4 An object located on the principle axis of a diverging lens. Use ray diagram to...
4 An object located on the principle axis of a diverging lens. Use ray diagram to determine the following properties of the image: (a) real or virtual, (b) upright or inverted, (c) magnified or shrunk, (d) on the same side or other side of the object?
Homework Answers
Here is the ray diagram. I have use random values for focal length and object distance. Note the focal length is negative for diverging lens
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As we can see that, the image is on the same side as the object, image is upright , shrunk (smaller than object) , virtual.
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4 An object located on the principle axis of a diverging lens.
Use ray diagram to...
An object located on the principle axis of a converging lens and inside of focal point. Use ray diagram to determine the...
An object located on the principle axis of a converging lens and inside of focal point. Use ray diagram to determine the following properties of the image: (a) real or virtual, (b) upright or inverted, (c) magnified or shrunk, (d) on the same side or other side of the object?
A 20 cm tall object is located 70 cm away from a diverging lens that has...
A 20 cm tall object is located 70 cm away from a diverging lens
that has a focal length of 20 cm. Use a scaled ray tracing to
answer parts a-d.
a. Is the image real or virtual?
b. Is the image upright or inverted?
c. How far from the lens is the image?
d. What is the height of the image?
e. Now use the thin lens equation to calculate the image
distance and the magnification equation to determine...
A 4-cm tall object is placed 59.2 cm from a diverging lens having a focal length...
A 4-cm tall object is placed 59.2 cm from a diverging lens having a focal length of -29.5 cm. a) Is the image produced by this lens virtual or real? b) Is the image inverted or upright? c) Is the image on the same side of the lens as the object or on the opposite side as the object? d) Where is the image located? (Please provide the magnitude of the position, no negative numbers) e) How tall is the...
11. An object is 20 cm in front of a diverging lens with a focal length...
11. An object is 20 cm in front of a diverging lens with a focal length of 10 cm. a. Use ray tracing to determine the location of the image. b. Is the image upright or inverted? c. Is it real or virtual
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a fo...
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. (As viewed from the side, from left to right, the sequence is object - diverging lens - converging lens - observer. Rays then travel from left to right through the system.) (a) Use ray tracing to draw image 1 and image...
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 diverging lens located in the y-z plane at x = 0 forms an image of...
A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -14.1 cm. The image of the tip of the arrow is located at y = y2 = 6.3 cm. The magnitude of the focal length of the diverging lens is 28.8 cm. light image х 1 Ay 3) A converging lens of focal length fconverging = 9.02 cm is now inserted at x = x3 = -14.36...
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 lens focal length f=+35cm is located on an optical bench. An object is located 50.0cm...
A lens focal length f=+35cm is located on an optical bench. An object is located 50.0cm in front of the lens. Determine the image position for this system. State whether the image is real or virtual. Upright or inverted. Magnified or demagnified.
A 20 cm tall object is located 70 cm away from a diverging lens
that has a focal length of 20 cm. Use a scaled ray tracing to
answer parts a-d.
a. Is the image real or virtual?
b. Is the image upright or inverted?
c. How far from the lens is the image?
d. What is the height of the image?
e. Now use the thin lens equation to calculate the image
distance and the magnification equation to determine...
11. An object is 20 cm in front of a diverging lens with a focal length of 10 cm. a. Use ray tracing to determine the location of the image. b. Is the image upright or inverted? c. Is it real or virtual
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. (As viewed from the side, from left to right, the sequence is object - diverging lens - converging lens - observer. Rays then travel from left to right through the system.) (a) Use ray tracing to draw image 1 and image...
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 diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -14.1 cm. The image of the tip of the arrow is located at y = y2 = 6.3 cm. The magnitude of the focal length of the diverging lens is 28.8 cm. light image х 1 Ay 3) A converging lens of focal length fconverging = 9.02 cm is now inserted at x = x3 = -14.36...
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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