Question

A diverging lens, f= 20, is placed 10cm to the left of a converging lens, f=20 cm. The final image of an object placed to the left of the diverging lens is real, and 70 cm away from the diverging lens. Draw a complete ray diagram to scale showing four rays leaving the object and arriving at the final image

Answer #1

One diverging and converging lenses are separated by 12cm. The
first lens on the left is divergent with focal point of 25cm while
the second lens on the right is f=10cm. The object is real and
located 10cm in front of the divergent lens. a) Find the position
of the image due to the first lens, b) Find the final position of
the image, c) calculate the total magnification.

A 1.8 cm tall object is placed 5 cm to the left of a converging
lens (lens #1) with a focal length of 1.3 cm. To the right of this
converging lens is a diverging lens (lens #2) that has a focal
length of 2.6 cm. The diverging lens is placed 22.4 cm from the
converging lens. Where is the final image, is it real or virtual,
and is it upright or inverted?

An
object is placed 30 cm to the left of a converging lens of focal
length 10 cm. A diverging lens of focal length -15 cm is 20 cm to
the right of the converging lens.
a) Where is the image created by the first lens relative to
the first lens?
b) Is it real or virtual?
c) Where is the final image relative to the second lens?
d) Is it real or virtual?
e) What is the final lateral...

A converging lens with a focal length 16cm is placed 30cm to the
left of a diverging lens with focal length 10cm. A small object
that is 2.2cm tall is placed 40cm to the left of the converging
lens. Where is the image formed by the diverging lens? Give your
answer as a distance from the diverging lens.

a converging lens (f=40cm) and a diverging lens (f=40cm) are
placed 140cm apart, and an object is placed 80cm in front of the
converging lens. Objects height is 2cm. Find the LOCATION and
distance of final image?

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.

. Diverging lens 1 (focal length f1 = −20.0 cm) is placed a
distance of d = 45.0 cm to the left of diverging lens 2 (focal
length f2 = −15.0 cm). An object is placed p1 = 25.0 cm to the left
of diverging lens 1 and has a height of h = 16.0 cm.
(a) Where is the image formed by diverging lens 1 (i.e. what
is q1)? What is this image’s height and is it upright or...

A converging lens (f1 = 24.0 cm) is located
56.0 cm to the left of a diverging lens (f2 =
−28.0 cm). An object is placed to the left of the converging lens,
and the final image produced by the two-lens combination lies 22.1
cm to the left of the diverging lens. How far is the object from
the converging lens?

A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left
of a diverging lens (f2 = -28.0 cm). An object is placed to the
left of the converging lens, and the final image produced by the
two-lens combination lies 21.1 cm to the left of the diverging
lens. How far is the object from the converging lens?

A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left
of a diverging lens (f2 = -28.0 cm). An object is placed to the
left of the converging lens, and the final image produced by the
two-lens combination lies 19.4 cm to the left of the diverging
lens. How far is the object from the converging lens?

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