Question

A converging lens (f = 13.3 cm) is located 33.0 cm to the left of a diverging lens (f = -5.88 cm). A postage stamp is placed 36.6 cm to the left of the converging lens.

(a) Locate the final image of the stamp relative to the diverging lens. (Include sign to indicate which side of the lens the image is on.)

(b) Find the overall magnification.

Answer #1

1/13.3 = 1/si + 1/36.6

si = 1/ ( 1/13.3 - 1/36.6) = 20.89

the magnification of this first image is

m_1 = - (36.6/20.98) = - 1.744

For the second lens, the image is at

-1/5.88 = 1/si + 1/(33 - 20.89) = 1/si + 1/12.11

si = 1/( -1/5.88 -1/12.11) = -3.958

So, the image is virtual and located -145 cm from the diverging
lens. The magnification of the image is

m_2 = - ( -3.958/12.11) = 0.3267

The overall magnification is then

m = m_1 m_2 = - 0.5701

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.

A converging lens (f = 14.2 cm) is located 23.2 cm to the left
of a diverging lens (f = -4.59 cm). A postage stamp is placed 37.1
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.

A converging lens (f = 11.0 cm) is located 23.6 cm to the left
of a diverging lens (f = -4.51 cm). A postage stamp is placed 52.7
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.

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 (f1 = 11.9 cm) is located 27.6 cm
to the left of a diverging lens (f2 = -5.94 cm). A
postage stamp is placed 34.8 cm to the left of the converging lens.
Find the overall magnification.

A converging lens (f1 = 22.0 cm) is located 43.0 cm
to the left of a diverging lens (f2 = -26.0 cm). An
object is placed 38.0 cm to the left of the converging lens.
Relative to the diverging lens, where is the final image
located?
Is it real or virtual?
How does the size of the image compare to the size of the
object?

A converging lens with focal length 10.0 cm is
located 28.0 cm to the left of a diverging lens
with focal length - 8.0 cm. An object is placed 36
cm to the left of the converging lens. Where is the final image
located relative to the diverging lens?
Select one:
a) at infinity
b) 13.9 cm to the right
c) 14.2 cm to the right
d) 5 cm to the left

An object is placed 33.0 cm to the left of a diverging lens with
a focal length of -20.0 cm. A converging lens of focal length of
33.0 cm is placed a distance d to the right of the diverging lens.
Find the distance d that the final image is at infinity. ______
cm

A converging lens is placed at
x = 0,
a distance d = 11.0 cm to the left of a diverging lens
as in the figure below (where FC and
FD locate the focal points for the converging
and the diverging lens, respectively). An object is located at
x =
−1.10
cm to the left of the converging lens and the focal lengths of
the converging and diverging lenses are
3.50 cm
and
−8.60 cm,
respectively.
HINT
(a)
Determine the...

A converging lens with a focal length of 4.8 cm is
located 24.8 cm to the left of a diverging lens having a
focal length of -15.0 cm. If an object is located 9.8
cm to the left of the converging lens, locate and describe
completely the final image formed by the diverging lens.
Where is the image located as measured from the diverging
lens?
What is the magnification?
Also determine, with respect to the original object whether the
image...

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