Help and explain please A converging lens has a focal length of 6.0 cm. a) Locate...
A converging lens has a focal length of 6.0 cm. a) Locate the position of the images for object distance of 4.0 cm. b) Find the height of the image of an object 1.0 cm high. c) State whether the image is real or virtual and upright or inverted. d) Draw on the paper that you will upload a detailed and proportional ray diagram for this situation. Do that and than enter "OK" in the blank space below to indicate...
A converging lens has a focal length of 6.0 cm A.) Locate the position of the images for the object distance of 4.0cm B.) Find the height of the image of an object 1.0 cm high C.) State whether the image is real, virtual, upright, or inverted D.) Draw a detailed and proportional ray diagram for this situation
A converging lens with a focal length of 6.0 cm is located 24.0 cm to the left of a diverging lens having a focal length of -13.0 cm. If an object is located 11.0 cm to the left of the converging lens, locate and describ completely the final image formed by the diverging lens. Where is the image located as measured from the diverging lens? 63.81 cm Submit Answer Incorrect. Tries 3/10 Previous Tries What is the magnification? Submit Answer...
A converging lens has a focal length of 23.0 cm. Locate the images for the following object distances, if they exist. Find the magnification. (Enter 0 in the q and M fields if no image exists.) (a) 23.0 cm cm M= Select all that apply to part (a). real virtual upright inverted no image (b) 2.56 cm cm M= Select all that apply to part (b). real virtual upright inverted no image (c) 173 cm q cm M= Select all...
A converging lens has a focal length of 65 cm. Locate the images for the following object distances if they exist. (Enter O in the 9 and M fields if no image exists.) (a) 65.00 cm 9 = m= Select all that apply to part (a). upright inverted virtual O no image Oreal (b) 13.00 cm m = Select all that apply to part (b). upright inverted no image o virtual Oreal (c) 195.00 cm 9 = m = Select...
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. (As viewed from the side, from left to right, the sequence is object - diverging lens - converging lens - observer. Rays then travel from left to right through the system.) (a) Use ray tracing to draw image 1 and image...
A converging lens has a focal length of 53.0 on. Locate the images for the following object distances, if they exist. Find the magnification (Enter O in the 9 and M fields if no image exists.) Welcome, Samantha - Blackboard Learn Select all that apply to part (a). real virtual upright inverted no image (h) 11.8 cm Select all that apply to part (b). real virtual upright Inverted no image (c) 292 cm Select all that apply to part (c)....
NOTES A converging lens has a focal length of 100 cm. Locate the images for the following object distances, if they exist. Find the magnification (Enter in the g and M fields if no image exists.) (a) 100 cm cm M Select all that apply to part (a). real virtual upright inverted no image x (b) 22.2 cm G- N- am Select all that apply to part (b) a real virtual upright inverted no image X (c) 250 cm Q-...
A converging lens has a focal length of 15.0 cm. Locate the images for the following object distances, if they exist. Find the magnification. (Enter 0 in the q and M fields if no image exists.) (a) 15.0 cm q = cm M = Select all that apply to part (a). real virtual upright inverted (b) 1.88 cm q = cm M = Select all that apply to part (b). real virtual upright inverted (c) 97.5 cm q = cm M = Select all that apply...
A converging lens has a focal length of 10 cm. Locate the images for the following object distances if they exist. (Enter O in the q and M fields if no image exists.) (a) 10.00 cm 9 = 20 m= 2 Select all that apply to part (a). upright virtual no image Oreal inverted (b) 2.50 cm 9 = -3.24 m= 1.3 Select all that apply to part (b). Oreal no image virtual inverted upright (c) 20.00 cm 9 =...