Physics 212 Lecture 26: Lenses

1. Physics 212 Lecture 26: Lenses

2. Main Point 1 First, we used two principal rays to derive the lens equation , that the sum of the inverse object and image distances is equal to the inverse of the focal length of the lens. We determined that this equation holds for both converging and diverging lenses, if we follow the convention that converging lenses have positive focal lengths while diverging lenses have negative focal lengths. Real images, corresponding to positive image distances, are produced by converging lenses when the object distance is greater than the focal length. Virtual images, corresponding to negative image distances, are always produced by diverging lenses and also by converging lenses when the object distance is less than the focal length.

3. Main Point 2 • Second, we defined  the magnification in terms of the ratio of the image size to the object size. Using the lens equation, we obtained the result that the magnification is equal to minus the focal length divided by the difference of the object distance and the focal length. The sign convention used is that a negative magnification indicates an inverted image while a positive negative magnification indicates an upright image. For all single lenses, real images are inverted and virtual images are upright.

4. Main Point 3 Finally, we used Snell’s law to derive the lensmaker’s formula  that relates the focal length of the lens to the index of refraction and the radii of curvature of the surfaces.

5. Refraction Snell’s Law n1sin(q1) = n2sin(q2) q1 n1 n2 q2 That’s all of the physics – everything else is just geometry!

6. q2 q2 air water qi qi 1.3 glass glass 1.5 1.5 CaseI CaseII In Case I light in air heads toward a piece of glass with incident angleqi In Case II, light in water heads toward a piece of glass at the same angle. In which case is the light bent most as it enters the glass? I or II or Same (A) (B) (C)

7. Checkpoint 1a A converging lens is used to project the image of an arrow onto a screen as shown above.The image is A. real and upright B. real and invertedC. virtual and upright D. virtual and inverted

9. Checkpoint 1b A converging lens is used to project the image of an arrow onto a screen as shown above. A piece of black tape is now placed over the upper half of the lens. Which of the following is true? A. Only the lower half of the object (i.e. the arrow tail) will show on the screenB. Only the upper half of the object (i. e. the arrow head) will show on the screenC. The whole object will show on the screen

10. Checkpoint 2 What happens to the focal length of a converging lens when it is placed under water? A. increases B. decreases C. stays the same

13. 1) Draw ray parallel to axis refracted ray goes through focus 2) Draw ray through center refracted ray is symmetric image You now know the position of the same point on the image Recipe for finding image: object f

15. f Diverging Lens: Consider the case where the shape of the lens is such that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance fin front of the mirror:

16. image is: virtual upright smaller object f image f<0 S S’<0

17. Executive Summary - Lenses: real inverted smaller S > 2f converging f f real inverted bigger 2f > S > f virtual upright bigger f > S > 0 virtual upright smaller S > 0 diverging

19. We can also see this from the “Lensmaker’s Formula” nlens nair This is why you can make thinner glasses out of special materials that have n greater than nglass. Recall glass rod in oil demo – n was the same and you saw no refraction at all

20. Calculation A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted? • Conceptual Analysis • Strategic Analysis