hw7sol - HW7 solution March 3, 2008 From the given...

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Unformatted text preview: HW7 solution March 3, 2008 From the given conditions, ZABC = 90° [ADC = 90° 50— = A1 A7? = A2 E = h fl=fi 112 A1 (1) (2) (3) (4) (5) (6) We will look at two triangles, AABC and AADC to prove Snell’s law. 9i + AEAD = 90° 0 + ZEAD = 90° 01: a (7) , and 6; + (BAG = 90° [3 + ABAC' = 90° 9t = fl We can write :4? in two different ways, (9) = (10) _ )‘i AC = sin(0i) __ _ )‘t AC — sin(0¢) Ai At sin(9i) = sin(6t) A; Sin(0i) 1' At Sin(0t) Ai n, since — = -- /\t m n; sin(0i) = nt sin(0¢) m < n; 01 > 92 A1 > A2 n1 sin(01) = M sin(02) (9) (10) (11) (12) (13) (14) nair Sin(9i) = “water Sin(9t) sin(6t) = “a” sin(0,-) (22) “water 0: = (23) f = 5 x 10“ Hz (15) 0 "air = 1 = 'Uair Uair = c = 3 x 108 m/sec (16) A I _ 2 We need to prove that incident angle is equal to a" _ f reflected angle. From AABC and AABD, = 6 x 10-7 __ -m AB is a common line (24) = m (17) 7E=E= ,\ (25) (LADB = AACB = 90° (26) b ) Thus, AABC and AABD are congruent (RHS). The frequency does not change in different media. fwater = 5 x 1014 Hz c . 5 vwater = ("water IS nwmr W dt tht'ATB" uni—D ‘ e nee o prove a is para e o . given = 2.26 x 108 m/sec (19) conditions are, A = vwater ___ — water f = 452 x 10‘7m (we assume that two surfaces are parallel) [ABG = 0,- = (20) ACBE = t9t From Snell’s law, a) . _1 “glass - When refracted angle is 90 degree, the incident angle 1% = sm (—-—— sm 19¢) (27) is the critical angle. “glass Sin(0t) = flair sin(0¢) nair Also, ncabze Sin 9m = ’nblaod Sin(90°) (36) = = 9¢(because FE— acri = sin—1(nblood) TLM‘T sin(£DCH) = ngla“ sin(0t) (29) “cable a .s = 3 ADCH = sin—1("—9‘-5— sin 0,) (30) ( 8) fl nai'r plug into the eq. b) = sin—1(ngla” "a" sin 01.) The angle between the axis and a light ray is, "air "glass 0 t (31) 45 = 90 — 6 (39) = 9i (32) Then, the maximum ¢ is, = [ABC 33 ( ) ¢mam = 90° _ aori (40) Since [FBE is a vertical angle of [ABC and ZFCE = (41) is a vertical angles of ADCH, 7 AFBE = AFCE (34) 7— 3 9 a) From and ’ When an image is inverted and manification is 1, M value is -1.Then, [180 = AJCB (35) I __ __ M = —1 = —s— (42) Since ZIBC and AJCB are the same, AB II CD. I s s = s (43) Thus, the direction does not change. From the thin lense equation, 1 1 1 7 — g + E (44) use (36) 1-3 f _ 5’ f I — .— 5—2 = 20 cm W (a 10 0 From the ray diagrams, the image is smaller and it is closer to the focus as the object is moved far away from the lense. note: you can see the same result from the lense equa— tion. M:—% Ma %=§+§ an s' = siff (48) From the equations, when 3 z f, 3’ becomes a large number. Thus, magnification is large as well. When 5 is infinity, 3’ z f and magnification is zero. This is consistent with our result from the diagrams. Figure 1: an object close to a focus Figure 2: an object far way from a focus N H ('0 | N \ U1 The magnification is, 5/3 M- "23 = —2/3 From the diagram, 3' = 5.66 — 4 = 1.66 M = h’/h = —0.66/1 = ~0.66 (51) (52) (53) (54) (55) results are consistent with the thin lense equation. 0) From this diagram, we can see that an image is not formed when an object is at the focus(the image is formed at infinity.). d) l __ 1 + 1 f — s’ s 1 _ 1 + 1 1 " s’ 0.5 1 g = 1 — 2 s’ = The magnification is, —1 M = —— 0.5 = 2 From the diagram, 3' =3—4 — —1 M =h’/h = 1/0.5 = 2 (56) (57) (58) (59) (60) (61) (62) results are consistent with the thin lense equation. 1 __ 1 1 7—; g 1 __ 1 1 37% §=—1—0.5 —-2/3 M - ""2— = —1/3 From the diagram, 5’ = 3.34 — 4 = —0.66 M = h’/h = —0.66/1 = —Q.66 (63) (64) (65) (66) (67) (68) (69) results are consistent with the thin lense equation. b) From the thin lense equation, l=l+l f s’ s s’ = fs (70) (71) f > 0 for a converging lense. Thus when 3 = f, the 3' will diverge. On the other hand, f < O for a diverging lense, so the denominator of s’ is always bigger than f as long as s > 0. Thus, 3’ value does not show any dramatical change for a diverging lense. ...
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hw7sol - HW7 solution March 3, 2008 From the given...

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