The object in the figure below is midway between the lens and the mirror, which
ID: 1263010 • Letter: T
Question
The object in the figure below is midway between the lens and the mirror, which are separated by a distance d = 24.9 cm. The magnitude of the mirror?s radius of curvature is 20.3 cm, and the lens has a focal length of -16.9 cm. (a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. 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
here ,
focal length of mirror , fm = 20.3/2
fm = 10.15 cm
fl = -16.9 cm
Now , for the mirror , do = 24.9/2 = 12.45 cm
using mirror equation ,
1/f = 1/di + 1/do
1/10.15 = 1/di + 1/12.45
di = 54.94 cm
Now , for the lens ,
do = 54.94 - 24.9
do = 30 cm
Using len's formula
-1/16.9 = 1/30 + 1/di
di = -10.81
therefore , image distance = 24.9 - 10.81
image distance = 14.1 cm
the image is at -14.1 cm to the right of mirror
b)
image is virtual ,
c)
as the image is inverted by the mirror(real image) , and it is erect for lens , overall for the imge
it is inverted
d)
overall magnification ,
m = 10.81 * 54.94/(30 * 12.45)
m = 1.59
the overall magnification is 1.59