converging lens (f = 30.0 cm) is used to project an image of an object onto a screen. The object and the screen are 150 cm apart, and between them the lens can be placed at either of two locations. What is the larger of the two object distances?

What is the focal length of the lens in your eye when you can focus on an object at your near point? Assume that the distance from your lens to the back of your eye is 2.0 cm, that your near point is 21 cm in front of your lens, and that the liquid behind the lens has n=1.00 just like the air in front of your lens. (a) 1.8 cm (b) 18 cm (c) 2.1 cm (d) 21 cm (e) None of the above.

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.…

A converging lens (f = 33.5 cm) is used to project an image of an object onto a screen. The object and the screen are 197 cm apart, and between them the lens can be placed at either of two locations. Find the two object distances, the smaller being the answer to part (a). (a) Number i (b) Number Units Units

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:

A 3 cm tall object is placed 2 cm to the left of a converging lens that has a focal length with a magnitude of 4 cm. A diverging lens with a focal length of magnitude 8 cm is placed 10 cm to the right of the first lens. What is the magnification of the final image produced by these two lenses? Make sure you say whether it is bigger/smaller and upright/inverted.

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 student's uncorrected near and far points are at 9.0 cm and infinity, respectively. She looks through a meniscus-shaped lens as shown in the figure. The absolute value of the radius of curvature is 2.4 cm on the side of the lens near her eye, and 2.6 cm on the side away from her eye. The index of refraction of the lens is 1.4. What is the closest an object can get to the lens and still be clearly visible to the student? [You may assume that the lens is very close to her eye.] eye 3.5 cm 78.0 cm 10.2 cm 8.1 cm 2.3 cm

A small object is placed to the left of a convex lens and on its optical axis. The object is 30 cm from the lens, which has a focal length of 15 cm. If the object is moved to a position 22 cm away, what will the image position be? (A positive value for p, or position, means the image is on the right side of the lens.)