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Lecture 14 Images Chp. 35

Lecture 14 Images Chp. 35. Opening Demo Topics Plane mirror, Two parallel mirrors, Two plane mirrors at right angles Spherical mirror/Plane mirror comparison in forming image Spherical refracting surfaces Thin lenses Optical Instruments Magnifying glass, Microscope, Refracting telescope

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Lecture 14 Images Chp. 35

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  1. Lecture 14 Images Chp. 35 • Opening Demo • Topics • Plane mirror, Two parallel mirrors, Two plane mirrors at right angles • Spherical mirror/Plane mirror comparison in forming image • Spherical refracting surfaces • Thin lenses • Optical Instruments Magnifying glass, Microscope, Refracting telescope • Warm-up problem

  2. Geometrical Optics:Study of reflection and refraction of light from surfaces using the ray approximation. The ray approximation states that light travels in straight lines until it is reflected or refracted and then travels in straight lines again. The wavelength of light must be small compared to the size of the objects or else diffractive effects occur. Reflection Refraction: Snells Law Using Fermat’s Principle you can prove the above two laws. It states that the path taken by light when traveling from one point to another is the path that takes the shortest time compared to nearby paths. n1 n2

  3. Plane mirrors Real side Virtual side Angle of incidence i = - p Normal Virtual image Angle of reflection

  4. eye object 2 1 3 Problem: Two plane mirrors make an angle of 90o. How many images are there for an object placed between them? mirror mirror

  5. mirror 2 4 object mirror 5,6 1 3 Problem: Two plan mirrors make an angle of 60o. Find all images for a point object on the bisector.

  6. How tall must be a mirror be for a person to see his entire reflection in it. Let H be the height of a man and let l be the height of the mirror?

  7. i = - p magnification = 1 What happens if we bend the mirror? Concave mirror. Image gets magnified. Field of view is diminished Convex mirror. Image is reduced. Field of view increased.

  8. Thin Lenses: thickness is small compared to object distance, image distance, and radius of curvature. Converging lens Diverging lens

  9. Thin Lens Equation Lensmaker Equation What is the sign convention?

  10. p Sign Convention Real side - R Virtual side - V Light r1 r2 i Real object - distance p is pos on V side (Incident rays are diverging) Radius of curvature is pos on R side. Real image - distance is pos on R side. Virtual object - distance is neg on R side Incident rays are converging) Radius of curvature is neg on the V side. Virtual image- distance is neg o the V side.

  11. Rules for drawing rays to locate images • A ray initially parallel to the central axis will pass through the focal point. • A ray that initially passes through the focal point will emerge from the lens • parallel to the central axis. • A ray that is directed towards the center of the lens will go straight through the lens undeflected.

  12. Show Fig 35-14 figure overhead Then go to example

  13. Virtual side Real side . . F1 p F2 24(b). Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Image is real, inverted.

  14. . . r1 r2 F1 p F2 24(e). Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays. Real side Virtual side Image is virtual, upright.

  15. Lens 1 Lens 2 +20 -15 f1 f2 f1 f2 40 10 27. A converging lens with a focal length of +20 cm is located 10 cm to the left of a diverging lens having a focal length of -15 cm. If an object is located 40 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. Treat each lens Separately.

  16. Lens 1 Lens 2 30 40 +20 -15 f1 f2 f1 f2 40 10 Ignoring the diverging lens (lens 2), the image formed by the converging lens (lens 1) is located at a distance Since m = -i1/p1= - 40/40= - 1 , the image is inverted This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.

  17. Lens 1 Lens 2 30 40 +20 -15 f1 f2 f1 f2 40 10 Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i2 < 0). The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image has the same size orientation as the object.

  18. Optical Instruments Magnifying lens Compound microscope Refracting telescope

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