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Unformatted text preview: 1 University of California, Berkeley Physics H7C Spring 2011 ( Yury Kolomensky ) SOLUTION TO PROBLEM SET 4 Geometrical Optics Maximum score: 200 points Due: February 23, 2011 (note date !) Composed and formatted by Yu. Kolomensky, and T.R. Beals 1. (25 points) ( Hecht 5.6 ) Show that, in the paraxial domain, the magnification produced by a single spherical interface between two continuous media, as shown in Fig. P.5.6 (Hecht), is given by M T = n 1 s i n 2 s o 1. Solution From Hecht Eq. (5.24), we have M T ≡ y i y o . From the diagram below, we see that y i y o = s i R s o + R , since the tri angles are similar. Hecht Eq. (5.8) gives us n 1 s o + n 2 s i = n 2 n 1 R . Combining these equations, we can find the transverse magnification: M T ≡ y i y o = s i R s o + R = s i h 1 s o ( n 2 n 1 ) n 1 s i + n 2 s o i s o h 1 + s i n 1 s i + n 2 s o i = s i s o n 1 ( s i + s o ) n 2 ( s i + s o ) = s i n 1 s o n 2 s o s i R y o y i n 1 n 2 There’s also another way to do this problem, as hinted at by the diagram in Hecht (Figure P.5.6). 2. (25 points) ( Hecht 5.8 ) Locate the image of the object placed 1.2 m from the vertex of a gypsy’s crystal ball, which has a 20 cm di ameter ( n = 1 . 5 ). Make a sketch (of the rays, not the gypsy). 2. Solution It’s helpful to use Hecht’s Figure 5.14 as a reference. We then take Hecht’s Eq. (5.11) and Eq. (5.13), with s o 1 = 1 . 2 m , R 1 = R 2 = R = 0 . 1 m , n m = 1 . , and n ` = 1 . 5 . We will measure the distance to the image x from the vertex of the ball (the common point) and assume that it is right of the vertex, so s i 2 = x 2 R . This gives us the following relations: 1 s o 1 + n s i 1 = n 1 R (1) n s i 1 + 2 R + 1 x 2 R = n 1 R (2) We now have two equations with two unknowns: x , which is the distance from the front of the crystal ball to the real image, and s i 1 , which is the distance from the front the ball to the virtual image. From Eq. (1), we obtain s i 1 = Rs o 1 n ( n 1) s o 1 R = 36 cm The virtual image is behind the crystal ball. Plugging this into Eq. (2), we find the location of the real image: x = 2 R + R ( s i 1 2 R ) nR + ( n 1)( s i 1 2 R ) = 27 cm 2 So the image is just behind the ball (7 cm from the op posite end). It’s demagnified ( s i 2 < s o 1 ) and inverted. s o1 s i2 R s i1 3. (20 points) ( Hecht 5.15 ) Prove that the minimum separation between conjugate real object and image points for a thin positive lens is 4 f . 3. Solution From the ThinLens Equation (Hecht Eq. (5.17)), we know that s o = fs i s i f . Thus, d sep = s i + s o = s i 1 + f s i f = s 2 i s i f . To minimize, we take the derivative of d sep and set it equal to zero: 0 = d ds i s 2 i s i f = s i ( s i 2 f ) ( s i f ) 2 This tells us that the minimum occurs when s i = 2 f , and by plugging this into our expression for d sep , we see the minimum is 4 f ....
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This note was uploaded on 03/28/2012 for the course PHYSICS H7c taught by Professor Staff during the Spring '08 term at Berkeley.
 Spring '08
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