please write neatly. will thumbs up An object of height 9.00 en is placed 33.0 cm to the left of a converging lens with a focal length of 12.0 cm. Determine the image location in em, the magnification, and the Image height in cm HINT (a) the image location in cm cm (b) the magnification (c) the image height in cm Om (d) is the image real or virtual Oreal (el is the image pright or inverted? inverted An object...
11.87 A 1.00-cm-high object is placed 4.85 cm to the left of a converging lens of focal length 8.20 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 Take the image formed by the first lens to be the object for the second lens and apply the lens equation to each lens to locate the final image. cm 8.442...
A small object is placed 25.0 cm from a converging lens of focal length 40.0 cm. The object is to the left of the lens. Where is the image? A- 17cm to the right of the lens B- 25cm to the left of the lens C- 17 cm to the left of the lens D-25 cm to the right of the lens E-67cm to the right of the lens F-67 cm to the left of the lens A small object...
9. -15 points KatzPSE1 38.P.076 My Notes Ask Your The figure below shows an object placed a distance doi from one of two converging lenses separated by s 1.00 m. The first lens has focal length fi 23.0 cm, and the second lens has focal length 2 47.0 cm. An image is formed by light passing through both lenses at a distance = 12.0 cm to the left of the second lens. Include the sign of the value in your...
A 20 cm tall object is located 70 cm away from a diverging lens that has a focal length of 20 cm. Use a scaled ray tracing to answer parts a-d. a. Is the image real or virtual? b. Is the image upright or inverted? c. How far from the lens is the image? d. What is the height of the image? e. Now use the thin lens equation to calculate the image distance and the magnification equation to determine...
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...
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?
1.) An object is placed in front of a diverging lens with a focal length of 17.7 cm. For each object distance, find the image distance and the magnification. Describe each image. (a) 35.4 cm location _____cm magnification _____ nature real virtual upright inverted (b) 17.7 cm location _____ cm magnification _____ nature real virtual upright inverted (c) 8.85 cm location _____ cm magnification _____ nature real virtual upright inverted 2.) An object is placed in front of a converging lens...
A diverging lens has a focal length of magnitude 21.2 cm. (a) Locate the images for each of the following object distances. 42.4 cm distance cm location in front of A 21.2 cm distance location cm in front of 10.6 cm distance location in front of cm 4 (b) Is the image for the object at distance 42.4 real or virtual? o real o virtual Is the image for the object at distance 21.2 real or virtual? o real o...
A real object is 13.6 cm to the left of a thin, diverging lens having a focal length of magnitude 24.5 cm. (a) is the sign of the focal length negative or positive? negative positive (b) Find the image distance. (c) Find the magnification. (d) State whether the image is real or virtual. real virtual (e) State whether the image is upright or inverted. upright inverted