A converging lens with a focal length of +6.0 cm is placed 4.0 cm in front of a diverging lens with a focal length of -2.5 cm in the telephoto lens. Calculate

A 2.1-cm-tall object is 30 cm to the left of a lens with a focal length of 15 cm. A second lens with a focal length of -5 cm is 45 cm to the right of the first lens. Calculate the distance between the first image and the second lens. Calculate the image height.

A diverging lens has a focal distance of f=0.2 cm. If an object is placed 0.1cm away, What is the image distance? Is the image real or virtual? What is the magnification?

A lens of focal length 42 mm is used as a magnifier. The object being viewed is 6.8 mm long, and is positioned at the focal point of the lens. The lens is moved closer to the object, so that the image is now 15 cm from the lens. The distance the lens has been moved, in cm, is closest to:

An object is 20.0 cm to the left of the convex lens which is 50.0 cm from a concave lens as shown below. The focal lengths of the convex and concave lenses are 10.0 cm and -12.5 cm. respectively. (a) Use the thin lens equation to determine the location of the image made by the convex lens. (hi Treat the image made by the convex lens as the object of the concave lens and determine the location of the final image. (c) Characterize the final image completely. (Reduced or enlarged, real or virtual, Inverted or upright).

(numbers 8-9) An object is placed 12 cm from a diverging lens with a focal length of -6 cm. 8. How far is the image from the lens? 4 cm 4.5 cm 6 cm 8 cm 12 cm 9. The image is: real and smaller real and larger virtual and smaller virtual and larger none of the above answers are correct