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Q. A parallel beam of sodium light of wavelength $6000\, \mathring{A}$ is incident on a thin glass plate of refractive index $1.5$ such that the angle of refraction in the plate is $60^{\circ}$. The smallest thickness of the plate which will make it dark by reflection.

Wave Optics

Solution:

The condition for minimum thickness corresponding to a dark band in reflection
$2 \mu t \cos r-\lambda$
$\therefore t=\frac{\lambda}{2 \mu \cos r}=\frac{6000 \times 10^{-10}}{2 \times 1.5 \times \cos 60^{\circ}}$
$=4000\,\mathring{A}$