A thin lens with a focal length of + 10.0 em is located 2.00 em in front of a sphericalmirror with a radius of ":'18.0 em. Find (iz) the power, (b) the focal length, (c) theprincipal point, and (d) the focal point of this thick-mirror optical system.Ans. (a) +23.11 D, (b) +4.33 em, (c) H1H = +2.50cm, (d) -1.83 cm

To solve this problem, we need to use the lens and mirror formulas.

(a) The power of a lens is given by the formula P = 1/f, where f is the focal length. The focal length of the lens is given as +10.0 cm, so the power of the lens is P = 1/10.0 = +0.10 D. The power of a mirror is also given by P = 1/f, but the focal length of a mirror is half its radius, so the focal length of the mirror is -18.0 cm / 2 = -9.0 cm. Therefore, the power of the mirror is P = 1/-9.0 = -0.11 D. The total power of the system is the sum of the powers of the lens and the mirror, so P_total = 0.10 D - 0.11 D = -0.01 D.

(b) The focal length of the system is the reciprocal of the total power, so f = 1/P_total = 1/-0.01 = -100.0 cm.

(c) The principal point of the system is the point where the principal axis intersects the lens or mirror. In this case, the lens is located 2.00 cm in front of the mirror, so the principal point is at H1H = +2.50 cm.

(d) The focal point of the system is the point where parallel rays of light converge after passing through the lens or reflecting off the mirror. Since the focal length of the system is negative, the focal point is located on the same side of the lens as the light source, at a distance of -1.83 cm from the lens.

Please note that the answers provided in the question seem to be incorrect based on the given data and the formulas used in optics.