p 1 Lens 1 f 1 d Lens 2 f 2 i 2 m 12.4 converging 8.0 30.9 diverging 8.0 1 2 Des

Question

p1 Lens 1 f1 d Lens 2 f2 i2 m 12.4 converging 8.0 30.9 diverging 8.0 1      2          Describe the image. (Select all that apply.)
3 same sideuprightinvertedrealopposite sideimaginary   I cant seam to get this even if its same side or inverted or opposite.... PLEASE HELP!           3 same sideuprightinvertedrealopposite sideimaginary   I cant seam to get this even if its same side or inverted or opposite.... PLEASE HELP!           3 same sideuprightinvertedrealopposite sideimaginary   I cant seam to get this even if its same side or inverted or opposite.... PLEASE HELP!           p1 Lens 1 f1 d Lens 2 f2 i2 m 12.4 converging 8.0 30.9 diverging 8.0 1      2      The figure below, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p1. Lens 2 is mounted within the farther boxed region, at distance d. The combination of lenses and value for distance are given in centimeters.

Explanation / Answer

f1 = 8 cm
f2 = -8 cm

distance between two lenses, d = 30.9 cm

image distance for the second lense, v2 = -10 cm

we know,

1/p1 + 1/i1 = 1/f1

1/i1 = 1/f1 - 1/p1

1/i1 = 1/8 - 1/12.4

==> i1 = 22.55 cm

oject distance for the second lense, p2 = 30.9 - 22.55 = 8.35 cm



1/p2 + 1/i2 = 1/f2

1/i2 = 1/f2 - 1/p2

1/i2 = -1/8 - 1/8.35

==>i2 = -4.086 cm

m = -i2/p1 = -(-4.086/12.4) = 0.329

inverted and imaginary

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