InterviewSolution
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InterviewSolution
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A transparent diffraction grating has a period `d = 1.50mum`. Find the angular disperison `D` (in angular minutes per nanometers) corresponding to the maximum of highest order for a spectral line of wavelength `lambda = 530mm` of light falling on the grating (a) at right angles, (b) at the angle `theta_(0) = 45^(@)` to the normal. |
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Answer» For normal incidence, the maxima are given by `d sin theta= n lambda` so `sintheta = n(lambda)/(d) = nxx (0.530)/(1.500)` Clearly `nle 2` as `sin theta gt 1` for `n = 3`. Thus the highest order is `n = 2`. Then `D=(d theta)/(d lambda) = (k)/(d cos theta) = (k)/(d) (1)/(sqrt(1-((k lambda)/(d))^(2)))` Putting `k =2, lambda =0.53mum, d= 1.5mum = 1500 nm` we get `D = (2)/(1500) (1)/(sqrt(1-((1.06)/(1.5))^(2))) xx (180)/(pi) xx 60 =6.47ang.min//nm`. (b) We write the diffraction formula as `d(sin theta_(0) + sin theta) = k lambda` so `sin theta_(0) + sin theta = k (lambda)/(d)` Here `theta_(0) = 45^(@)` and `sin theta_(0) = 0.707` so `sin theta_(0)+sintheta le 1.707`. Since `(lambda)/(d) = (0.53)/(1.5) = 0.353333`, we see that `k le4` Thus highest order corresponding to `k = 4`. Now as before `D = (d theta)/(d lambda)`so `D = (k)/(d cos theta) = (k//d)/(sqrt(1-((k lambda)/(d)-sintheta_(0))^(2)))` `= 12.948ang. min//nm`, |
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