A convex lens forms a clear, focused image of some small, fixed object on a screen. If the screen is moved closer to the lens, will the lens have to be (a) moved closer to the object, (b) moved farther from the object, or (c) left at the same location to produce a clear image on the screen at its new location? Explain your answer.

The lens formula for a convex lens is given by: 1/f = 1/u + 1/v where f = focal length of the lens u = object distance (distance between the object and the lens) v = image distance (distance between the image on the screen and the lens)

Moving the screen closer to the lens means that the image distance (v) will be shorter. Since the object is fixed, object distance (u) remains constant. We want to know the effect of these changes on the position of the lens.

Let's think of two scenarios: one when the screen is at its initial position (v1) and the other when the screen is moved closer to the lens (v2). Scenario 1: 1/f = 1/u + 1/v1 Scenario 2: 1/f = 1/u + 1/v2 Since v2 < v1 (the screen is moved closer to the lens), the term 1/v2 will be greater than the term 1/v1. Now, since the lens hasn't changed and the object is fixed, 1/f must remain constant. Therefore, 1/u must decrease (get smaller) to compensate for the increase in 1/v2 (from the decreased image distance). This means that the object distance u will actually increase (get bigger), which means that the lens needs to be moved farther away from the object to maintain a clear focused image on the screen. So, the correct answer is (b) the lens will have to be moved farther from the object when the screen is moved closer to the lens in order to produce a clear focused image.