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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Spring &09 Solutions to Problem Set #3 Due Wednesday, March 4, 2009 Problem 1: Wanda&s world a) The geometry for this problem is shown in Figure 1. For part (a), the object (Wanda) is located inside the bowl and we are interested to &nd where the image is formed. We start by using the matrix formulation to analyze the given system, & & s ¡ ) i x i ¡ = & 1 0 1 & 1 & (1 & n 1 & R 1 ¡& R n 1 n ¡& & o x o ¡ (1) = " 1 (1 & n ) n R R + s n 1 + s (1 & n ) n R # & n& o : x o ¡ To &nd the imaging condition, we note that all the rays, regardless of their departure angle & o , arrive at the image point x i (i.e. @x i =@& o = 0 ), R s + n = 0 ) (2) n s = & R: We see that the image is formed at the center of the bowl and is virtual . Using this result in equation 1, & & 1 i x i ¡ = & (1 & n ) n o R ¡ n& n & x o ¡ : (3) The lateral magni&cation is, x i M L = = n; (4) x o and therefore, the image is erect . b) For this part, the object (Olive) is located outside the bowl and we are interested to &nd where the image is formed. Again, we solve this part using the matrix formulation, & n& i x i ¡ = & 1 s 1 1 n ¡& & ( n & 1) 1 R 1 ¡& s 1 ¡& & o x o ¡ (5) = " 1 & s ( n & 1) ( R & n & 1) R s + s n & ss ( n & 1) 1 nR & s ( n & 1) & nR # & o x o ¡ : 1 From the imaging condition ( @x i =@& o = 0) ; we solve for s , s ss ( n + s & 1) n & = 0 nR ) (6) snR s = : s ( n & 1) & R The lateral magni&cation is (for an onaxis ray, & o = 0 ), x i M L = s ( n = 1 & 1) x o & (7) nR snR ( n = 1 & 1) & R = ( s ( n & 1) & R ) nR & : s ( n & 1) & R From equations 6 and 7, the following cases arise: 1. If R > s ( n & 1) ! s < ; the image is virtual, erect and is located at a distance j s j outside the bowl. 2. If R > s ( n & 1) ! s > ; the image is real, inverted and is located at a distance j s j inside the bowl. c) If we were to consider the glass container of thickness t as well as the inner, R 1 ; and outer, R 2 , radii, the matrix formulation becomes, & n& i x i ¡ = & 1 s 1 1 n ¡& & ( n & n g ) 1 R 1 1 ¡& t 1 1 n g ¡& & ( n g & 1) 1 R 2 1 ¡& s 1 ¡& & o x o ¡ (8) = & 1 s 1 1 n ¡ " & t ( n & n g ) P R 1 n g t 1 n g & t ( n g & 1) 1 R 2 n g # & s 1 ¡& n& o x o ¡ = " 1 + Ps & t ( n & n g ) P R 1 n g s 1 n ¢ & t ( n & n g ) + R 1 n g £ t + s n g ¢ s P + 1 n & t ( n g & 1) 1 R 2 n g £ + s P t n & ( n g & 1) & R 2 n g # & o x o ¡ ; where, P = & & ( n & n g ) ( n g + & 1) R 1 t R 2 & ( n & n g ) ( n g & 1) ¡ : (9) n g R 1 R 2 For the case of uniform glass: R 1 = R 2 = R . The imaging condition is, s ss ( n + s & 1) n & s t ( n & n nR & g ) t + Rn g n st ( n n g & g & 1) ss + Rn g t nR ¤ ( n & n g ) ( n g & 1) = Rn g ¥ 2 Figure 1: Wanda&s world problem s ss ( n + s & 1) n & + & g = 0 ; (10) nR where, s t ( n & g = & n & g ) t + Rn g n st ( n n g & g & 1) ss + Rn g t nR & ( n & n g ) ( n g & 1) 1 R ¡ : (1 ) n g Comparing equations 10 and 6, we see that in order to neglect the aquarium walls we require that &...
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.710 taught by Professor Sebaekoh during the Spring '09 term at MIT.
 Spring '09
 SeBaekOh

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