Assignment 19 Query

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course Phy 122

3/10 12 pm

019. `Query 17

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Question: `qGeneral College Physics and Principles of Physics Problem 24.2: The third-order fringe of 610 nm light created by two narrow slits is observed at 18 deg. How far apart are the slits?

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Your solution:

Path difference for a third-order fringe is 3 wavelengths. Therefore, light from one slit travels 3 * 610 nm = 1830 nm

To find the split spacing, we take a sin(18 deg) = 1830 nm

Now we have to solve for a.

a = 1830 nm / sin(18 deg) = 5922.3 m

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Given Solution:

`aThe path difference for a 3d-order fringe is 3 wavelengths, so light from one slit travels 3 * 610 nm = 1830 nm further.

The additional distance is equal to slit spacing * sin(18 deg), so using a for slit spacing we have

a sin(18 deg) = 1830 nm.

The slit spacing is therefore

a = 1830 nm / sin(18 deg) = 5920 nm, or 5.92 * 10^-6 meters.

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