Two-lens systems. The stick figure O (the object) stands on the common central a
ID: 1519705 • Letter: T
Question
Two-lens systems. The stick figure O (the object) stands on the common central axis of two thin, symmetric lenses. Lens 1 is mounted within the boxed region closer to O, which is at object distance do 1. Lens 2 is mounted within the farther boxed region, at distance d.
The table refers to the focal distances f1 and f2 for lens 1 and lens 2 respectively, the image distance di 2 for the final image produced by lens 2 (the final image produced by the system) and the overall lateral magnification mtot for the system. All distances are in centimeters. Fill in the missing information, including signs.
* signs not given
Describe the image. (Select all that apply to receive credit.)
real
virtual
upright
inverted
on the same side of lens 2 as object O
on the opposite side of lens 2
do 1 Lens 1 f1 d Lens 2 f2 di 2 mtot +12 converging *7.5 32 converging *6.3 help here HELP HEREExplanation / Answer
. To analyze two-lens systems, we first ignore lens 2, and apply the standard procedure used for a single-lens system. The object distance p1, the image distance i1, and the focal length f1 are related by:
1/f1= 1/p1 + 1/i1
Next, we ignore the lens 1 but treat the image formed by lens 1 as the object for lens 2. The object distance p2 is the distance between lens 2 and the location of the first image. The location of the final image, i2, is obtained by solving
1/f2= 1/p2 + 1/i2
Since lens 1 is converging, f1 = +7.5 cm, and we find the image distance to be
i1=p1*f1/ (p1-f1)= 12*7.5/(12-7.5) = 20 cm
This serves as an “object” for lens 2 (which has f2 = +6.3 cm) with an object distance given by p2 = d – i1 = 32 cm- 20 cm= 12 cm. The negative sign means that the “object” is behind lens 2. Solving the lens equation, we obtain
i2=p2*f2/ (p2-f2)= 12*6.3/(12-6.3) = 13.26 cm
The overall magnification is M = m1m2 = (-i1/p1)(-i2/p2)= (-20/12)(- 13.26/10)= 2.21
The fact that the (final) image distance is a positive value means the image is real (R).
The fact that the magnification is a positive value means the image is upward (U).
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