A converging lens that for which f = 55mm forms an image 4 times larger than...
A converging lens that for which f = 55mm forms an image 4 times larger than the object. How far is the object if the image is virtual?
Homework Answers
The general formula for a converging (convex) lens is
------------------------------ (1)
Where
is
the object distance
is
the image distance and
is
the focal length.
Given 
Here the image is virtual. This is because the object distance is smaller than the focal length.
It is given that the linear magnification 
.
That is
Substituting all these in equation (1), we will get
That is
Or
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A converging lens that for which f = 55mm forms an image 4 times
larger than...
A converging lens for which f = 50 mm forms an image that is three times...
A converging lens for which f = 50 mm forms an image that is three times larger than the object. How far from the lens is the object?
Problem 33.117 A converging lens for which f = 65 mm forms an image that is...
Problem 33.117 A converging lens for which f = 65 mm forms an image that is three times larger than the object. Part A How far from the lens is the object if the image is real? Express your answer with the appropriate units. Part B How far from the lens is the object if the image is virtual? Express your answer with the appropriate units.
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...
A converging lens with a focal length of 12.0cm forms a virtual image 8.00mm tall. Image...
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A converging lens with a focal length of 87.0 cmcm forms an image of a 3.35...
A converging lens with a focal length of 87.0 cmcm forms an image of a 3.35 cmcm -tall real object that is to the left of the lens. The image is 4.75 cmcm tall and inverted. Part A Where are the object and image located in relation to the lens? Express your answers in centimeters separated by a comma. |s||s|, |s′||s′| = nothing cmcm SubmitRequest Answer Part B Is the image real or virtual? Is the image real or virtual?...
1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm...
1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm tall, 11 2emto right of the lens. a. Determine the position of the object. b. Determine the size of the object. Is the image upright or inverted? Are the object and image on the same side or opposite sides of the lens? c. d. 2 You want to use a lens with a focal length of magnitude 36cm with the image twice as long...
A small object is placed 25.0 cm from a converging lens of focal length 40.0 cm....
A small object is placed 25.0 cm from a converging lens of focal length 40.0 cm. The object is to the left of the lens. Where is the image? A- 17cm to the right of the lens B- 25cm to the left of the lens C- 17 cm to the left of the lens D-25 cm to the right of the lens E-67cm to the right of the lens F-67 cm to the left of the lens A small object...
A converging lens with a focal length of 12.2 cm forms a virtual image 8.10 mm...
A converging lens with a focal length of 12.2 cm forms a virtual image 8.10 mm tall, 18.0 cm to the right of the lens. Determine the position of the object. Determine the size of the object.
An object is 150mm to the left of a converging lens with a focal length f=...
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A converging lens for which f = 50 mm forms an image that is three times larger than the object. How far from the lens is the object?
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
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1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm tall, 11 2emto right of the lens. a. Determine the position of the object. b. Determine the size of the object. Is the image upright or inverted? Are the object and image on the same side or opposite sides of the lens? c. d. 2 You want to use a lens with a focal length of magnitude 36cm with the image twice as long...
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An object is 150mm to the left of a converging lens with a focal length f= 200mm. How far is the image from the lens? Answer: mm Which side of the lens is the image (right or left)? Answer: Is the image real or virtual? Answer: What is the magnification? Answer: Is the image upright or inverted? Answer:
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