Diverging & Converging LENS
The same object (height = y) is placed at several different distances s to the left of the same lens (focal length = f). For each case, draw the 3 principal rays to locate the image. Then use the “thin lens” equations to calculate:
• the image height y' (in terms of y)
• whether the image is real or virtual
• the image distance from the lens s' (in terms of s and f)
• whether the image is upright or inverted
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Diverging & Converging LENS
The same object (height = y) is placed at several different
distances...
please provide explaination because i have hard time understand this problem The same object (height =...
please provide explaination because i have hard time understand
this problem
The same object (height = y) is placed at several different distances s to the left of the same lens (focal length = f). For each case, draw the 3 principal rays to locate the image. Then use the "thin lens" equations to calculate: • the image height y' (in terms of y) • the image distance from the lens s' (in terms of s and f) • whether...
2. Two thin lenses, one a converging lens and the other a diverging lens, are arated...
2. Two thin lenses, one a converging lens and the other a diverging lens, are arated by 1.00 m along the same principal axis, as shown in the figure. The magnitude of the focal length of the converging lens is 25 cm, while the magnitude of the focal length of the diverging lens is 40 em. An object 8,25 cm tall is placed 35 cm to the left of the converging lens. (a) Where is the final image produced by...
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 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
A 1.00-cm-high object is placed 4.85 cm to the left of a converging lens of focal...
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 cm height cm Is the image inverted or upright? O upright inverted Is the image real or virtual? Oreal 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 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...
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...
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 of height 3.6 cm is placed at 24 cm in front of a diverging...
An object of height 3.6 cm is placed at 24 cm in front of a diverging lens of focal length, f = -18 cm. Behind the diverging lens, there is a converging lens of focal length, f = 18 cm. The distance between the lenses is 5 cm. In the next few steps, you will find the location and size of the final image. Where is the intermediate image formed by the first diverging lens? Image distance from first lens...
please provide explaination because i have hard time understand
this problem
The same object (height = y) is placed at several different distances s to the left of the same lens (focal length = f). For each case, draw the 3 principal rays to locate the image. Then use the "thin lens" equations to calculate: • the image height y' (in terms of y) • the image distance from the lens s' (in terms of s and f) • whether...
2. Two thin lenses, one a converging lens and the other a diverging lens, are arated by 1.00 m along the same principal axis, as shown in the figure. The magnitude of the focal length of the converging lens is 25 cm, while the magnitude of the focal length of the diverging lens is 40 em. An object 8,25 cm tall is placed 35 cm to the left of the converging lens. (a) Where is the final image produced by...
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 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
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 cm height cm Is the image inverted or upright? O upright inverted Is the image real or virtual? Oreal 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 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...
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 of height 3.6 cm is placed at 24 cm in front of a diverging lens of focal length, f = -18 cm. Behind the diverging lens, there is a converging lens of focal length, f = 18 cm. The distance between the lenses is 5 cm. In the next few steps, you will find the location and size of the final image. Where is the intermediate image formed by the first diverging lens? Image distance from first lens...
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