Question

An object is 20.0 cm from a converging lens, and the image falls on a screen. When the object is moved 3.00 cm closer to the lens, the screen must be moved 3.10 cm farther away from the lens to register a sharp image. Determine the focal length of the lens.

Answer #1

An object is placed 53.5 cm from a screen. (a) Where should a
converging lens of focal length 9.0 cm be placed to form an image
on the screen?
shorter distance_________cm from the screen
farther distance___________cm from the screen
(b) Find the magnification of the lens.
magnification if placed at the shorter distance ________
magnification if placed at the farther distance__________

When an object is placed 70.0 cm from a certain converging lens,
it forms a real image. When the object is moved to 32.0 cm from the
lens, the image moves 9.00 cm farther from the lens. Find the focal
length of the lens.

Distance between object and screen is fixed and equal 120 cm.
The converging lens of focal lengths 35 cm is positioned between
object and screen in attempts to get clear image on the screen.
Based on the data given we may conclude that:
There are two positions of lens which will provide sharp image
on the screen.
There is one position of the lens which will provide sharp image
on the screen.
No position of the lens will provide sharp...

An object is placed 20.0 cm to the left of a converging lens
with focal length of 10.0cm. A second identical, lens is placed to
the right of the first lens and moved until the final image
produced is identical in size and orientation to the object. What
is the separation between the two lenses?

For a particular converging lens, an object a certain distance
to the left of the lens produces a clear image on a a screen a
distance 30.2 cm to the right of the lens. When a second lens is
placed 15.4 cm to the right of the first lens, the screen has to be
moved 18.7 cm to the right of its previous location to produce a
clear image.
Determine the focal length of the second lens.

A converging lens of focal length f is placed between an
object and a screen located a distance d to the right of the
object. Given that 4f < d, find the largest of two possible
locations in cm for the lens as measured from the object so that a
sharp image is formed on the screen. (Input your answer in 2
significant figures without unit)

An object 2.00 cm high is placed 31.1 cm to the left of a
converging lens having a focal length of 26.1 cm. A diverging lens
having a focal length of −20.0 cm is placed 110 cm to the right of
the converging lens. (Use the correct sign conventions for the
following answers.)
(a) Determine the final position and magnification of the final
image. (Give the final position as the image distance from the
second lens.)
Find
final position in...

A thin convex lens is used to form a real image of an object
placed between the focal point and
the lens. The object is moved closer to the lens. The new
image:
Is closer to the lens and larger.
Is farther away from the lens and smaller.
Is closer to the lens and smaller.
Is farther away from the lens and lager.
More information is needed to work out the answer.

A converging lens of focal length 7.97 cm is 20.0 cm to the left
of a diverging lens of focal length -6.00 cm. A coin is placed 11.7
cm to the left of the converging lens.
a) Calculate the location of the coin's final image.
b) Calculate the magnification of the coin's final image.

A converging lens of focal length 8.080 cm is 20.0 cm to the
left of a diverging lens of focal length -6.91 cm . A coin is
placed 12.1 cm to the left of the converging lens.
di = cm to the right of the diverging
lens
Find the magnification of the coin's final image.
m=

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