A converging lens and a diverging lens, separated by a distance of 30.0 cm, are used in combination. The converging lens has a focal length of 15.3 cm. The diverging lens is of unknown focal length. An object is placed 19.5 cm in front of the converging lens; the final image is virtual and is formed 12.0 cm in front of the diverging lens. What is the focal length of the diverging lens?

You use a lens to form an image. The object is 50 cm from the lens and the image is formed at a position of -14 cm. What is the focal length of your lens? f = cm What is the magnification of the lens? m = Is the image formed real or virtual? --- Is the image upright or inverted? Are you using a converging lens or a diverging lens?

A 1.00-cm-high object is placed 4.30 cm to the left of a converging lens of focal length 8.45 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 height cm ---Select--- cm

The distance between the eyepiece and the objective lens in a certain compound microscope is 10.4 cm. The focal length of the objective is 0.430 cm, and that of the eyepiece is 1.40 cm. Find the overall magnification of the microscope. (The near point of the eye is 25 cm. Assume that the object is placed at the focal point of the objective lens, and one places the eyepiece at the near point of the eye.)

For safety reasons, you install a rear-window lens with a -0.299 m focal length in your van. Before putting the van in reverse, you look through the lens and see the image of a person who appears to be 0.339 m tall and 0.243 m behind the van. Determine the following. (a) actual distance of the person behind the van m (b) height of the person

An object is 6 cm in front of a converging lens with a focal length of 10cm. Draw a ray diagram (to scale with a ruler) to find the location of the image. Is the image upright or inverted, and Is the image real or virtual? Then I want to use the thin lens formula to find the image distance and the magnification. I got stuck in the middle of this problem and am confused. Thank you for the help!

A converging lens of focal length f=2 cm is used to focus the image of an object onto a screen. The object and screen are separated by 15 cm. The lens is placed between the object and the screen, at a distance from the screen such that the image of the object is focused into the screen. At what distance from the screen must the lens be placed in order to have an image magnification < 1? Provide your answer to two significant figures. distance = cm.

A system of two lenses forms an image of an arrow at x = x3 = 57.4 cm. The first lens is a diverging lens located at x = 0 and has a focal length of magnitude f₁ = 11.5 cm. The second lens is located at x = x₂ = 25.4 cm and has an unknown focal length. The tip of the object arrow is located at (x,y) = (xo, Yo) = (-36 cm, 20.6 cm). (x,y) 1) What is x₁, the x-coordinate of image of the arrow formed by the first lens? cm Submit 2) What is y₁, the y-coordinate of the image of the tip of the arrow formed by the first lens? cm Submit X₂ Real and Inverted Real and Upright Virtual and Inverted Virtual and Upright cm Submit (+) 3) What is f2, the focal length of the second lens. If the lens is a converging lens, f2 is positive. If the lens is a diverging lens, f2 is negative. cm Submit + 4) What is y3, the y-coordinate of the image of the tip of the arrow formed by the two lens system? (+ 5) The positions of the two lenses are now interchnaged (i.e., the second lens is moved to x = 0 and the…