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solutions_chapter26

D sin u 5 m l u sin u l d y 5 r l dn 5 1 0750 m 21

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d sin u 5 m l . u sin u l D y 5 R l 0 dn 5 1 0.750 m 21 500 3 10 2 9 m 2 1 0.450 3 10 2 3 m 21 1.333 2 5 6.25 3 10 2 4 m 5 0.625 mm. y m 1 1 2 y m 5 R l d . D y 5 y m 5 R A m 1 1 2 B l d u l 5 l 0 n . n 5 1.333. D y 5 1 0.750 m 21 500 3 10 2 9 m 2 1 0.450 3 10 2 3 m 2 5 8.33 3 10 2 4 m 5 0.833 mm D y 5 y m 1 1 2 y m 5 R l d . y m 5 R A m 1 1 2 B l d u l d 5 500 3 10 2 9 m 0.450 3 10 2 3 m 5 1.11 3 10 2 3 . 6 54.3°. 6 35.5°, 6 20.4°, u 5 6 6.66°, sin u 5 A m 1 1 2 B l d 5 A m 1 1 2 B 1 0.232 2 . 6 68.1°. 6 44.1°, 6 27.6°, u 5 6 13.4°, sin u 5 m l d 5 m 1 2.78 m 12.0 m 2 5 m 1 0.232 2 . l 5 c f 5 3.00 3 10 8 m / s 107.9 3 10 6 Hz 5 2.78 m. l 5 c f . d 5 12.0 m. 6 2, c 6 1, d sin u 5 A m 1 1 2 B l , m 5 0, 6 2, c 6 1, m 5 0, d sin u 5 m l , 2.60 cm. y 5 R tan u 5 1 85.0 cm 2 tan 1.75° 5 u 5 1.75°. sin u 5 A 2 1 1 2 B l d 5 1 2.5 2 1 4.75 3 10 2 7 m 3.89 3 10 2 5 m 2 5 0.0305 d 5 m l R y m 5 3 1 4.75 3 10 2 7 m 21 0.850 m 2 0.0311 m 5 3.89 3 10 2 5 m 5 0.0389 mm Interference and Diffraction 26-3
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26.14. Set Up: The interference of these two coherent sources obeys the same equation as in the Young two-slit experiment. Solve: (a) Constructive interference occurs when Since can’t be larger than 1.00, the equation isn’t satisfied for (b) Destructive interference occurs when 26.15. Set Up: For both reflections, both from the bottom and from the top surfaces of the film, the waves reflect from a material of greater n and undergo a half-cycle reflection phase shift. The condition for destructive interference therefore is where is the wavelength in the film. The thinnest film is for Solve: Reflect: Since the path difference 2 t occurs in the film we must compare 2 t to the wavelength in the film. 26.16. Set Up: The interference is destructive since the film appears black. The waves reflected from the front surface of the film (waves in air reflecting from water) undergo a half-cycle reflection phase shift. The waves reflected from the back surface of the film (waves in water reflecting from air) do not undergo a reflection phase shift. Therefore, the condition for destructive interference is The thinnest film is for Solve: 26.17. Set Up: For both reflections, from the top and bottom surfaces of the film, the waves reflect from a material of greater n and undergo a half-cycle reflection phase shift. A nonreflective coating requires destructive interference and the condition for destructive interference is The minimum thickness is for Solve: 26.18. Set Up: Consider light reflected at the inner and outer surfaces of the film. At the inner surface of the film, light in air reflects from the film and there is a phase shift due to the reflection. At the outer surface of the film, light in the film reflects from glass and there is also a phase shift. Both rays have a phase shift due to reflection, so there is no net phase difference produced. The path difference for these two rays is 2 t , where t is the thickness of the film. The wavelength in the film is Solve: (a) Since the reflections produce no net phase difference, for constructive interference of the reflected light, The minimum thickness is 236 nm.
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