()1112221112sinsin11tancos1sin1/sin1/110.89.900nm/600nm1dθ====−−λ−−64. (a) From the expression for the half-width ∆θ hw(given by Eq. 36-28) and that for the resolving power R(given by Eq. 36-32), we find the product of ∆hwand Rto be ∆θθhwRNdNmmddd=FHGIKJ===λλcoscossincos
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