Question

‘When an object is 15.5 cm from a lens, an image is formed 11.0 cm from the lens on the same side as the object. What is the focal length of the lens? Is the lens converging or diverging? If the object is 8.00 mm tall, how tall is the image? Express your answer in millimeters to three significant figures.

Answer #1

A lens forms an image of an object. The object is 16cm from the
lens. The image is 12 cm from the lens on the same side as the
object. 1) what is the focal length of the lens 2) is the lens
converging or diverging 3)if the object is 8.50mm tall how tall is
the image 4) is it erect or inverted

A converging lens forms an image of an 8.50 mm -tall real
object. The image is 12.5 cm to the left of the lens, 3.50 cm tall,
and upright.
A) What is the focal length of the lens?
Express your answer in centimeters to three significant
figures.
B) Where is the object located?
Express your answer in centimeters to three significant
figures.

A converging lens with a focal length of 60 cm and a diverging
lens with a focal length of -70 cm are 310 cm apart. A 2.7-cm-tall
object is 80 cm in front of the converging lens.
Calculate the distance between the final image and the diverging
lens.
Express your answer to two significant figures and include the
appropriate units.
Calculate the image height.
Express your answer to two significant figures and include the
appropriate units.

A diverging lens with a focal length of -13 cm is placed 12 cm
to the right of a converging lens with a focal length of 22 cm . An
object is placed 34 cm to the left of the converging lens.
Where will the final image be located?
Express your answer using two significant figures.
d =
cm
to the left of the diverging lens
Where will the image be if the diverging lens is 58 cm from...

A converging lens with a focal length of 94.0 cm forms an image
of a 3.05 cm -tall real object that is to the left of the lens. The
image is 4.65 cm tall and inverted.
Where are the object and image located in relation to the
lens?
Express your answers using three significant figures separated by a
comma.

A diverging lens with a focal length of -13 cm is placed 12 cm
to the right of a converging lens with a focal length of 19 cm . An
object is placed 31 cm to the left of the converging lens.
A.Where will the image be if the diverging lens is 39 cm
from the converging lens?
Express your answer using two significant figures. di = ? cm
from the diverging lens

A converging lens with a focal length of 91.5 cm forms an image
of a 3.15 cm -tall real object that is to the left of the lens. The
image is 4.40 cm tall and inverted.
a) Where are the object and image located in relation to the
lens? Express your answers using three significant figures
separated by a comma.
b) Is the image real or virtual?

Two thin lenses with a focal length of magnitude 11.0 cm, the
first diverging and the second converging, are located 8.25 cm
apart. An object 1.60 mm tall is placed 18.3 cm to the left of the
first (diverging) lens. Part A How far from this
first lens is the final image formed? Part B Is
the final image real or virtual? Part C What is
the height of the final image? Part D Is it
upright or inverted?

A 1.5-cm-tall object is 90 cm in front of a converging lens that
has a 32 cm focal length.
Calculate the image position.
Calculate the image height.
Express your answer to two significant figures and
include the appropriate units.

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?

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 39 minutes ago

asked 43 minutes ago

asked 45 minutes ago

asked 49 minutes ago

asked 50 minutes ago

asked 56 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago