Question

In a projector, an object (an image on an LCD screen) is 5.1 cm from a lens with focal length 5 cm. Draw the ray diagram for this situation and calculate the distance to the image. What is the magnification? Is the image real or virtual?

Answer #1

The lens formula gives us

So, the image distance is 255 cm.

Magnification is given by

The negative sign signifies the **inverted**
image.

As the image is another side of the lens (than the object), the
image is **real**.

The Ray Diagram:

An object is placed 3 cm from a concave lens of focal length 4
cm. (a) Draw a ray diagram for the situation. (b) Calculate the
size and position of the image. (c) If the image is real or virtual
and explain from part b above how you know this.

1. A converging lens has a focal length of 14 cm. An object is
18 cm to the left of the lens.
Find the image position
The magnification
Real or virtual?
Erect of inverted?
e. Draw the ray diagram using principal rays and a ruler.
f. If the object was located 7 cm to the left of the lens, draw
the ray diagram.

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...

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...

An object is 17.5 cm to the left of a lens of focal length 8.5
cm. A second lens of focal length -30 cm is 5 cm to the right of
the first lens.
Find the distance between the object and final image formed by
the second lens.
What is the overall magnification?
Is the final image real or virtual? Upright or inverted?

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__________

An object with a height of 8 cm is placed 10 cm in front of a
converging lens with a focal lenth of 30 cm.
a) Is the resulting image real or virtual?
b) Draw a ray tracing diagram including all three principle
rays.
c) Using the thin-lens equation and the definition of
magnification, determine both the image distance and the height of
the image. Express your anser in cm, and include any algebraic
signs. S’= ____________________________ H’ =
____________________________

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.

3) An object is 17.5 cm to the left of a lens of focal length
8.5 cm. A second lens of focal length -30 cm is 5 cm to the right
of the first lens. a) Find the distance between the object and
final image formed by the second lens. b) What is the overall
magnification? c) Is the final image real or virtual? Upright or
inverted?

Principal Ray Diagrams and Equations
1.) An object is 6 cm in front of a convex mirror with a focal
length of 10 cm.
a.) Use ray tracing alone to determine the location and
magnification of the image. Is the image upright or inverted? Is it
real or virtual?
b.) Use equations alone to determine the location and
magnification of the image. Is the image upright or inverted? Is it
real or virtual?
2.) A 2.0-cm-tall object is 20 cm...

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