n_l) thickness (d) = 6 cm radii of curvature R_1 = 8 cm R_2 = 10 cm (a) Focal length = (h_l - 1) (1/R_1 - 1/R_2 + (h_l - 1)d/n l R_1 R_2) f = (1.5 - 1) (1/8 - 1/10 + (1.5 - 1)6/1.5 times 8 times 10) f = 0.5 (0.125 - 0.1 + 0.025) f = 0.025 cm Principle points = -f(n_l - 1)d/n_l R_2 = -0.025(1.5 - 1)6/1.5 times -10 +ve means right to vertex V_1 = 5 times 10^-3 cm Prici point(H_2) = -f(nl - 1)d/n_2 R_1 = -0.025 (1.5 - 1)6/1.5 times 8 -ve means left to vertex V_2 = -6.25 times 10^-3 cm Then we already know that f = Front focal point + H, 0.025 = f_1 + 5 times 10^-3 f = 0.02 cm f = Back focal point + H_2 0.025 = f_2 + 6.25 times 10^-3 = 0.01875 cm \r\n