A converging lens makes an image of an object located a finite distance away. Which of...
A converging lens makes an image of an object located a finite distance away. Which of the following statements is true? A. The image is virtual B. The image is real C. the object distance is positive D. the object distance is negative E. the image can be either real or virtual
Homework Answers
The following statements is true
B. The image is real
D. the object distance is negative
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A converging lens makes an image of an object located a finite
distance away. Which of...
An object is placed to the right of a converging lens and an image is formed...
An object is placed to the right of a converging lens and an image is formed to the left of the lens. lens object image 1) Following the standard sign convention for the lens equation 5 + 5 = }, the image distance is: positive negative Submit 2) The image is: real virtual Submit 3) For this lens, the sign of the focal length is: positive negative Submit Help 4) An object is placed to the left of a converging...
A converging lens makes an image of an object that is located outside of its focal...
A converging lens makes an image of an object that is located outside of its focal length. Which of the following statements is true? A. The image is smaller than the object. B. The image is upright. C.The image is inverted. D.The image is virutal
An object is on the left side of a thin converging lens. The object is located...
An object is on the left side of a thin converging lens. The object is located at a distance of 6 cm away from a thin converging lens with focal length of 2 cm. Use the thin lens equation (1/f = 1/s' + 1/s) to predict the following: (a) Location of the image? (b) Magnification of the image (including inverted versus non-inverted)? (c) Real or virtual? Draw diagram please!
(2) An object is located at a distance of 4 cm from a thin converging lens...
(2) An object is located at a distance of 4 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 3 cm from the converging lens and 7 cm from the object. The diverging lens has a focal length of -2 cm. Use the thin lens equation to predict the following (a) Location of the final image? to the object? (c) Is the final image real or virtual?
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 located at a distance of 6 cm from a thin converging lens with...
An object is located at a distance of 6 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 4 cm from the converging lens and 10 cm from the object. The diverging lens has a focal length of -3 cm. Note: To handle a multiple lens system, we treat them independently. We first find the image created by the first lens. We then use the image from the first lens to act...
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...
Question 18 (8 points) An object is placed 50.0cm in front of a lens. The image...
Question 18 (8 points) 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) upright, converging, negative, real inverted, converging, positive, virtual upright, converging, negative, virtual upright, converging, positive, virtual upright, converging, positive, real O inverted, converging, positive, real inverted, diverging, positive, real inverted, diverging,...
A converging lens with a focal length of 4.9 cm is located 20.9 cm to the...
A converging lens with a focal length of 4.9 cm is located 20.9 cm to the left of a diverging lens having a focal length of -11.0 cm. If an object is located 9.9 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. a) Where is the image located as measured from the diverging lens? b) What is the magnification? c) Also determine, with respect to the original object...
For a converging lens, if it is observed that the image is upright for an object...
For a converging lens, if it is observed that the image is upright for an object located at the left side of a lens. Answer the following questions by choosing the correct choice: 1). The image is (real or virtual) 2). The image is on the (left or right ) side of the lens. 3). The image is (enlarged or diminished).
An object is placed to the right of a converging lens and an image is formed to the left of the lens. lens object image 1) Following the standard sign convention for the lens equation 5 + 5 = }, the image distance is: positive negative Submit 2) The image is: real virtual Submit 3) For this lens, the sign of the focal length is: positive negative Submit Help 4) An object is placed to the left of a converging...
(2) An object is located at a distance of 4 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 3 cm from the converging lens and 7 cm from the object. The diverging lens has a focal length of -2 cm. Use the thin lens equation to predict the following (a) Location of the final image? to the object? (c) Is the final image real or virtual?
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...
Question 18 (8 points) 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) upright, converging, negative, real inverted, converging, positive, virtual upright, converging, negative, virtual upright, converging, positive, virtual upright, converging, positive, real O inverted, converging, positive, real inverted, diverging, positive, real inverted, diverging,...
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