A beam of light has a small wavelength spread | vedclass

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Question

(C) According to Snell's Law,the relationship between the angle of incidence $	heta$ and the angle of refraction $	heta'$ is given by: $\sin 	heta

Vedclass · Accepted Answer

$\delta 	heta' = \left| \frac{	an 	heta'}{n} \frac{dn(\lambda)}{d\lambda} \delta \lambda ight|$

Vedclass · Answer

(C) According to Snell's Law,the relationship between the angle of incidence $	heta$ and the angle of refraction $	heta'$ is given by: $\sin 	heta = n(\lambda) \sin 	heta'$.Since the angle of incidence $	heta$ is constant,we differentiate both sides with respect to $\lambda$:$0 = \frac{d}{d\lambda} [n(\lambda) \sin 	heta']$$0 = \frac{dn}{d\lambda} \sin 	heta' + n(\lambda) \cos 	heta' \frac{d	heta'}{d\lambda}$Rearranging the terms to solve for $\frac{d	heta'}{d\lambda}$:$n(\lambda) \cos 	heta' \frac{d	heta'}{d\lambda} = -\frac{dn}{d\lambda} \sin 	heta'$$\frac{d	heta'}{d\lambda} = -\frac{1}{n(\lambda)} \frac{dn}{d\lambda} 	an 	heta'$For a small spread $\delta \lambda$,the angular spread $\delta 	heta'$ is given by $\delta 	heta' = |\frac{d	heta'}{d\lambda}| \delta \lambda$.Substituting the expression for $\frac{d	heta'}{d\lambda}$:$\delta 	heta' = \left| \frac{	an 	heta'}{n} \frac{dn(\lambda)}{d\lambda} \delta \lambda ight|$.

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