The object in Figure P23.50 is midway between the lens and the mirror. The mirro
ID: 1272235 • 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?
Is it upright or inverted?
What is the overall magnification of the image?
PLEASE INPUT EXACT ANSWERS TO EACH
Explanation / Answer
for mirror,
f = -r/2 = -20.8/2 = -10.4
for mirror we have 1/f + 1/0 = 1/i
we have 0 =distance of object from mirror = -25/2 = -12.5
so, 1/(-10.4) + 1/(-12.5) = 1/i
i=-5.676855 cm
we have magnification = - i/o = -5.676855/12.5 = -0.4541484
the imge is inverted after the object is reflected from mirror.
Noe for lens, we have f = -16.8
1/f = 1/u + 1/v
we have u = distance of object from the lens = 25 -5.676855 = 19.323145 cm
so, 1/-16.8 = 1/19.323145 + 1/v
v=-8.986726 cm
the image is real ,
magnification = -v/u = 8.986726/19.323145 = 0.4650757
the image will remain inverted.
the answers are :
loacation of final image = 25 + 8.986726 = 33.986726 cm infront of the mirror.
the image is real.
the image is inverted.
Overall magnification = 0.4541484 * 0.4650757 = 0.211213 is the answer.