Phy2005 Applied Physics II Spring 2017 Announcements: Today: catch up w/ syllabus and practice Exam 3 April 10 in-class

Q2. A parallel beam of light is sent through an aquarium. If a convex lens is held in the water, it focuses the beam (…….. ……………………. ) than outside the water nair = 1, nwater = 1.33 closer to the lens (b) at the same position as (c) farther from the lens

off the optical axis as a point light source. Rules for Images Trace principal beams considering one end of an object off the optical axis as a point light source. A beam passing through the focal point runs parallel to the optical axis after a lens. A beam coming through a lens in parallel to the optical axis passes through the focal point. A beam running on the optical axis remains on the optical axis. A beam that pass through the geometrical center of a lens will not be bent. Find a point where the principle beams or their imaginary extensions converge. That’s where the image of the point source will be formed.

two focal points: f1 and f2 Parallel beams: no image!!

Virtual image Magnifying glass object object Virtual image

Glass lens (nG = 1.52) screen D x A candle is a distance D from a screen. A converging lens of focal length f is placed in between. There are 2 positions x of the lens for which a real image will be formed on the screen. Where are they? b) What is the ratio of the image sizes to the two lens positions?

screen D D-x ho f f hi x similar ho+hi ho divide: f x

hi ho f D screen x D-x Another similarity: quadratic equation x2-xD+DF=0 so

hi ho f D screen x D-x Magnification so plug in values of x for each solution

New definitions: we found: object distance p = D-x hi ho f D screen x D-x we found: New definitions: object distance p = D-x image distance q = x magnification M = hi/ho Rewrite: Or:

Exactly the same as the mirror eq.!!! Now let’s think about the sign. Lens equation and magnification 1/p + 1/q = 1/f M = -q/p Exactly the same as the mirror eq.!!! Now let’s think about the sign. positive negative p real object virtual object (multiple lenses) q real image (opposite side of object) virtual image (same side of object) f for converging lens for diverging lens M erect image inverted image

Ex. 1 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 40 cm from the lens, on the opposite side from the object, what is the focal length of the lens? 26.6 cm (2) 45.3 cm (3) 90.9 cm (4) 120.0 cm (5) 21.4 cm

Ex. 2 It is desired to use a 60-cm focal length diverging lens to form a virtual image of an object. The image is to be one- third as large as the object. Where should the object be placed and what will be the image distance in cm? (1) (55, 22.5) (2) (-165, -45.2) (3) (55, -22.5) (4) (-155, 41.3) (5) (120, -40)

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Q1 In a slide projector, the slide is 12cm from the lens Q1 In a slide projector, the slide is 12cm from the lens. The image appears on the screen 4.0 m away. What is the focal length of the lens? (1) 40 cm (2) 31 cm (3) 1.3 m (4) 11.7 cm (5) 0 cm

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