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
$\delta r=\left|\frac{2b\tan r}{a{\lambda }^3+b\lambda }.\delta \lambda \right|$