The object in the figure below is midway between the lens and the mirror, which
ID: 1546891 • Letter: T
Question
The object in the figure below is midway between the lens and the mirror, which are separated by a distance d = 24.7 cm. The magnitude of the mirror's radius of curvature is 20.5 cm, and the lens has a focal length of -17.0 cm. (a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. 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
the mirror : O1 has f1 = R/2 = 10.25 cm
the lens : O2 has f2 = -17.0 cm
The object : A.
The distance between the lens and the mirror is l
.........O1(f1) ..... ....... O2(f2)
A ------- ----------> A1 ------- ----------> A2
d1........ ........ d'1 d2 ...... ........ d'2
d1 = 24.7/2 = 12.35 cm
d'1 = d1f / (d1 - f) = 12.35*10.25 / (12.35 -10.25) = 60.3 cm
The distance between A1 and O2 is: | d2 |
d2 = l - d'1 = 24.7 - 60.3 = -35.6 cm
a) Location of the final image formed by this system is :
=> d2' = d2f / (d2 - f) = (-24.7)*(-17) / [ -24.7 - ( -17) ] = - 54.5 cm
b) Image is virtual
c) Since total magnification is > 0, upright.
d) the overall magnification of the image is :
K = k1*k2 = (-d1' /d1) * (-d2' /d2) = ( d1'd'2 )/ (d1d2)
K = ( 60.3 / 12.35 ) * ( -54.5 / -35.6)
K = 7.475