An object is placed in front of a diverging lens whose focal length is - 30.7 cm. A virtual image is formed whose height is 0.532 times the object height. How far is the object from the lens?
determine how far in front of the lens the object should be placed.. You place an object 19.6 cm in front of a diverging lens which has a focal length with a magnitude of 14.2 cm. Determine how far in front of the lens the object should be placed in order to produce an image that is reduced by a factor of 4.35.
The distance between an object and its image formed by a diverging lens is 46.9 cm. The focal length of the lens is -244.2 cm. (a) Find the image distance. (b) Find the object distance. An object is 20 cm in front of a diverging lens that has a focal length of -13 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 1.0?
An object is placed 10.0 cm in front of a lens. An upright, virtual image is formed 30.0 cm away from the lens. What is the focal length of the lens? Is the lens converging or diverging?
1.) The distance between an object and its image formed by a diverging lens is 46.1 cm. The focal length of the lens is -237.3 cm. (a) Find the image distance. cm (b) Find the object distance. cm 2.) An object is 19 cm in front of a diverging lens that has a focal length of -11 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by...
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)...
Use the thin lens equation to solve problems 14 –18. 14. An object is 10 cm high and is placed 20 cm in front of a converging lens of focal length 20 cm. Determine the image distance, the image height and the magnification. 15. An object is 10 cm high and is placed 16 cm in front of a converging lens of focal length 20 cm. Determine the image distance, the image height and the magnification. 16. An object is...
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...
You place an object 20.2 cm in front of a diverging lens which has a focal length with a magnitude of 10.2 cm. Determine how far in front of the lens the object should be placed in order to produce a new image that is 4.20 times smaller than the original image.
A 0.49 cm high object is placed 8.5 cm in front of a diverging lens whose focal length is -7.5 cm. What is the height of the image?
A 0.44 cm high object is placed 8.4 cm in front of a diverging lens whose focal length is -7.5 cm. What is the height of the image?