Question

A 6.61-mm-high firefly sits on the axis of, and 14.1 cm in front of, the thin lens A, whose focal length is 5.21 cm. Behind lens A there is another thin lens, lens B, with focal length 21.1 cm. The two lenses share a common axis and are 60.3 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?

Answer 1

A)

image due to first lense A

object distance u=14.1 cm

focal lemgth f1=5.21 cm

let image distance be v

1/u+1/v=1/f1

1/14.1+1/v=1/5.21

===>

v=8.26cm, this is the image due to first lense A

separation between lensA and lens B is =60.3 cm

now,

object distance for the second lense is,

u'=60.3-8.26

u'=52.04cm

let image distnace be v'

and

focal length of lense B is, f2=21.1cm

now,

1/u'+1/v'=1/f2

1/52.04+1/v'=1/(21.1)

===>

v'=35.49 cm

image distance due to second lense is v'=35.49 cm

image is

now,

final image distance from the actual object is=(14.1+60.3-35.49)=38.91cm

or

final image position from the first lense is =60.3+35.49= 95.79cm

B)

total magnification M=m1*m2=h'/h

here,

object height h=6.61 mm

final image height is h'

and

magnificatin of first lense is m1=v/u=8.26/14.1=0.586

magnificatin of second lense is m2=v'/u'=35.49/52.04=0.682

now,

M=m1*m2=h'/h

===>

h'=h*(m1*m2)

h'=6.61*(0.586*2*0.682)

h'=5.3 mm

there fore image height h'=5.3 mm cm (image is upright)