14_Ch 24 College Physics ProblemCH24 Wave Optics

14_Ch 24 College Physics ProblemCH24 Wave Optics - 316...

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316 CHAPTER 24 24.24 ±²³´µ¶·¸³¹¸º¸¹·¶» ´µ¶·¸³³¹¸º¸¹·¶» ¶¼¹ ± ² ³ ´ ³½³ ³ ³¾³ ² { G»¶·· G»¶·· From the geometry shown in the figure, () 2 22 RR tr = −+ , or 2 3 5 3.0 m 3.0 m 9.8 10 1.6 10 m tR R r =− × 2 m With a phase reversal upon reflection at the lower surface of the air layer, but no reversal with reflection from the upper surface, the condition for a bright fringe is 11 2 n air tm m n λ 1 2 m  =+   m , where 0, 1, 2, = At the bright fringe, , and the wavelength is found to be th 50 49 m = ( ) 5 2 1.6 10 m 2 6.5 12 49 .5 t m × == + 72 10 nm × 10 m 6.5
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