Three-lens systems. In the figure, stick figure O (the object) stands on the com

Question

Three-lens systems. In the figure, stick figure O (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to O, which is at object distance p1. Lens 2 is mounted within the middle boxed region, at distance d12 from lens 1. Lens 3 is mounted in the farther boxed region, at distance d23 from lens 2. For this problem, p1 = 7.1 cm, lens 1 is diverging, d12 = 8.5 cm, lens 2 is diverging, d23 = 6.5 cm, and lens 3 is converging. The distance between the lens and either focal point is 7.8 cm for lens 1, 17 cm for lens 2, and 8.3 cm for lens 3. (You need to provide the proper sign). Find (a) the image distance i3 for the (final) image produced by lens 3 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real (enter 1) or virtual (enter 0), (d) inverted (enter 1) from object O or noninverted (enter 0), and (e) on the same side of lens 3 as object O (enter 1) or on the opposite side (enter 0).

Explanation / Answer

lense equation in general

1/f = 1/di + 1/do

for lense 1

p1 = d0= 7.1 cm, lens 1 is diverging or we can say that concave

f= -7.8cm

1/(-7.8) = 1/7.1 + 1/di 1

vi1 = -3.7167cm

m1 = - (di/do) = 0.5234

now

do for 2nd lens = 3.7167 + d12 = 12.2167 cm

1/(-17) = 1/12.2167 + d i by 2nd lense

d i by 2nd lense = -7.10839 cm

m2 = -(-7.10839/12.2167) = 0.58185

now for lens 3 converging or convex means f is positive

do 3rd = 7.10839+ d23 =13.6084 cm

1/8.3 = 1/13.6084 + 1/di 3 final image

di 3 final image = 21.2775 cm from thrid lens towards right side

m3 = - 21.2775/13.6084 = -1.56355

moverall =m1*m2*m3 =0.5234*0.58185 *( -1.56355 ) = -0.476165

so fianl image is real so anwere of part c is 1

part d

inverted image because moverall is negative

part e

opposite side of lense

ans 0

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