52. (a) From the expression for the half-width ∆ θ hw (given by Eq. 37-25) and that for the resolving power R (given by Eq. 37-29), we ±nd the product of ∆ θ hw and R to be ∆ θ hw R = µ λ Nd cos θ ¶ Nm = mλ d cos θ = d sin θ d cos θ =tan θ, where we used mλ = d sin θ (see Eq. 37-22). (b) For ±rst order m = 1, so the corresponding angle
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.