The angles of incidence and refraction of a m | vedclass

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(C) From Snell's law,we have $\sin i = \mu \sin r$.Substituting $\mu(\lambda) = a + \frac{b}{\lambda^2}$,we get $\sin i = (a + \frac{b}{\lambda^2}) \s

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$\left|\frac{2 b 	an r}{a \lambda^3+b \lambda} \delta \lambdaight|$

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(C) From Snell's law,we have $\sin i = \mu \sin r$.Substituting $\mu(\lambda) = a + \frac{b}{\lambda^2}$,we get $\sin i = (a + \frac{b}{\lambda^2}) \sin r$.Differentiating both sides with respect to $\lambda$,keeping $i$ constant (since the incident angle is fixed):$0 = \frac{d}{d\lambda} [(a + \frac{b}{\lambda^2}) \sin r]$$0 = (a + \frac{b}{\lambda^2}) \cos r \frac{dr}{d\lambda} + \sin r (-\frac{2b}{\lambda^3})$.Rearranging the terms to solve for $\frac{dr}{d\lambda}$:$(a + \frac{b}{\lambda^2}) \cos r \frac{dr}{d\lambda} = \frac{2b}{\lambda^3} \sin r$.$\frac{dr}{d\lambda} = \frac{2b \sin r}{\lambda^3 (a + \frac{b}{\lambda^2}) \cos r} = \frac{2b 	an r}{\lambda^3 (a + \frac{b}{\lambda^2})} = \frac{2b 	an r}{\lambda(a\lambda^2 + b)}$.Thus,the angular spread $\delta r$ is given by $|\frac{dr}{d\lambda}| \delta \lambda = |\frac{2b 	an r}{a\lambda^3 + b\lambda}| \delta \lambda$.

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