An object is placed 10.0 cm in front of a lens. An upright, virtual image is...
An object is placed 10.0 cm in front of a lens. An upright, virtual image is formed 30.0 cm away from the lens. What is the focal length of the lens? Is the lens converging or diverging?
Homework Answers
The expression for the focal length of the lens is,
Here, P is the object distance and q is the image distance.
The object distance is 
The image distance is 
An image distance is taken as negative because the image is virtual.
From lens equation
Substitute 10 cm for p and -30 cm for q in the above equation.
Rewrite the above expression for f.
Hence, the focal length of the lens is 
.
The expression for the magnification of the image is,
Substitute 10 cm for p and -30 cm for q in the above equation.
As the image is magnified, upright and virtual the lens is convex lens.
Add Answer to:
An object is placed 10.0 cm in front of a lens. An upright, virtual image is...
7. An object is placed 10.0 cm in front of a lens. An upright, virtual image...
7. An object is placed 10.0 cm in front of a lens. An upright, virtual image is formed 30 cm away from the lens. (a) What is the focal length of the lens? (b) Find the magnification of the image.
An object is placed 13.5 cm in front of a lens. An upright, virtual image is formed 25.1 cm from the lens.
An object is placed 13.5 cm in front of a lens. An upright, virtual image is formed 25.1 cm from the lens.(a) What is the magnification?(b) What is the focal length of the lens?
A 2.0 cm tall object is placed 10.0 cm in front of converging lens of focal...
A 2.0 cm tall object is placed 10.0 cm in front of converging lens of focal length 8 cm. What can you say about the image formed by the lens? de the image is real, magnified and inverted the image is virtual, diminished and inverted O the image is virtual, magnified and upright the image is real, magnified and upright the image is real, diminished and inverted
1. A 5.0 cm high object is placed 10.0 cm from a converging lens of focal...
1. A 5.0 cm high object is placed 10.0 cm from a converging lens of focal length 25.0 cm. What is the height of the image formed by the lens? a. 1.67 cm b. 8.35 cm c. 5.38 cm d. 3.85 cm 2. A 2.0 cm tall object is 30.0 cm in front of a diverging lens of focal length 15.0 cm. What is the magnification produced by the lens? a. 0.23 b. 0.33 c. 1.0 d. 0.25
an object is placed 30.0 cm from a concave lens. If a virtual image appears 10.0...
an object is placed 30.0 cm from a concave lens. If a virtual image appears 10.0 cm from the lens on the same side as the object what is the focal length of the lens? explain
1.) An object is placed in front of a diverging lens with a focal length of...
1.) An object is placed in front of a diverging lens with a focal length of 17.7 cm. For each object distance, find the image distance and the magnification. Describe each image. (a) 35.4 cm location _____cm magnification _____ nature real virtual upright inverted (b) 17.7 cm location _____ cm magnification _____ nature real virtual upright inverted (c) 8.85 cm location _____ cm magnification _____ nature real virtual upright inverted 2.) An object is placed in front of a converging lens...
2) A diverging lens is placed 20. cm away from an object, 80. cm from the...
2) A diverging lens is placed 20. cm away from an object, 80. cm from the diverging lens is a converging lens whose focal length is 25 cm. 35. cm from the converging lens is a screen showing an image of the object. a) What is the focal length of the diverging lens? f = 25. cm o *20. cm → 80. cm + 35 cm → b) What type of image is it? Circle one Real or Virtual c)...
An object is placed 50.0cm in front of a lens. The image forms on the same...
An object is placed 50.0cm in front of a lens. The image forms on the same side of the lens and is larger than the object. The image is (upright or inverted), the lens is (converging/diverging), image distance is (positive/negative), the image is (real, virtual) O inverted, diverging, positive, real upright, converging, positive, virtual inverted, converging, positive, virtual upright, converging, positive, real upright, converging, negative, virtual upright, converging, negative, real inverted, converging, positive, real O inverted, diverging, negative, real
An object is placed 50.0cm in front of a lens. The image forms on the same...
An object is placed 50.0cm in front of a lens. The image forms on the same side of the lens and is larger than the object. The image is (upright or inverted), the lens is (converging/diverging), image distance is (positive/negative), the image is (real, virtual) O upright, converging, positive, virtual O inverted, converging, positive, real inverted, diverging, negative, real O upright, converging, negative, virtual O inverted, diverging, positive, real O inverted, converging, positive, virtual O upright, converging, positive, real O...
ASAP please. 3. An object is placed at 10 cm from a thin converging lens of...
ASAP please.
3. An object is placed at 10 cm from a thin converging lens of focal length f. Its image is formed 30 cm from the lens on the same side as the object. (20 points) (a) What is the focal length? (4 points) (b) What are the magnification of the image, the orientation of the image (upright/inveredy (2+2 (c what kind ofimage is formed (reduced/magnified, real/virtual, and uprigh?1nvertedy? (2+2 (d) If the lens is replaced by a thin...
7. An object is placed 10.0 cm in front of a lens. An upright, virtual image is formed 30 cm away from the lens. (a) What is the focal length of the lens? (b) Find the magnification of the image.
A 2.0 cm tall object is placed 10.0 cm in front of converging lens of focal length 8 cm. What can you say about the image formed by the lens? de the image is real, magnified and inverted the image is virtual, diminished and inverted O the image is virtual, magnified and upright the image is real, magnified and upright the image is real, diminished and inverted
2) A diverging lens is placed 20. cm away from an object, 80. cm from the diverging lens is a converging lens whose focal length is 25 cm. 35. cm from the converging lens is a screen showing an image of the object. a) What is the focal length of the diverging lens? f = 25. cm o *20. cm → 80. cm + 35 cm → b) What type of image is it? Circle one Real or Virtual c)...
An object is placed 50.0cm in front of a lens. The image forms on the same side of the lens and is larger than the object. The image is (upright or inverted), the lens is (converging/diverging), image distance is (positive/negative), the image is (real, virtual) O inverted, diverging, positive, real upright, converging, positive, virtual inverted, converging, positive, virtual upright, converging, positive, real upright, converging, negative, virtual upright, converging, negative, real inverted, converging, positive, real O inverted, diverging, negative, real
An object is placed 50.0cm in front of a lens. The image forms on the same side of the lens and is larger than the object. The image is (upright or inverted), the lens is (converging/diverging), image distance is (positive/negative), the image is (real, virtual) O upright, converging, positive, virtual O inverted, converging, positive, real inverted, diverging, negative, real O upright, converging, negative, virtual O inverted, diverging, positive, real O inverted, converging, positive, virtual O upright, converging, positive, real O...
ASAP please.
3. An object is placed at 10 cm from a thin converging lens of focal length f. Its image is formed 30 cm from the lens on the same side as the object. (20 points) (a) What is the focal length? (4 points) (b) What are the magnification of the image, the orientation of the image (upright/inveredy (2+2 (c what kind ofimage is formed (reduced/magnified, real/virtual, and uprigh?1nvertedy? (2+2 (d) If the lens is replaced by a thin...
Most questions answered within 3 hours.
-
exercise on VSEPR and molecular structrue.
octahedral
SeCl62-
TeCl62-
ClF62-
distorted
SeF62–
IF6–
asked 35 minutes ago -
A regression equation that describes the relationship between
the amount of the bill ($) at a...
asked 6 minutes ago -
284 mL of a 0.52 M potassium hydroxide solution is added to 467
mL of a...
asked 34 minutes ago -
Little’s Law: Val d’Costa is a world famous ski village in the
French Alps. Because of...
asked 1 hour ago -
Find the absolute error D for the calculation if A + B/C=D A=
9.4 +/- 0.4...
asked 1 hour ago -
New Air Heating and Cooling, manufactures furnaces and central
air units. The company pride itself on...
asked 1 hour ago -
A coach uses a new technique to train gymnasts. Seven
gymnasts were randomly selected and their...
asked 3 hours ago -
While rotating the tires on your car you notice a rock [mass =
0.1 Kg] stuck...
asked 5 hours ago -
Using MARS simulator, write MIPS programs according to
the following scenarios: Receive a positive integer number...
asked 7 hours ago -
An object in front of a concave mirror has a real image that is
11.5 cm...
asked 7 hours ago -
Consider the reaction, C3 H8 + O2 --> CO2 + H2O. How many
moles of O2...
asked 9 hours ago -
You and your opponent both roll a fair die. If you both roll the
same number,...
asked 9 hours ago