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An object of height 2.9 cm is placed at 22 cm in front of a diverging lens of focal length, f = -17 cm. Behind the diverging lens, there is a converging lens of focal length, f = 17 cm. The distance between the lenses is 3 cm. In the next few steps, you will find the location and size of the final image. Hint a. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is di= cm. (Use the sign to indicate which side the image is on; positive sign means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) b. Where is the final image formed by the second converging lens? Image distance from second lens is di₂ = cm. (Use the sign to indicate which side the image is on.) c. How large is the intermediate image formed by the first diverging lens? Intermediate image height is h₂ = cm. (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) d. How large is the final image formed by the second converging lens? Final image height is h₂2 = cm. (Use the sign to indicate whether the image is upright or inverted.)

An object of height 2.9 cm is placed at 22 cm in front of a diverging lens of focal length, f = -17 cm. Behind the diverging lens, there is a converging lens of focal length, f = 17 cm. The distance between the lenses is 3 cm. In the next few steps, you will find the location and size of the final image. Hint a. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is di= cm. (Use the sign to indicate which side the image is on; positive sign means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) b. Where is the final image formed by the second converging lens? Image distance from second lens is di₂ = cm. (Use the sign to indicate which side the image is on.) c. How large is the intermediate image formed by the first diverging lens? Intermediate image height is h₂ = cm. (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) d. How large is the final image formed by the second converging lens? Final image height is h₂2 = cm. (Use the sign to indicate whether the image is upright or inverted.)

College Physics
College Physics
10th Edition
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter25: Optical Instruments
Section: Chapter Questions
Problem 10P: A patient cant see objects closer than 40.0 cm and wishes to clearly see objects that are 20.0 cm...
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An object of height 2.9 cm is placed at 22 cm in front of a diverging lens of focal length, f = -17 cm. Behind the diverging lens, there is a converging lens of focal length, f = 17 cm. The distance between the lenses is 3 cm. In the next few steps, you will find the location and size of the final image. Hint a. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is di= cm. (Use the sign to indicate which side the image is on; positive sign means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) b. Where is the final image formed by the second converging lens? Image distance from second lens is di₂ = cm. (Use the sign to indicate which side the image is on.) c. How large is the intermediate image formed by the first diverging lens? Intermediate image height is h₂ = cm. (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) d. How large is the final image formed by the second converging lens? Final image height is h₂2 = cm. (Use the sign to indicate whether the image is upright or inverted.)
Transcribed Image Text:An object of height 2.9 cm is placed at 22 cm in front of a diverging lens of focal length, f = -17 cm. Behind the diverging lens, there is a converging lens of focal length, f = 17 cm. The distance between the lenses is 3 cm. In the next few steps, you will find the location and size of the final image. Hint a. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is di= cm. (Use the sign to indicate which side the image is on; positive sign means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) b. Where is the final image formed by the second converging lens? Image distance from second lens is di₂ = cm. (Use the sign to indicate which side the image is on.) c. How large is the intermediate image formed by the first diverging lens? Intermediate image height is h₂ = cm. (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) d. How large is the final image formed by the second converging lens? Final image height is h₂2 = cm. (Use the sign to indicate whether the image is upright or inverted.)
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