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Physics 112 Homework 12 (Ch24) 1. If an f 135 -mm telephoto lens is designed to cover object distances from 1.2 m to , over what distance must the lens move relative to the plane of the film? Solution We locate the image from equation: 1 1 d o di 1 . f a) For an object at infinity, the image will be in the focal plane, so we have di f 135mm. b) When the object is at 1.2 m, 1 1 120 cm di 1 , which gives di 152 mm. 135cm Thus the distance from the lens to the film must change by dib dia 152mm 135mm 17mm. 2. A person’s left eye is corrected by a 3.50-diopter lens, 2.0 cm from the eye. (a) Is this person’s left eye near- or farsighted? (b) What is this eye’s far point without glasses? Solution (a) Since the diopter is negative, the lens is diverging, so it produces images closer than the object; thus the person is nearsighted. (b) We find the far point by finding the image distance for an object at infinity: 1 1 1 P; d o di f 1 1 3.50D, which gives di 0.286 m 28.6cm. di Because this is the distance from the lens, the far point without glasses is 28.6cm 2.0cm 30.6cm. Physics 112 Homework 12 (Ch24) 3. A 3.30-mm-wide beetle is viewed with a 9.50-cm-focal-length lens. A normal eye views the image at its near point. Calculate (a) the angular magnification, (b) the width of the image, and (c) the object distance from the lens. Solution (a) The angular magnification with the image at the near point is M 25.0cm 1 3.63 . N 1 f 9.50cm hi (b) Because the object without the lens and the image with the lens are at the near point, the angular magnification is also the ratio of widths: ' h M i hi Mh0 ; h0 hi 3.63 3.30mm , which gives hi 12.0mm. ho O I N ho O (c) We find the object distance from 1 1 1 ; d o di f 1 1 1 cm, do 25.0cm 9.50 Method 2: ' 25cm c) M d0 d0 which gives d o 6.88 cm from the lens. 25cm 25cm 6.88cm M 3.632 4. A magnifying glass with a focal length of 8.5 cm is used to read print placed at a distance of 7.5 cm. Calculate (a) the position of the image; (b) the angular magnification. Solution (a) We find the image distance from 1 1 1 ; d o di f 1 1 1 cm, which gives di 64cm. 7.5cm di 8.5 (b) The angular magnification is 25cm 25cm = 3.3 . M do 7.5cm Physics 112 Homework 12 (Ch24) 5. An astronomical telescope has its two lenses spaced 75.2 cm apart. If the objective lens has a focal length of 74.5 cm, what is the magnification of this telescope? Assume a relaxed eye. Solution For both object and image far away, we find the focal length of the eyepiece from the separation of the lenses: L fo fe ; f e L f 0 75.2cm-74.5cm = 0.7cm The magnification of the telescope is given by M 75.2cm 110 . fo fe 0.7cm 6. A microscope has a 1.8-cm-focal-length eyepiece and a 0.80-cm objective lens. Assuming a relaxed normal eye, calculate (a) the position of the object if the distance between the lenses is 16.0 cm, and (b) the total magnification. Solution (a) Because the image from the objective is at the focal point of the eyepiece, the image distance for the objective is dio l f e 16.0cm 1.8cm 14.2cm. We find the object distance from the lens equation for the objective: 1 1 1 ; doo dio f o 1 1 1 , which gives doo 0.85cm. doo 14.2 cm 0.80cm (b) With the final image at infinity (relaxed normal eye), the magnification of the eyepiece is Me N 25.0cm 13.9 . f 1.8cm The magnification of the objective is d 14.2cm mo io 16.7 . d oo 0.85cm The total magnification is M= mo M e 16.7 13.9 230 Method 2: 25cm l f e 25cm 16cm 1.8cm M 232 f e d0 1.8cm 0.85cm Physics 112 Homework 12 (Ch24) 7. Two stars 15 light-years away are barely resolved by a 55-cm (mirror diameter) telescope. How far apart are the stars? Assume 550 nm and that the resolution is limited by diffraction. Solution The resolution of the telescope is 9 1.22 1.22 550 10 m 1.22 106 rad. D 0.55m The separation of the stars is d L 15ly 9.46 1015 m ly 1.22 106 rad 1.7 1011 m.