Question

A 3 cm. "tall" object is placed 25 cm. from a convex lens with a focal length of 50 cm. Determine at what distance a focused image will appear. Determine the height of this image. Create a sketch (ray diagram) showing the scenario described above labeled as “A” and a scenario where everything is the same except the focal length of the lens is changed to 12.5 cm.

Answer #1

Thin lens formula ( real positive sign convention)

Focal length of lens is

Magnification

Height of the image is

Image is virtual and upright.

Size of image is double that of object.

-------------------------------

Focal length of lens is , object distance

( positive sign of v indicates real image)

Magnification

Height of the image is

(negative sign of hi indicates inverted image) and size of image is same as that of object.

Consider a convex lens with a focal length of 5.00 cm. An object
is located at 12.50 cm. The object is 5.50 cm tall. Draw the ray
diagram to scale. Determine the image distance, the height of the
image, the magnification and the characteristices of the image.

An object is placed 4 cm away from a concave lens of focal
length 3 cm. A convex lens of focal length 5 cm is placed 4 cm away
from the concave lens in the same direction (so the distance
between convex lens and the object is 8 cm). Use any method to
figure out if the final image is real/virtual, upright/inverted and
find its distance from the convex lens.
Explain in thorough detail.
draw a proper ray diagram for...

The focal length of a convex lens is 15 cm. An object 30 cm tall
is placed 60 cm from the lens. Which--if any--of the following
image attributes are correct?
A. upright, virtual, 20 cm tall
B. inverted, real, 10 cm tall
C. upright, real, 12 cm tall
D. inverted, virtual, 16 cm tall
E. None of these

A 2.00-cm-tall object is located 18.0 cm in front of a
converging lens with a focal length of 30.0 cm. (a) Use the lens
equation and (b) a ray diagram to describe the type, location, and
height of the image that is formed.

A 3.0-cm-tall object is placed 45.0 cm from a diverging lens
having a focal length of magnitude 20.0 cm.
a) What is the distance between the image and the lens?
b) Is the image real or virtual?
c) What is the height of the image?

An object is placed in front of a converging lens (convex lens)
with a radius of curvature of 4.88 cm and focal length of 2.44 cm.
An image is formed the object. (a) Calculate how far is the lens
from the object if the image is real. (b) How far is image from the
lens? (c) is the image upright or inverted, and is the image
magnified or diminished? (d) Draw the ray diagram to confirm your
result prediction.

An object 6 cm high is placed 12 cm in front of a 4 cm focal
length convex lens.
(a) Draw a ray diagram to produce an image.
(b) Calculate the image location and magnification using the
lens equation.
(c) How close is your image position and height to your
calculated value (% difference)?

A convex lens with a focal length of magnitude 4.0 cm and a
concave lens with a focal length of magnitude 10 cm are separated
by a distance of 18 cm as shown in the diagram below. The convex
lens is on the left, and the concave lens is on the right. An
object is then placed 6.0 cm to the left of the convex lens.
Locate the final image. State the distance of the final image
to the right...

A 1.0 cm tall object is 2.0 cm in front of a converging lens
with a focal length of 3.0 cm.
A. Determine the image position and height by using ray tracing
to find image
B. Calculate the image position and height

A 3.00-cm tall object is placed 18.0 cm from the center of a
converging lens with a 24.0 cm center of curvature.
a. Determine a scale, then use graph paper and color pencils or
pens to neatly draw a complete
ray diagram to scale.
b. Based on the scale of your ray diagram determine the
image distance, height, and magnification.
c. Based on your drawing state whether the image is
real or virtual, upright, or inverted.
d. Use the mirror...

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