Question

A 5.95-mm-high firefly sits on the axis of, and 12.9 cm in front of, the thin lens A, whose focal length is 6.21 cm. Behind lens A there is another thin lens, lens B, with local length 29.5 cm. The two lenses share a common axis and are 57.1 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

Note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    6.21   cm
do = object distance =    1.29E+01   cm
di = image distance      
      
Thus, solving for di,      
      
di =    11.97443946   cm
      
Thus, the magnification, m = -di/do,      
      
m =    -0.928251121  
      
Thus, as      
hi =    5.95E+00   mm
      
And hf = m * hi, then      
      
hf =    -5.52309417   mm

Thus, for the second interface,

Note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    29.5   cm
do = object distance =    4.51E+01   cm
di = image distance      
      
Thus, solving for di,      
      
di =    85.19386136   cm   [IT IS REAL, PART A and B]
      
Thus, the magnification, m = -di/do,      
      
m =    -1.887927504  
      
Thus, as      
hi =    -1.12E+01   mm
      
And hf = m * hi, then      
      
hf =    21.20746723   mm   [ANSWER, PART C]

IT IS UPRIGHT AS hf > 0.   [ANSWER, PART D]