The object in Figure P23.50 is midway between the lens and the mirror. The mirro
ID: 1272059 • Letter: T
Question
The object in Figure P23.50 is midway between the lens and the mirror. The mirror's radius of curvature is 20.8 cm, and the lens has a focal length of -16.8 cm.
Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system.
cm ---Location--- behind the mirror in front of the lens in front of the mirror
Is this image real or virtual?
virtual real
Is it upright or inverted?
invertedupright
What is the overall magnification of the image?
?
Explanation / Answer
1/f1 = 1/u1 + 1/v1
f1 = R/2 = 20.8/2 = 10.4 cm
u1 = 25/2 = 12.5 cm
1/10.4 = 1/12.5 + 1/v1
v1 = 61.9 cm
now for the lens,
image of the mirror is the object for th elens
u2 = 61.9-25 = 36.9 cm
f2 = -16.8 cm
1/f2 = 1/u2 + 1/v2
1/-16.8 = 1/36.9 + 1/v2
v2 = -11.544 cm
the final image formed is 11.544 cm infront of th elens
b) image is virtual
c) image is inverted
d) overall magnification = m1*m2
M = (v1/u1)(v2/u2) = (61.9/12.5)(-11.544/36.9) = -1.55
negative sign means image is inverted