Question

Does a positive s’ value correspond to a real or a virtual image? Does a positive...

Does a positive s’ value correspond to a real or a virtual image? Does a positive s’ value correspond to an image to the left or to the right of the lens?

• Does a negative s’ value correspond to a real or a virtual image? Does a negative s’ value correspond to an image to the left or to the right of the lens?

• Which kind of image (real or virtual) appears to the right of the lens? Which kind of image (real or virtual) appears to the left of the lens?

• Which kind of lens (converging/diverging) creates real images? Which kinds of lenses (converging/diverging) create virtual images?

• Does a positive value for M correspond to an upright or an inverted image? What about a negative M value?

0 0
Add a comment Improve this question Transcribed image text
Answer #1

According to cartesian sign convention- (Remind it )

  • positive value of s' represents the image is real. also to the right of the lens ( if object is left side ).
  • negative value of s' represents the image is virtual , also to the left side (if object is left side )
  • IF object is left side then right side image is real and left side image is virtual .
  • converging lens creates Real and virtual image for real image it is converging lens. diverging lens always creates virtual image.
  • positive value of the magnification m represents the image is upright . and negative value of m represents the image is inverted.
Add a comment
Know the answer?
Add Answer to:
Does a positive s’ value correspond to a real or a virtual image? Does a positive...
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
  • 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

  • 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 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...

  • A charge, q=91.0000 microCoulombs on a particle with mass m=1.00000 milli- grams, moves through a pipe...

    A charge, q=91.0000 microCoulombs on a particle with mass m=1.00000 milli- grams, moves through a pipe from the origin to a point at coordinate x=1.40000m and y=1.8000m. All space is filled with a uniform electric field E=1,900.00000N/C and pointing parallel to the x axis. What is the change in electric potential as the mass moves from initial to final positions (in VOLTS) An object is placed 50.0cm in front of a lens. The image forms on the same side of...

  • Could you please show all of the steps taken in solving the problem and explain the process fully...

    Could you please show all of the steps taken in solving the problem and explain the process fully, using sentences/phrases (if possible) to understand why those steps were taken and why you got that answer. Also could you please draw a diagram with any applicable variables/values to help visualize the problem better. Thank you in advance for all the help! A 1.20-cm-tall object is 50.0 cm to the left of a diverging lens (lens 1) of focal length of magnitude...

  • Starting with a real object, answer the following statements (True or False) about the image formed...

    Starting with a real object, answer the following statements (True or False) about the image formed by a single lens? B A converging lens can produce a virtual, upright, enlarged image. e A diverging lens can produce a real, inverted, reduced image. A diverging lens always produces a virtual, upright, reduced image. False B A converging lens cannot a A converging lens can never produce a virtual, upright, reduced image. Suggestion: Make a table or a set of drawings showing...

  • Now, a diverging lens with focal length having a magnitude of 20 cm is placed 10...

    Now, a diverging lens with focal length having a magnitude of 20 cm is placed 10 cm to the right of the converging lens in problem that has a 2 cm tall object placed 12 cm to the left of a converging lens with focal length of magnitude 15 cm. Determine the location of the final image formed by both lenses (in relation to the diverging lens) and the magnification of the final image. State whether the final image is...

  • I only need A,B,C What kind of image will be produced (real or virtual? Upright or...

    I only need A,B,C What kind of image will be produced (real or virtual? Upright or inverted?) for a converging lens under the following four conditions? Use the thin lens equation and then check using a ray tracing picture. s > 2f s = 2f s = f 0 < s <f

  • 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...

  • 1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm...

    1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm tall, 11 2emto right of the lens. a. Determine the position of the object. b. Determine the size of the object. Is the image upright or inverted? Are the object and image on the same side or opposite sides of the lens? c. d. 2 You want to use a lens with a focal length of magnitude 36cm with the image twice as long...

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