An object of height 2.9 cm is placed at 26 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 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 is d(i1) = . (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.) Where is the final image formed by the second converging lens? Image distance from second lens is d(i2) = (Use the sign to indicate which side the image is on.) How large is the intermediate image formed by the first diverging lens? Intermediate image height is h(i1) = (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) How large is the final image formed by the second converging lens? Final image height is h(i2) . (Use the sign to indicate whether the image is upright or inverted.)

University Physics Volume 3
17th Edition
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
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Chapter2: Geometric Optics And Image Formation
Section: Chapter Questions
Problem 13CQ: Use a ruler and a protractor to find the image by refraction in the following cases. Assume an...
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An object of height 2.9 cm is placed at 26 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 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 is d(i1) =
. (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.)
Where is the final image formed by the second converging lens? Image distance from second lens is d(i2) =
(Use the sign to
indicate which side the image is on.)
How large is the intermediate image formed by the first diverging lens? Intermediate image height is h(i1) =
(Use the sign to
indicate whether the image is upright (positive) or inverted (negative).)
How large is the final image formed by the second converging lens? Final image height is h(i2)
. (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 26 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 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 is d(i1) = . (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.) Where is the final image formed by the second converging lens? Image distance from second lens is d(i2) = (Use the sign to indicate which side the image is on.) How large is the intermediate image formed by the first diverging lens? Intermediate image height is h(i1) = (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) How large is the final image formed by the second converging lens? Final image height is h(i2) . (Use the sign to indicate whether the image is upright or inverted.)
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