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(a) How many rulings must a 4.00-cm-wide diffraction grating have to resolve the wavelengths 415.496 and 415.487 nm in the second order? (b) At what angle are the second-order maxima found?

Short Answer

Expert verified
  1. The number of rulings is 23100.
  2. The second-order maxima can be found at θ=28.7°.

Step by step solution

01

The resolving power

It is known the the resolving power of a grating is given byR=Nm=λavg∆λ , where Nis the number of rulings in the grating andm is the order of the lines, λavgis the average of wavelengths and∆λ is the separation.

02

The number of rulings

It is known that theNm=λavgΔλ . From the given values in the question, we have

N2=415.496+415.4872415.496-415.487N2=830.98320.009N=23100

Thus, the number of rulings is 23100.

03

The second-order maxima

The width of the single grating, d, is given byd=LN. So, the width can be calculated as follows:

d=LN=0.0423100=1.732×10-6 m

The angle of maxima can be calculated as follows:

sinθ=mλdsinθ=2415.496×10-91.73.×10-6θ=sin-1830.9921730θ=28.7°

Thus, the second-order maxima can be found atθ=28.7°

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