A converging lens with a focal length of 7.00 cm forms an image of a 4.00 mm tall real object that is to the left of the lens. The image is 1.30 cm tall and upright. Where are the object and image located (in relation to the lens)? Is the image real or virtual? Draw a principal ray diagram
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- A converging lens with a focal length of 7.00 cm forms an image of a 4.00 mm tall real object that is to the left of the lens. The image is 1.30 cm tall and upright.
- Where are the object and image located (in relation to the lens)?
- Is the image real or virtual?
- Draw a principal ray diagram.
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Solved in 3 steps with 2 images
- An object 10.0 mm high is placed 20.0 cm to the left of a biconvex lens of focal length 10.0 cm. A biconcave lens, of focal length 24.0 cm, is placed 30.0 cm to the right of the first lens. a) Sketch the ray diagram for this situation. b) What are the image characteristics? Location Type size orientation c) What is themagnification of the system?In a darkened room, a burning candle is placed 1.63 m from white wall. A lens is placed between candle and wall at a location that causes a larger, inverted image to form on the wall. When the lens is moved 86.6 cm toward the wall, another image of the candle is formed. K 1.63 m How far apart are the two object distances that produce these images? Answer in units of m. Find the focal length of the lens. Answer in units of m.Consider a converging lens whose focal length is 6.51 cm. An object is placed on the axis of the lens at a distance of 10.1 cm from the lens. How far is the object's image from the lens? image distance: cm If it can be determined, is the image real or virtual? virtual cannot be determined real If it can be determined, is the image upright or inverted with respect to the object? inverted upright cannot be determined
- An object is placed a distance do = 6.0 cm in front of a positive (converging) lens of focal length f = 13.0 cm. Where is the image formed relative to the lens position (positive = to the right of the lens)? d₁ = What is the image magnification? m = cm Is the image upright or inverted? inverted upright no image is formed Is the image real or virtual? real O no image is formed O virtual o doConsider the optical system shown in the figure. The lens and mirror are separated by d = 1.00 m and have focal lengths of f₁ = +87.5 cm and f₂= = -53.4 cm, respectively. An object is placed p = 1.00 m to the left of the lens as shown. Lens LL -1.00 m Object distance location (a) What is the distance (in cm) and location of the final image formed by light that has gone through the lens twice? (Give the magnitude of the distance and select its location with respect to the lens.) ---Select--- -1.00 m- cm Mirror (b) Determine the overall magnification of the image. (c) State whether the image is upright or inverted. upright invertedA 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 7.80 cm. A diverging lens of focal length –16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position cm ---Select--- height cm Is the image inverted or upright? O upright O inverted Is the image real or virtual? O real O virtual
- An object that is 2.0 cm tall is located 5.0 cm behind the vertex of a converging lens with a focal length of 10 cm as shown in the diagram below. Which of the lettered choices below gives the correct image characteristics? Specifically, what willI be the height, orientation, and nature of the final image? 10 cm Object İ50 5.0 cm f Converging Lens a. 3.0 cm, inverted, real b.4.0 cm, upright, virtual c. 5.0 cm, upright, real d. 4.0 cm, inverted, virtual e. 2.5 cm, upright, virtual O O O O OA lens forms an image of an object. The object is 15.4 cm from the lens. The image is 11.7 cm from the lens on the same side as the object. a. What is the absolute value of the focal lens of the lens? cm b. Is the lens converging or diverging? converging diverging c. If the object is 8.48 mm tall, how tall is the image? mm d. Is the image upright or inverted? upright O inverted e. To receive full credit for this problem you will need to draw a principal ray diagram, including three principal rays. yes, got it - I will attach the diagram in Canvas with the rest of my work. no, I don't have a diagram to attach.A 15.0 cm tall object is 75.0 cm to the left of a diverging lens (with focal length of magnitude 40.0 cm). A second diverging lens (with a focal length of magnitude 100.0 cm) is 57.2 cm to the right of the first lens. The object and both lenses are on the same optic axis. What is the height of the final image that is produced by light passing through both lenses?
- A converging lens (f₁ = 24.0 cm) is located 56.0 cm to the left of a diverging lens (f2 = -28.0 cm). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies 20.9 cm to the left of the diverging lens. How far is the object from the converging lens? do1 = iA 2.0-cm-tall object is 23 cm in front of a converging lens that has a 40 cm focal length. Calculate the image position. Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.In the following three scenarios, an object is located on one side of a converging lens. In each case, you must determine if the lens forms an image of this object. If it does, you also must determine the following. whether the image is real or virtual whether the image is upright or inverted the image's location, q the image's magnification, M The focal length is f = 60.0 cm for this lens. Set both q and M to zero if no image exists. Note: If q appears to be infinite, the image does not exist (but nevertheless set q to 0 when entering your answers to that particular scenario). (a) The object lies at position 60.0 cm. (Enter the value for q in cm.) q= cmM= Select all that apply to part (a). realvirtualuprightinvertedno image (b) The object lies at position 7.06 cm. (Enter the value for q in cm.) q= cmM= Select all that apply to part (b). realvirtualuprightinvertedno image (c) The object lies at position 300 cm. (Enter the value for q in cm.) q= cmM= Select all that…