Question

The angles of incidence an d re fraction of a monochromatic ray of light of wave length $\lambda$ at an air-glass interface are $i$ and $r,$ respectively. $A$ parallel beam of light with a small spread $\delta\lambda$ in wavelength about a mean wavelength $\lambda$ is refracted at the samenair-glass interface. The refractive index $\mu$ of glass depends on the wavelength $\lambda$ as $\mu (\lambda) = a + b /\lambda^{2}$, where $a$ and $b$ are constants. Then, the angular spread in the angle of refraction of the beam is

• A. $\left|\frac{\sin i}{\lambda^{3} \cos r}\delta\lambda\right|$
• B. $\left|\frac{2b}{\lambda^{3}}\delta\lambda\right|$
• C. $\left|\frac{2b\, \tan\,r}{a\lambda^{3}+b\lambda}\delta\lambda\right|$
• D. $\left|\frac{2b \left(a+b/ \lambda^{2}\right)\sin i}{\lambda^{3}}\delta\lambda\right|$

Answer 1

Answer: (C)

From Snell’s law, we have
sin $i=\mu \sin\, r$
$\Rightarrow \sin i=\left(a+\frac{b}{\lambda^{2}}\right) \sin r $
Differentiating with respect to $\lambda,$ we get
$\Rightarrow \frac{d}{d\lambda}\sin i =\frac{d}{d\lambda}\left(a+\frac{b}{\lambda^{2}}\right). \sin r $
$\Rightarrow 0=\left(a+\frac{b}{\lambda^{2}}\right) ,\cos r. \frac{dr}{d\lambda}+\sin r \left(\frac{-b}{\lambda^{3}}\right)$
$\Rightarrow \frac{b}{\lambda^{3}}\sin r =\left(a+\frac{b}{\lambda^{2}}\right) \cos r \frac{dr}{d\lambda}$
$\Rightarrow dr=\frac{2bd\lambda}{\lambda}.\frac{\tan r}{\left(a\lambda^{2}+b\right)} $
Angular spread of angle of refraction is
$\delta r=\left|\frac{2b\tan r}{a\lambda^{3}+b\lambda}. \delta \lambda \right|$