1016_PartUniversity Physics Solution

1016_PartUniversity Physics Solution - 35-16 Chapter 35 At...

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35-16 Chapter 35 EXECUTE: At a point on the screen located by the angle θ the difference in path length is sin . d This introduces a phase difference of 0 2 (s i n) , d π φ λ ⎛⎞ = ⎜⎟ ⎝⎠ where 0 is the wavelength of the light in air or vacuum. In the thickness L of glass the number of wavelengths is 0 . L nL = A corresponding length L of the path of the ray through the lower slit, in air, contains 0 / L wavelengths. The phase difference this introduces is 00 2 nL L φπ =− and 0 2( 1 ) (/ ) . nL πλ The total phase difference is the sum of these two, 0 2 i n )2 ( 1 ) ( /)( 2 /) i n ) ) . dn L d L n θπ +− = + Eq.(35.10) then gives 2 0 0 cos ( sin ( 1)) . II d L n ⎡⎤ =+ ⎢⎥ ⎣⎦ (c) Maxima means cos / 2 1 and / 2 , 0, 1, 2, mm = =±± 0 i n ) ) dL nm λθ = 0 sin ( 1) n m = 0 (1 ) sin mL n d −− = EVALUATE: When 0 L or 1 n the effect of the plate goes away and the maxima are located by Eq.(35.4). 35.60. IDENTIFY: Dark fringes occur because the path difference is one-half of a wavelength. SET UP: At the first dark fringe, d sin = /2. The intensity at any angle is given by 2 0 sin cos d = . (a) At the first dark fringe, we have d sin = /2 d/ = 2/(2 sin 15.0&) = 1.93 (b) 2 0 0 sin cos 10 dI πθ == sin 1 cos 10 d = sin 1 arccos 10 d = = 71.57& = 1.249 rad Using the result from part (a), that d / = 1.93, we have π (1.93)sin = 1.249. sin = 0.2060 and = –11.9& EVALUATE: Since the first dark fringes occur at –15.0&, it is reasonable that at 12& the intensity is reduced to only 1/10 of its maximum central value. 35.61. IDENTIFY: There are two effects to be considered: first, the expansion of the rod, and second, the change in the rod±s refractive index. SET UP: 0 n = and 51 0 (2.50 10 (C ) ) nn T Δ= × Δ ° . 61 0 (5.00 10 (C ) ) L LT Δ Δ ° . EXECUTE: The extra length of rod replaces a little of the air so that the change in the number of wavelengths due to this is given by: glass glass 0 air 1 0 22 ( 1 ) 2 n L T N α λλ Δ− Δ Δ = and 6 1 7 2(1.48 1)(0.030 m)(5.00 10 C )(5.00 C ) 1.22. 5.89 10 m N −× ° ° = × The change in the number of wavelengths due to the change in refractive index of the rod is: 5 glass 0 2 7 0 2 2(2.50 10 C )(5.00 C min)(1.00 min)(0.0300 m) 12.73. 5.89 10 m N Δ ×° ° = = × So, the total change in the number of wavelengths as the rod expands is 12.73 1.22 14.0 N Δ = fringes/minute. EVALUATE: Both effects increase the number of wavelengths along the length of the rod. Both L Δ and glass n Δ are very small and the two effects can be considered separately. 35.62. IDENTIFY: Apply Snell’s law to the refraction at the two surfaces of the prism. 1 S and 2 S serve as coherent sources so the fringe spacing is R y d , where d is the distance between 1 S and 2 S.
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Interference 35-17 SET UP: For small angles, sin θ , with expressed in radians.
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1016_PartUniversity Physics Solution - 35-16 Chapter 35 At...

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