University of Rochester Department of Physics and Astronomy Physics123, Spring 2012 Homework 5 - Solutions Problem 5.1 An optometrist finds that a farsighted person has a near point at 125 cm. a) If the eye is approximately 2.0 cm long (i.e. the distance between retina and iris), how much power does the refractive system have when the eye is relaxed (i.e. focused on an object at infinity) and when the eye is accommodated to focus at 125 cm? b) What power will be required for contact lenses if they are effectively to move the near point inward to 25 cm, the near point of a normal eye? Solution 5.1 a) By applying the thin lens equation, when d0 → 0 we have di → f and 1 1 1 D = = = = 50.0 m-1 = 50.0 D . f di 0.02 When the eye is imaging an object at 125 cm, 1 1 1 1 1 D = = + = + = 50.8 m-1 = 50.8 D . f d0 di 1.25 0.02 b) The correction is done by a converging lens. According to the adopted sign convention (see Giancoli 33-2), the image distance is negative if it is on the same side of the lens from where the light is coming. In other words, the image distance is negative for virtual image. In our case we want to form a virtual image at 125 cm when the object is placed at 25 cm: 1 1 1 1 1 D = = + = + = 3.2 m-1 = 3.2 D . f d0 di 0.25 (−1.25) (See also Example 33-12 on Giancoli.) Problem 5.2 The refracting telescope at the Yerkes Observatory has a 1.0 m diameter object lens of focal length 20.0 m. Assume it is used with an eyepiece of focal length 2.5 cm. (a) Determine the magnification of Mars as seen through this telescope. (b) Are the Martian polar caps right side up or upside down? 1 Solution 5.2 (a) The angular magnification produced by this telescope is � = − !! = -800 !! (b) Since m < 0, the image is inverted. Problem 5.3 A composed microscope in air includes an objective lens with focal length fo = 8 mm and an eyepiece (or ocular) with focal length fe = 18 mm. They are separated by l = 16.8 cm and both lenses can be considered thin. If an object is observed with a relaxed (i.e. without accommodation) normal eye, a) What is the distance (do) between object and objective lens? b) What is the overall magnification of the microscope? A normal human eye can distinguish two separate point sources if their distance is subtended by an angle ≥ 2π 21600 rad . (You can take this as the definition of the resolving power of the eye in ray optics.) c) Using the composite microscope above, what is the minimum distance that a normal human eye can resolve? Solution 5.3 a) Since the final image is at infinity (relaxed eye) the image from the objective is at the focal point of the eyepiece (or ocular). We subtract this distance from the separation between the lenses to calculate the objective image distance. Then using Eq. 33-2, Giancoli, we calculate the object distance. di1 = l − fe = 0.168m − 0.018 m = 0.150m 0.80cm 15.0cm 1 1 1 fodi1 ( )( ) −3 = + → d = = = 8.5×10 m f d d o1 d − f 15.0cm − 0.80cm o o1 i1 i1 o b) With the final image at infinity, the magnification of the eyepiece is given by Eq. 33- 10a, Giancoli. 25.0cm N ⎛ l − fe ⎞ ( ) ⎛ 16.8cm −1.8cm ⎞ M = ⎜ ⎟ = ⎜ ⎟ = 247 × f ⎝ d ⎠ 1.8cm ⎝ 0.85cm ⎠ e o ( ) c) The near point of a normal eye is N = 0.25 m. Thus, two points will appear separated if their distance is equal or larger than 2π 2π h = N = 0.25 ≅ 73 × 10−6 m . min 21600 21600 2 The composite microscope above has magnification M equal to 247 × , thus the minimum distance becomes h 73 × 10−6 m min = ≅ 0.30 × 10−6 m . M 247 Problem 5.4 Two lenses made of kinds of glass having different indices of refraction n1 and n2 are cemented together to form an optical doublet. Optical doublets are often used to correct chromatic aberrations in optical devices and are called achromatic doublet (or color- corrected lens). A concave lens of flint glass is commonly combined with a convex lens of crown glass to produce an achromatic doublet lens because of their compensating optical properties, which reduces chromatic aberration. The first lens of a certain achromatic doublet has index of refraction n1, one flat side, and one concave side with a radius of curvature of magnitude R. The second lens has index of refraction n2 and two convex sides with radii of curvature also of magnitude R. Show that the doublet can be modeled as a single thin lens with a focal length described by 1 2� − � − 1 = ! ! � � Solution 5.4 (See below.) 3 .