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In Fig. , the two light waves that are represented by the rays have wavelength 550.0 nm before entering media 1 and 2. They also have equal amplitudes and are in phase. Medium 1 is now just air, and medium 2 is a transparent plastic layer of index of refraction 1.600 and thickness 2.600 mum. What is the phase difference of the emerging waves in wavelengths, radians, and degrees? What is their effective phase difference (in wavelengths)? |
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Answer» Solution :KEY IDEA The phase difference of TWO light waves can change if they travel through different media, with different indexes of refraction. The reason is that their wavelengths are different in the different media. We can calculate the change in phase difference by counting the number of wavelengths that fits into each medium and then subtracting those numbers. Calculations: When the path lengths of the waves in the two media are IDENTICAL, Eq. 35-9 gives the result of the SUBTRACTION. Here we have `n_(1)= 1.000` (for the air), `n_(2) = 1.600`, L = 2.600 MM, and `lambda = 550.0 nm`. Thus, Eq. 35-9 yields `N_(2)-N_(1)=(L)/(lambda)(n_(2)-n_(1))` `=(2.600 xx 10^(-6)m)/(5.500 xx 10^(-7) m)(1.600-1.000)` = 2.84. (Answer) Thus, the phase difference of the emerging waves is 2.84 wavelengths. Because 1.0 wavelength is equivalent to `2pi` rad and `360^(@)`, you can show that this phase difference is equivalent to phase difference = 17.8 rad `~~ 1020^(@)` (Answer) The effective phase difference is the DECIMAL part of the actual phase difference expressed in wavelengths. Thus, we have effective phase difference = 0.84 wavelength. (Answer) You can show that this is equivalent to 5.3 rad and about `300^(@)`. Caution: We do not find the effective phase difference by taking the decimal part of the actual phase difference as expressed in radians or degrees. For example, we do not take 0.8 rad from the actual phase difference of 17.8 rad. |
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