Question

An object of height 2.90 cm is placed 11.0 cm from a diverging lens of focallength magnitude 11.0 cm. Find the height of the image. 3. 145 b. -1.45 cm c. +3.22 cm d. +1.76 cm e. -2.82 cm
0 0
Add a comment Improve this question Transcribed image text
Answer #1

2 Lu 5s cm 0

Add a comment
Know the answer?
Add Answer to:
An object of height 2.90 cm is placed 11.0 cm from a diverging lens of focallength...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • . A diverging lens (f = –11.0 cm) is located 19.0 cm to the left of...

    . A diverging lens (f = –11.0 cm) is located 19.0 cm to the left of a converging lens (f = 30.0 cm). A 2.90-cm-tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

  • An object of height 2.8 cm is placed 27 cm in front of a diverging lens of focal length 16 cm

    An object of height 2.8 cm is placed 27 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length.Part (a) Find the location of the final image, in centimeters beyond the converging lens. Part (b) What is the magnification of the final image? Include its sign to indicate its orientation with respect to the object.

  • 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...

  • 2) A diverging lens is placed 20. cm away from an object, 80. cm from the...

    2) A diverging lens is placed 20. cm away from an object, 80. cm from the diverging lens is a converging lens whose focal length is 25 cm. 35. cm from the converging lens is a screen showing an image of the object. a) What is the focal length of the diverging lens? f = 25. cm o *20. cm → 80. cm + 35 cm → b) What type of image is it? Circle one Real or Virtual c)...

  • A 4.0 cm tall object is placed 50.0 cm from a diverging lens having a focal...

    A 4.0 cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 20.0cm. a. What is the location of the image? b. What is the magnification of the image? c. Is the image inverted or upright? d. Is the image virtual or real?

  • An object 10 cm tall is placed 50 cm in front of a diverging lens with...

    An object 10 cm tall is placed 50 cm in front of a diverging lens with a focal length of 10 cm. a. find the image location b. what is the magnifications? c. what is the height of the image?

  • 14. An object is 2 cm from a concave (diverging) lens. The resulting height of the...

    14. An object is 2 cm from a concave (diverging) lens. The resulting height of the virtual image is half as large as the the height of the object. What is the focal length of the lens? (Hint: Use the information given to determine the lateral magnification, and from that determine the image distance. The use the thin lens equation.) (A) - cm (B) - cm (C) -1 cm (D) –2 cm

  • 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...

  • 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...

  • A diverging lens (f = –11.0 cm) is located 23.0 cm to the left of a...

    A diverging lens (f = –11.0 cm) is located 23.0 cm to the left of a converging lens (f = 34.0 cm). A 2.50-cm-tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT