White light (400 nm - 700 nm) diffraction pattern lights up a screen that is 3.60 m away. The diffraction grating has 503 slits per mm, which produces a rainbow of diffracted light. What is the width of the first order rainbow on the screen?
The distance between slits is from the information given.
d =1 mm / 503
=19.88 ×10-4 mm
0r 19.8 × 10−7 m. Let us call the two angles θv for violet (400 nm) and θR for red (700 nm).
Solving the equation d sin θV = mλ for sin θV,
sinθv = mλV/d
sinθV = mλV/d, where m = 1 for first order and λV = 400 nm = 4.00 × 10−7 m. Substituting these values gives
sinθV = 4.0 ×10-7 m / 19.88 ×10-7 m
sinθV = 0 .201
Thus the angle θV is θV = sin−1 0.210 = 12.12 º.
Similarly,
sinθR = 7.00×10-7 m / 19.88 ×10-7 m
sinθR = 0 .35
Thus the angle θR is θR = sin−1 0. 35 = 20.48 º.
Solution for Part 2
The distances on the screen are labeled yV and yR
Noting that tanθ= y / x
tanθ solve for yV and yR.
That is, yV = x tan θV = ( 3.60 m)(tan 12.12º) = 0.756 m
and yR = x tan θR = ( 3.6 m)(tan 20.48º) = 1.34 m.
The distance between them is therefore yR − yV = 1.34 - 0 .756 m.
the width of first order rainbow = 0.59 m
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