A 6.87-mm-high firefly sits on the axis of, and 12.1 cm in front of, the thin le
ID: 1415558 • Letter: A
Question
A 6.87-mm-high firefly sits on the axis of, and 12.1 cm in front of, the thin lens A, whose focal length is 6.51 cm. Behind lens A there is another thin lens, lens B, with focal length 22.5 cm. The two lenses share a common axis and are 58.5 cm apart. Is the image of the firefly that lens B forms real or virtual?
How far from lens B is this image located (expressed as a positive number)?
What is the height of this image (as a positive number)?
Is this image upright or inverted with respect to the firefly?
Explanation / Answer
height of the object ho=6.87*10^-2 cm
object distance u=12.1 cm
focal length of the first lens f1=6.51 cm
focal length of the second lens f2=22.5 cm
distance d=58.5 cm
now we first find the image formed b/w the lens
1/v1-1/u1=1/f1
1/v1=1/u1+1/f1
=1/12.1+1/6.51
V1=78.8/18.61=4.23 cm
now we find the finial image
object distance u2=d-v1=58.5-4.23=54.27 cm
the magnifaction of lens A=Ma=-v1/u1=-4.23/12.1=0.35
the image distance of lens B
1/v2-1/u2=1/f2
1/v2-1/54.27=1/22.5
1/v2=1/54.27+1/22.5
V2=1221.075/76.77=15.91 cm
the image formed at lens B=V2=15.91 cm
the magnification of lens B=Mb=-V2/u2=-15.91/54.27=-0.3
therefore the total magnifaction m=Ma*Mb=0.3* 0.35=0.105
now we find the height of the image
m=hi/ho
0.105=hi/6.87
hi=0.72135 mm
the image is upright