Question

PRINIENVENSION CES Two-lens systems. In the figure, 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 baxed region closer to O, which is at object distance P. Lens 2 is mounted within the farther boxed region, at distance d. Each problem in the table refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance for the image produced by lens 2 (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 or virtual, (d) inverted from object O or noninverted, and (e) on the same side of lens 2 as object O or on the opposite side. BACK NEXT (а) (b) (c) (d) (e) Ps Lens 1 d Lens 2 M R/V /NI Side +6.0 C, 10 14 D. -10 (a) Number Units dy (b) Number Units (c) (d) Vers-on 4.24.17

Answer 1

For lens-1 :-

we have lens equation : (1/v) - (1/u) = 1/f

where v = lens-to-image distance,

u = lens-to-object distance = -6 cm ( sign convention applied )

f = focal length = 10 cm ( converging lens, +ve )

hence we have, (1/v) + (1/6) = 1/10 or    v = -15

First image is formed left side of lens-1 at a distance 15 cm from lens-1.

For lens-2 :-

we have lens equation : (1/v) - (1/u) = 1/f

where v = lens-to-image distance,

u = lens-to-object distance = 14+15 cm = 29 cm ( sign convention applied for virtual object)

f = focal length = -10 cm ( diverging lens, -ve )

hence we have, (1/v) - (1/29) = -(1/10) or    v = -290/19 = - 15.26

Second image is formed left side of lens-2 at a distance -15.26 cm from lens-2.

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(a) image distance = 15.26 cm from lens-2 at left side of lens-2

or 1.26 cm from lens-1 at left side of lens-2

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(b) magnification m1 for lens-1 :- v/u = -15/-6 = 2.5

magnification m2 for lens-2 :- v/u = -15.26/ 29 = -0.53

overall magnification :- m1 $\times$ m2 = 2.5 $\times$ (-0.53) = - 1.325

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(c) final image is virtual

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(d) inverted

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(e) same side of lens-2 as object O