280CHAPTER 23 23.23Since the center of curvature of the surface is on the side the light comes from, 0R<giving . Then, 4.0 cmR=−122nnnn1pqR−+=becomes 1.001.004.q−=−1.501.500 cm4.0 cm−, or 4.0 cmq= −Thus, the magnification 12qpnhMhn′==−, gives ()()()121.504.0 cm3.8 mm1.004.0 cmn qhhn p−′ = −= −2.5 mm=23.24Light scattered from the bottom of the plate undergoes two refractions, once at the top of the plate and once at the top of the water. All surfaces are planes ()R→∞, so the image distance for each refraction is ()21qnn=−p. At the top of the plate, ()1B1B1.3338.00 cm6.42 cm1.66waterglassnqpn= −= −= −or the first image is 6.42 cm below the top of the plate. This image serves as a real object for the refraction at the top of the water, so the final image of the bottom of the plate is
This is the end of the preview.
access the rest of the document.