Question

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?

Answer 1

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

sin⁡θV = 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

sin⁡θV = 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

sin⁡θR = 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

tan⁡θ 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