Physics Solution Manual for 1100 and 2101

# Y l grating we will use the above relations to obtain

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Unformatted text preview: flected from the top surface of the plastic is one-half of a wavelength out of phase with the light reflected 1444 INTERFERENCE AND THE WAVE NATURE OF LIGHT from the bottom surface of the glass. This half-wavelength difference means destructive interference occurs, resulting in the dark fringe at A. At the second dark fringe (the first is at A), the down-and-back distance traveled by the light in the air wedge is one wavelength in the air film. At the third dark fringe, the down-andback distance traveled by the light in the air wedge is two wavelengths. Therefore, at the seventh dark fringe at B, the down-and-back distance traveled by the light in the air wedge is six wavelengths. Since the down-and-back distance is six wavelengths, the thickness of the air wedge at B is one-half of this distance, or three wavelengths. SOLUTION If the thickness of the air wedge is t, the condition for destructive interference is 1λ = 1 λair , 3 λair , 5 λair , K 12t3 + 2 2 air 2 24 2 123 14444 244444 3 Extra distance traveled by wave in the air One-half wavelength net phase change due to reflection Condition for destructive interference ( ) = m + 1 λair 2 m = 0, 1, 2, 3, K The case m = 0 corresponds to the first dark fringe at A. Therefore, the dark fringe at B corresponds to m = 6. Solving this equation for the thickness of the air wedge at B gives ( 2t = m + 1 2 ) λair − 1 λair = mλair 2 t = 1 mλair = 1 ( 6 ) ( 520 nm ) = 1560 nm 2 2 19. SSM WWW REASONING To solve this problem, we must express the condition for constructive interference in terms of the film thickness t and the wavelength λ film of the light in the soap film. We must also take into account any phase changes that occur upon reflection. SOLUTION For the reflection at the top film surface, the light travels from air, where the refractive index is smaller (n = 1.00), toward the film, where the refractive index is larger (n = 1.33). Associated with this reflection there is a phase change that is equivalent to onehalf of a wavelength. For the reflection at the bottom film surface, the light travels from the film, where the refractive index is larger (n = 1.33), toward air, where the refractive index is smaller (n = 1.00). Associated with this reflection, there is no phase change. As a result of these two reflections, there is a net phase change that is equivalent to one-half of a wavelength. To obtain the condition for constructive interference, this net phase change must be added to the phase change that arises because of the film thickness t, which is traversed twice by the light that penetrates it. For constructive interference we find that 1 2 t + 2 λ film = λ film , 2 λ film , 3 λ film , ... Chapter 27 Problems or ch 1 2 t = m + 2 λ film , 1445 where m = 0, 1, 2, ... Equation 27.3 indicates that λ film = λ vacuum / n . Using this expression and the fact that m = 0 for the minimum thickness t, we find that the condition for constructive interference becomes ch c 1 1 2 t = m + 2 λ film = 0 + 2 or t= λ vacuum 4n = FJ hλ n I GK H vacuum 611 nm = 115 nm 4 1.33 bg 20. REASONING When the light strikes the film of oil from above, the wave reflected from the top surface of the film undergoes a phase shift that is equivalent to one-half of a wavelength, since the light travels from the smaller refractive index of air toward the larger refractive index of oil. On the other hand, there is no phase shift when the light reflects from the bottom surface of the film, since the light travels from the larger refractive index of oil toward the smaller refractive index of water. Thus, the net phase change due to reflection from the two surfaces is equivalent to one-half of a wavelength in the film. This half-wavelength must be combined with the extra distance 2t traveled by the wave reflected from the bottom surface, where t is the film thickness. Thus, the condition for destructive interference is + 12t3 2 Extra distance traveled by wave in the film 1λ 12 24 4 film 3 = 1λ , 3λ , 5λ , K 2 film 2 film 2 film 144444 244444 3 Half-wavelength net phase change due to reflection Condition for destructive interference Note that the left-hand side of this equation is greater than must also be greater than 3λ . 2 film Therefore, 1λ . 2 film we 1λ . 2 film Thus, the right-hand side The smallest value that is greater than have that 2t + 1 λfilm = 2 3λ , 2 film or 1λ 2 film is the term 2t = λfilm . Since λfilm = λvacuum / nfilm = ( 640.0 nm ) / nfilm (see Equation 27.3), the condition for destructive interference becomes 2t = λfilm = 640.0 nm nfilm The condition for constructive interference is (1) 1446 INTERFERENCE AND THE WAVE NATURE OF LIGHT 1 λ′ ′ = mλfilm 1 24 43 12 24 4 film 3 + 12t3 2 Extra distance traveled by wave in the film Half-wavelength net phase change due to reflection m = 1, 2, 3,K Condition for constructive interference where λf′ilm is the wavelength that produces constructive interference in the film. Solving this relation for 2t gives λ′ ′ 2t = m − 1 λfil...
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