Question

The angle of incidence and refraction of a monochromatic ray of light of wavelength $\lambda$ 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 same air-glass interface. The refractive index $\mu$ of glass depends on the wavelength $\lambda$ as $\mu \left(\lambda \right)=a+b/{\lambda }^2$ where a and b are constant. Then the angular spread in the angle of refraction of the beam is$?$

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

Answer 1

The correct option is C $\left|\begin{array}{c}\frac{2b\tan r}{a{\lambda }^3+b\lambda }\delta \lambda \end{array}\right|$

From Snell's law,

$sini=\mu sinr=(a+\frac{b}{{\lambda }^2})sinr$

On differentiating with respect to $\lambda$

$0=cosrdr(a+\frac{b}{{\lambda }^2})+sinr(-2\frac{b}{{\lambda }^3})d\lambda$

Angular spread in the angle of refraction of the beam is:

$\delta r=\frac{2btan\left(r\right)\delta \lambda }{a{\lambda }^3+b\lambda }$