The object in the figure below is midway between the lens and the mirror, which
ID: 1441727 • 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 19.63 cm, and the lens has a focal length of -17.0 cm. 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 Is this image real or virtual? Is it upright or inverted? What is the overall magnification? Your response differs from the correct answer by more than 10%. Double check your calculations.Explanation / Answer
given that
radius of curvature of mirror = 19.6 cm,so
focal length of mirror
f1 = R/2 = 19.6/2 = 9.8 cm
focal length of lens
f2 = -17 cm
distance b/w mirror and lens
d = 24.5 cm
Let the distance of object fron mirror is d1.
Let the distance of image from mirror is d1'.
the mirror image will be object for lens ,
Let the distance b/w lens and this object is d2.
Let the distance of image from lens is d2'.
now
d1 = d/2
d1 = 24.5/2 = 12.25 cm
d1' = d1*f1 / (d1-f1)
d1' = 12.25*9.8 /(12.25 - 9.8)
d1' = 49 cm
d2 = d - d1'
d2 = 24.5 - 49 = -24.5
Location of the final image
d2' = d2 * f2 /d2 - f2
d2' = (-24.5) * (-17) / (-24.5) -(-17)
d2' = 416.5 / -7.5
d2' = -55.53
final image is right to the mirror .
part(B)
the distance of final image from lens is negetive so the image will be virtual .
part(d)
magnification m
m = m1*m2
m = (-d1' /d1) * (-d2' /d2)
m = (49 / 12.25) * (55.53 / 24.5)
m = 4 * 2.26
m = 9.04
part (c)
the magnification is positive so the image is upright