Example: What kind of lens must be used, in order to give an

1. Example: What kind of lens must be used, in order to give an erect image 1/5 as large as an object placed 15 cm in front of it? M = -q/p  -q/p=1/5 So q = -p/5 = -15/5 = -3 cm 1/p + 1/q = 1/f  1/15 - 1/3 = 1/f 1/f = (1-5)/15 f = -15/4 = -3.75 cm Diverging lens

2. Magnifier • Consider small object held in front of eye • Height y • Makes an angle  at given distance from the eye • Goal is to make object “appear bigger”: ' >  y 

3. Magnifier • Single converging lens • Simple analysis: put eye right behind lens • Put object at focal point and image at infinity • Angular size of object is , bigger! Outgoingrays Rays seen comingfrom here y   f f Image atInfinity

4. Angular Magnification (Standard) • Without magnifier: 25 cm is closest distance to view • Defined by average near point. Younger people do better •   tan  = y / 25 • With magnifier: put object at distance p = f • '  tan ' = y / f • Define “angular magnification” m = ' /  • Note that magnifiers work better for older people because near point is actually > 25cm ~y/25 ’~y/f M= ’/  = 25/f

5. Example • Find angular magnification of lens with f = 5 cm

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

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

9. Combinations of Thin Lenses • The image produced by the first lens is calculated as though the second lens were not present • The light then approaches the second lens as if it had come from the image of the first lens • The image of the first lens is treated as the object of the second lens • The image formed by the second lens is the final image of the system

10. Combination of Thin Lenses, 2 • If the image formed by the first lens lies on the back side of the second lens, then the image is treated at a virtual object for the second lens • p will be negative • The overall magnification is the product of the magnification of the separate lenses

11. Combinations of Thin Lenses • The image produced by the first lens is calculated as though the second lens were not present • The light then approaches the second lens as if it had come from the image of the first lens • The image of the first lens is treated as the object of the second lens • The image formed by the second lens is the final image of the system

12. Combination of Thin Lenses, 2 • If the image formed by the first lens lies on the back side of the second lens, then the image is treated at a virtual object for the second lens • p will be negative • The overall magnification is the product of the magnification of the separate lenses

13. What is the combined focal length? 1/p1 +1/q1 =1/f1 1/p2 +1/q2 =1/f2 p2=-q1 !!! 1/p1 +1/q2 =1/f1 +1/f2 1/f = 1/f1 +1/f2

14. Combination of Thin Lenses, example

15. Q. Two converging lenses are placed 20.0 cm apart. If the first lens has focal length of 10.0 cm and the second has a focal length of 20.0 cm, locate the. final image formed of a object 30.0 cm in front of the first lens. Find the magnification of the system 1/30 + 1/q = 1/10 q=+15 cm M1 = -q/p= -0.5 For second lens: P= 20 –15 = 5 cm 1/5 +1/q = 1/20 q=-6.67 cm M2 = -q/p = 1.33 M=M1 M2 =-0.667 Virtual, inverted

16. objective eyepiece pe po qo qe Magnified inverted virtual image Compound Microscope (two converging lenses) 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)

17. 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 object lens.

18. 1/po + 1/qo = 1/fo 1/qo = 1/fo – 1/po = 1/8 – 1/8.4 qo = 168 pe = D – qe = 200 – 168 = 32 1/qe = 1/fe – 1/pe = 1/40 – 1/32 qe = -160 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

19. Telescope • View Distant Objects • (Angular) Magnification M=fobj/feye • Increased Light Collection • Large Telescope use Mirror

21. Spherical Aberration • Results from the focal points of light rays far from the principle axis are different from the focal points of rays passing near the axis • For a mirror, parabolic shapes can be used to correct for spherical aberration

22. Chromatic Aberration • Different wavelengths of light refracted by by a lens focus at different points • Violet rays are refracted more than red rays • The focal length for red light is greater than the focal length for violet light • Chromatic aberration can be minimized by the use of a combination of converging and diverging lenses

23. Multiple Lenses in Cameras • Multiple lenses correct for various aberrations • Spherical aberration (poor focus at edge of lens) • Chromatic aberration (index of refraction varies with ) • Gauss arrangement probably most common • Actual arrangements are compromises! • No perfect corrections for all factors • Balance of many factors, including cost