When a converging lens is used in a camera (as in the first drawing), the film must be placed at a distance of 0.190 m from the lens to record an image of an object that is 4.90 m from the lens. The same lens is then used in a projector (as in the second drawing), with the screen 0.590 m from the lens. How far from the projector lens should the film be placed? Number i Bulb Object -Object (upside down) Projector Units (b) -Film -Real image (inverted) Real image Screen

A mother sees that her child’s contact lens prescription is 0.750 D. What is the child’s near point?

A converging lens has a focal length of 10.0 cm. Locate the object if a real image is located at a distance from the lens of (a) 20.0 cm and (b) 50.0 cm. What If? Redo the calculations if the images are virtual and located at a distance from the lens of (c) 20.0 cm and (d) 50.0 cm.

Two thin lenses of focal lengths f1 = 15.0 and f2 = 10.0 cm, respectively, are separated by 35.0 cm along a common axis. The f1 lens is located to the left of the f2 lens. An object is now placed 50.0 cm to the left of the f1 lens, and a final image due to light passing though both lenses forms. By what factor is the final image different in size from the object? (a) 0.600 (b) 1.20 (c) 2.40 (d) 3.60 (e) none of those answers

Show that the magnification of a thin lens is given by M = di/do (Eq. 38.6). Hint: Follow the derivation of the lens makers equation (page 1233) and start with a thick lens.

The power for normal close vision is 54.0 D. In a vision-correction procedure, the power of a patient’s eye is increased by 3.00 D. Assuming that this produces normal close vision, what was the patient’s near point before the procedure?

The contact lens prescription for a mildly farsighted person is 0.750 D, and the person has a near point of 29.0 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?

Suppose you want to use a converging lens to project the image of two trees onto a screen. One tree is a distance x from the lens; the other is at 2%, as in Figure CQ23.7. You adjust the screen so that the near tree is in focus. If you now want the far tree to be in focus, do you move the screen toward or away from the lens? Figure CQ23.7

It has become common to replace the cataract-clouded lens of the eye with an internal lens. This intraocular lens can be chosen so that the person has perfect distant vision. Will the person be able to read without glasses? If the person was nearsighted, is the power of the intraocular lens greater or less than the removed lens?

What is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin?