Question

A converging lens (f_{1} = 11.9 cm) is located 27.6 cm
to the left of a diverging lens (f_{2} = -5.94 cm). A
postage stamp is placed 34.8 cm to the left of the converging lens.
Find the overall magnification.

Answer #1

first find image distance for converging lens

f = 11.9 cm

u = 34.8 cm

1/f = 1/u + 1/v

v = f*u/(u - f) = 11.9*34.8/(34.8 - 11.9) = 18.08 cm

Now this image will work like object for diverging lens, So now

f1 = -5.94 cm

u1 = 27.6 - 18.08 = 9.52 cm

1/f1 = 1/u1 + 1/v1

v1 = f1*u1/(u1 - f1)

v1 = 9.52*(-5.94)/(9.52 - (-5.94))

v1 = -3.66 cm

Now overall magnification will be

Mnet = M*M1

Mnet = (-v/u)*(-v1/u1)

Mnet = (-18.08/34.8)*(-(-3.66)/9.52)

Mnet = -0.1997

**Mnet = -0.20**

A converging lens (f1 = 12.7 cm) is located 30.6 cm
to the left of a diverging lens (f2 = -6.48 cm). A
postage stamp is placed 33.9 cm to the left of the converging lens.
Locate the final image of the stamp relative to the diverging
lens.

A converging lens (f = 11.3 cm) is located 24.0 cm to
the left of a diverging lens (f = -5.42 cm). A postage
stamp is placed 46.0 cm to the left of the converging lens.
(a) Locate the final image of the stamp relative
to the diverging lens. (b) Find the overall
magnification.

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of a diverging lens (f = -4.59 cm). A postage stamp is placed 37.1
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magnification.

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magnification.

A converging lens (f1 = 24.0 cm) is located
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two-lens combination lies 20.0 cm to the left of the diverging
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How does the size of the image compare to the size of the
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