Homework Answers
here
for a lens
when the magnification is greater than 1
the image is enlarged
when positive magnification , the image is upright
and
when negative magnification , the image is inverted
so, the correct option is THE IMAGE FORMED BY THE LENS WILL BE UPRIGHT
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Dealing with a lens and a real object, a positive magnification factor of the lens means...
In a converging (convex) lens, positive magnification m means the image is virtual, erect and lies...
In a converging (convex) lens, positive magnification m means the image is virtual, erect and lies in the same side of the object. positive magnification m means the image is real, inverted, and on the same side of the object. negative magnification m means the image is virtual, erect, and on the opposite side of the object.
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...
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
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,...
An object is a distance of 6 64f from a converging lens, where is the lens's...
An object is a distance of 6 64f from a converging lens, where is the lens's focal length. (Include the sign of the value in your answers) (a) What is the location of the image formed by the lens? d_i = f (b) Is the image real or virtual real virtual (c) What is the magnification of the image? (d) Is the image upright or inverted? upright inverted
An object is placed 15.0 cm to the left of a convex (converging) lens of focal...
An object is placed 15.0 cm to the left of a convex (converging) lens of focal length 20.0 cm. The image of this object is located (Figure out if the image is real or virtual, it will help to locate the image] O 60.0 cm to the left of the lens. O 60.0 cm to the right of the lens. O 8.57 cm to the right of the lens. O 8.57 cm to the left of the lens. Question 8...
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...
11.87 A 1.00-cm-high object is placed 4.85 cm to the left of a converging lens of...
11.87 A 1.00-cm-high object is placed 4.85 cm to the left of a converging lens of focal length 8.20 cm. A diverging lens of focal length - 16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position Take the image formed by the first lens to be the object for the second lens and apply the lens equation to each lens to locate the final image. cm 8.442...
An object is a distance 16.9cm to the left of a converging lens ("Lens 1") with...
An object is a distance 16.9cm to the left of a converging lens ("Lens 1") with a focal length of 20cm. Lens 1 is itself a distance 30cm to the left of a second converging lens ("Lens 2") with a focal length 20cm. The object is along the optical axis of the two-lens system. What is the magnification of the image due to Lens 1? If the image is inverted your answer should be negative, and if it is upright...
A converging lens has a focal length of 46.0 cm. If an object is at a...
A converging lens has a focal length of 46.0 cm. If an object is at a distance of 13.1 cm from the lens, determine the location d of the image, the magnification m, and the type of image, if it exists. d = -18.4 cm, m = 1.4, virtual and upright d = +18.4 cm, m = -1.4, real and inverted image does not exist d = 56.2 cm, m = -0.222, real and inverted d = -23.0 cm, m...
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...
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
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,...
An object is a distance of 6 64f from a converging lens, where is the lens's focal length. (Include the sign of the value in your answers) (a) What is the location of the image formed by the lens? d_i = f (b) Is the image real or virtual real virtual (c) What is the magnification of the image? (d) Is the image upright or inverted? upright inverted
An object is placed 15.0 cm to the left of a convex (converging) lens of focal length 20.0 cm. The image of this object is located (Figure out if the image is real or virtual, it will help to locate the image] O 60.0 cm to the left of the lens. O 60.0 cm to the right of the lens. O 8.57 cm to the right of the lens. O 8.57 cm to the left of the lens. Question 8...
11.87 A 1.00-cm-high object is placed 4.85 cm to the left of a converging lens of focal length 8.20 cm. A diverging lens of focal length - 16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position Take the image formed by the first lens to be the object for the second lens and apply the lens equation to each lens to locate the final image. cm 8.442...
An object is a distance 16.9cm to the left of a converging lens ("Lens 1") with a focal length of 20cm. Lens 1 is itself a distance 30cm to the left of a second converging lens ("Lens 2") with a focal length 20cm. The object is along the optical axis of the two-lens system. What is the magnification of the image due to Lens 1? If the image is inverted your answer should be negative, and if it is upright...
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