An object is placed a distance of 13.5 cm to the left of a thin converging lens
ID: 1519766 • Letter: A
Question
An object is placed a distance of 13.5 cm to the left of a thin converging lens of focal length f = 7.80 cm, and a concave spherical mirror with radius of curvature +16.0 cm is placed a distance of 41.2 cm to the right of the lens (see figure below). What is the location of the final image formed by the lens-mirror combination as seen by an observer positioned to the left of the object? cm to the left of the mirror What is the magnification of the final image as seen by an observer positioned to the left of the object? (Include the sign of the value in your answer.) Is the final image formed by the lens-mirror combination upright or inverted? upright invertedExplanation / Answer
A) Now consider image formation by the converging lens
(1/f) = (1/u) + (1/v)
(1/v) = (1/f) - (1/u)
here f =focal length =7.80 cm , u =object distance from lens =13.5 cm
and v = image distance , on substituting these values we get
(1/v) = (1/ 7.80cm ) - (1/13.5 cm)
(1/v) =0.054131 cm-1
v =18.47 cm (to the left of the lens)
Now concider the image formed by the concave spherical mirror
here focal length(fm) = (radius of curvature ) /2 = + 16.0 cm/2 =+ 8.0 cm
object distance (do) = image distance formed by the lens + distance between the lens and mirror
do= v + 41.2 cm =18.47 cm + 41.2 cm =59.67 cm
now from mirror equation we can find the image distance(di) from the mirror
(1/fm) = (1/do) + (1/di)
(1/di) = ( (1/fm) - ( (1/do)
(1/di) = (do - fm) / (fm do)
di =(fm do) /(do - fm) substitude all the values
di = (8.0 cm)(59.67 cm) / (59.67 cm -8.0 cm) =9.24 cm
Therefore the image formed at a distance of 9.24 cm to the left of the mirror
B) Magnification (m) is the ratio between the di and do
m = - (di /do )
here di =9.24 cm and do =59.67 cm (from part A)
m = - (59.67 cm / 9.24 cm) =-6.46
m = -6.46
C) The magnification of the final image is negative in sign so the image formed is inverted.