Two 18.7 cm focal length converging lenses are placed 18.9 cm apart. An object is placed 54.3 cm in front of one lens. Where will the final image formed by the second lens be located (in cm relative to the first lens)? In your written work, also calculate the total magnification.

A converging lens has a focal length of 4.0 cm A 1.5 cm object is placed 6.0 cm to the left of the lens (as shown in the diagram below) Part A The distance of the image will be ανα ΑΣφ d; = cm Submit Request Answer Part B The height of the image will be Πνη ΑΣφ ? h4 = cm Submit Request Answer Part C The image is and O real, upright O real, inverted O virtual, inverted O virtual, upright

Thank you so much in advance. The diagram shows a lens with a positive focal length 11 cm. (a) If we place an object at a distance of 25 cm from the lens, where will the resulting image position on the other side of the lens be found? Include units in answer, (b) With the object at 25 cm from the lens, what will the magnification be for the image at this position? (c) If we place an object at a distance of 4 cm from the lens, where will the resulting image position be found? (d) With the object at 4 cm from the lens, what will the magnification be for the image at this position?

A convex lens produces an image with a magnification of +5. Describe the locatior size, and orientation of the image produced. Circle one from each pair of answers Same side as object/ opposite side as object Upright/Inverted Real /Virtual Enlarged /Reduced

Please answer the question in the image making sure to answer parts a,b, and c.

A concave lens creates a virtual image at -47.0 cm and a magnification of +1.75. What is the focal length? (Mind your minus signs.) (Unit = cm)

Determine the distance from the second lens to the final image.