Reading: Lenses and Mirrors; Applications Key concepts: Focal points and lengths; real images; virtual images; magnification; angular magnification.

Transcription

1 Reading: Lenses and Mirrors; Applications Key concepts: Focal points and lengths; real images; virtual images; magnification; angular magnification. 1.! Questions about objects and images. Can a virtual image be seen on a projection screen? Can a virtual image be photographed? What is a virtual object? Give an example. 2.! Questions about the eye and corrective measures. Why is vision blurred under water? How do goggles solve the problem? Most eyeglass lenses have a convex outer surface (the one furthest from the eye) with a standard radius, for cosmetic reasons. Draw a cross section of such a lens that has a negative focal length. If you look closely through the eyeglasses at the eyes of the wearer, you can tell whether the person is nearsighted or farsighted. How? 3.! Questions about magnification. The magnification of instruments such as a magnifying glass increases when the focal length of the eyepiece is reduced. Why are lenses of focal length shorter than about 5 cm rarely used? What is angular magnification, and why is it the appropriate measure used to describe the power of optical instruments? The telescope and microscope both use two positive lenses. Why is the microscope s magnifying power so much larger? 4.! Make a drawing of an overhead projector of the type used in classrooms. Show why, if the speaker is facing the audience and projecting onto a screen behind her, the image of the slide she is reading from is upright and enlarged on the screen.. 5.! Most optical microscopes have systems to provide bright illumination of the object, yet the image seen through the instrument is not especially bright. Why is the extra illumination necessary, and why is the image not so bright?

2 6.! A device to spread out the beam from a laser is shown. The incoming beam from the laser has circular cross-section of radius a. The device spreads the beam so its cross-section radius becomes b, without changing the direction of the beam. light device! The device has lenses of positive focal length at each end, as shown. Indicated is a ray at the top of the incident beam, therefore at distance a from the axis, entering lens #1. This ray emerges from lens #2 at distance b from the axis. 1 d 2 Sketch the path of the ray as is passes through the device. [It must emerge from lens #2 parallel to the axis.] How are the focal lengths f 1 and f 2 of the two lenses related to the beam radii a and b? [Use your drawing.] Express the two focal lengths in terms of a, b and d. 7.! In the situation shown the lens has a focal length 10 cm and the mirror has curvature radius +20 cm. The distance between adjacent dots on the diagram is 10 cm. An object is placed as shown 15 cm in front of the mirror. Use the formulas to locate the image formed by the mirror and the image subsequently formed by the lens. (Ignore the image formed directly by the lens.) Calculate the magnification of the final image relative to the object. Draw principal rays (at least two) to locate the image formed by the mirror. Using this as the object for the lens, locate the final image by drawing at least two principal rays.! [In drawing rays, use the vertical lines to represent the mirror and the lens in the paraxial ray approximation.]

3 8.! Shown is a device that displays a real image. Identical parts of spherical mirrors of focal length f, are put together as shown, separated by distance d. An object (the arrow) is at the bottom; light is viewed through the hole in the top. A final image appears in the hole. d The object distance p 1 for the upper mirror is d. Find the image distance q 1 and the magnification m 1, in terms of d and f. This image forms the object for the lower mirror, so the object distance for it is p 2 = d q 1. We wish to determine d so that the final image distance is also d. Use the image location formula to derive an equation from which d can be determined in terms of f. (It is a quadratic equation.) Find the two values of d in terms of f. (The equation can be factored.) d.! Find the overall magnification for each of these values of d. [The device shown in class uses the lower value of d.] 9.! A Galilean telescope uses a negative lens as its eyepiece. The arrangement is as shown. The right focal points of the two lenses coincide. f o f e f o, f e Draw a ray from the top of a distant object, passing through the left focal point of the objective lens, making small angle α with the axis. Continue this ray between the lenses, and through the eyepiece. The emerging ray makes angle α with the axis. Determine the angular magnification of the instrument in terms of the focal lengths. What are the advantages of this telescope?

4 10.! The power of a lens in diopters (D) is 1/ f, in meters. Consider the two thin lenses shown, with focal lengths f 1 and f 2. Parallel rays are incident from the left. The point where the rays converge is the focal point of the system. If the second lens were not present, where would the light focus? This image acts as the virtual object for the second lens, with a negative object distance. Locate the final image, showing that for the combination!! 1 f = 1 f f 2, or D tot = D 1 + D ! The power of the cornea-lens system in a normal eye with the lens relaxed would be 60 D, if air were on both sides, which we will assume for simplicity. (The actual power, with liquid between lens and retina, is about 45 D). What would be the distance from the lens to the retina? (The actual distance is about 2.2 cm; explain why this is larger.) What additional power must its lens provide (by changing its shape) in order to focus clearly on an object at the standard near point, 0.25 m? Weakening of the lens muscles has increased an elderly person s near point distance to N > 0.25 m. A corrective lens of what power (in terms of N) will restore her near point to 0.25 m? 12! A nearsighted person has an excessively curved cornea, and cannot reduce the power of her eye (by relaxing the lens) below 62 D. The lens can add 4 D extra power for near vision. How far can an object be from this person and still allow her to focus on it clearly (the far point distance)? What focal length of corrective lens should this person use to make the far point distance infinite? Without corrective lenses, what is this person s near point distance?

5 13.! A slightly farsighted person can focus clearly on distant objects by using 3 D of the 4 D extra power of the lens to augment the insufficient power of his cornea. Without corrective lenses, between what distances can this person focus clearly? What power corrective lens is needed to let this person focus on an object at the standard near point of 25 cm? With this correction, can the person see clearly at all distances from 25 cm to infinity? Explain.