ch36-p064 - 64(a From the expression for the half-width...

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() 11 1 22 2 1 1 1 2 sin sin 1 1 tan cos 1s i n 1/sin 1 / 1 1 0.89. 900nm/600nm 1 d θ == = = 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 hw and R to be θθ hw R Nd Nm m d d d = F H G I K J === λλ cos cos sin cos
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