An object is placed 35 cm to the left of a converging lens with focal length f1=15 cm. A second converging lens with focal length f2=7 cm is placed a distance d=38 cm to the right of the first lens. What us the final image location relative to the second lens?

The focal length of a diverging lens is negative. If f = −16 cm for a particular diverging lens, where will the image be formed of an object located 45 cm to the left of the lens on the optical axis? cm to the left of the lensWhat is the magnification of the image?

An object is initially at rest, 100 cm from a converging lens with a focal length of 30.0 cm. The object accelerates toward the lens at a constant rate of 0.300 cm/s?. At what time will the magnification of the image produced be -2.00?

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance ss to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance s′s′ and then use the equation 1s+1s′=1f1s+1s′=1f, to calculate the focal length ff of the lens. But this procedure won'’t work with a diverging lens−−by itself, a diverging lens produces only virtual images, which can'’t be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object 20.0 cmcm to the left of the lens, the image is 29.7 cmcm to the right of the lens.…

An object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.140 m in front of a diverging lens. The focal length of the lens is -0.25 m. (Include the sign of the value in your answers.) (a) What is the magnification? (b) What is the image height? m (c) What is the image distance? m

A 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 7.80 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 7.5 cm in front of the second lens v 0.5466 height Calculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cm Is the image inverted or upright? upright O inverted Is the image real or virtual? real virtual

A nearsighted person has a far point of 25.3cm and is prescribed contact lenses to correct her vision. What lens strength (a.k.a., lens power), in Diopters, should be prescribed? Give your answer using the usual convention of positive values for converging lenses and negative values for diverging lenses.