The object in the figure below is midway between the lens and the mirror, which
ID: 1569432 • Letter: T
Question
The object in the figure below is midway between the lens and the mirror, which are separated by a distance d = 24.5 cm. The magnitude of the mirror's radius of curvature is 20.4 cm, and the lens has a focal length of -16.0 cm. (a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. Your response differs from the correct answer by more than 100%. to the right of the mirror (b) Is this image real or virtual? real virtual (c) Is it upright or inverted? upright inverted (d) What is the overall magnification?Explanation / Answer
d = 24.5 cm ; R = 20.4 cm ; f2 = -16 cm
a)o1 = 24.5/2 = 12.25 cm
f1 = 20.4/2 = 10.2 cm
from lens eqn,
1/f = 1/i + 1/o
i = o x f / ( o - f)
i1 = o1 x f1/ (o1 - f1)
i1 = 12.25 x 10.2/(12.25 - 10.2) = 60.95 cm
o2 = 24.5 - 60.95 = -36.45
i2 = o2 f2/(o2 - f2)
i2 = -36.45 x -16 / (-36.45 + 16) = -28.52 cm
Hence, final image will be formed at i2 = -28.52 cm
b)Its a virtual image
c)The image will be upright
d)m1 = -i1/o1 = -60.95/12.25 = -4.98
m2 = -i2/o2 = -(-28.52)/-36.45 = -0.782
M = m1 x m2 = -4.98 x -0.782 = 3.89
Hence, M = 3.89