Phy2005 Applied Physics II Spring 2017 Announcements: Today: compound lenses, micro-, telescope Exam 3 April 10 in-class

Last time Definitions: object distance p image distance q hi ho f D screen x D-x Last time Definitions: object distance p image distance q magnification M = hi/ho Lens equation:

Q2 A diverging lens has a focal length of -12 cm Q2 A diverging lens has a focal length of -12 cm. An object of height ho = 4 cm is placed 4 cm from the lens. Which is closest to the size of the image? (1) 4 cm (2) 3 cm (3) 1 cm (4) 2 cm (5) 0 cm

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#8 S11 Exam 3. An object is placed at a distance of 80 cm from a thin lens along the axis. If a real image forms at a distance of 40 cm from the lens, on the opposite side from the object, what is the focal length of the lens in cm? (1) 26.6 (2) 45.3 (3) 90.9 (4) 120.0 (5) 21.4

Optical Instruments Perfect Eye Nearsighted Eye Glasses Perfect Eye Nearsighted Nearsighted can be corrected with a diverging lens.  A far object can be focused on retina.

A Farsighted Power of lens: diopter = 1/f (in m) (+) diopter  converging lens (-) diopter  diverging lens Larger diopter  Stronger lens (shorter f)

Material n Cornea 1.38 Aqueous Humor 1.33-1.34 Lens 1.41-1.45 Vitreous Humor 1.34 Air 1.00 Water 1.33

(two converging lenses) Compound Microscope (two converging lenses) objective eyepiece po qo pe Magnified inverted virtual image qe Real image formed by the objective lens  an object for the eyepiece lens fo fe po pe qo qe Mo Me Each set follows lens Equations!!! M = MoMe = (-qo/po)(-qe/pe) = (qo/po)(qe/pe)

M = MoMe = (-qo/po)(-qe/pe) = (qo/po)(qe/pe) Magnification becomes larger qo >> po: 1/po + 1/qo = 1/fo 1/po ≈ 1/fo The object should be put near the focal point of the objective lens.

Virtual image (qe<0) Q. In a compound microscope, the obj. lens has a focal length of 8 mm, And the eyepiece has a focal length 40 mm. The distance between the lenses is 200 mm. If the object is placed 8.4 mm from the obj. lens, What would be the magnification? Inverted (M<0) but Virtual image (qe<0) M = (qo/po)(qe/pe) po = 8.4 qo = ? fo = 8 D = 200 pe = ? qe = ? fe = 40 = (168/8.4)(-160/32) = -100 1/po + 1/qo = 1/fo 1/qo = 1/fo – 1/po = 1/8 – 1/8.4 qo = 168 pe = D – qo = 200 – 168 = 32 1/qe = 1/fe – 1/pe = 1/40 – 1/32 qe = -160

Ex. 3 Two thin lenses with 10.0-cm focal lengths are mounted at opposite ends of a 30.0-cm long tube. An object is located 45.0 cm from one end of the tube. How far from the opposite end is the final image? 38.2 cm 12.8 cm 25.6 cm (4) 33.6 cm (5) 24.0 cm

#7 S11 Exam 3 Two thin lenses with focal lengths 50.0 cm and 60.0 cm are placed in contact in an orientation so that their optic axes coincide. What is the focal length of the two in combination? (1) 110 cm (2) 13.6 cm (3) 55.0 cm (4) 27.3 cm (5) 10.0 cm (see section 27.13: Virtual objects) (1) Object p1 and image q1 for lens 1 alone (2) Object p2 and image q2 for lens 1 alone Image of lens 1 is virtual object for lens 2 Add (1) and (2): effective focal length is

Telescope: angular magnification