Question

A parallel beam of sodium light of wavelength $5890\mathring{A}$ 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

• A. $3926\mathring{A}$
• B. $4353\mathring{A}$
• C. $1396\mathring{A}$
• D. $1921\mathring{A}$

Answer 1

Answer: (A)

The condition for minimum thickness corresponding to a dark band is,
$2\mu tcosr=\lambda$
$\therefore t=\frac{\lambda }{2\mu cosr}$
$=\frac{5890\times 10^{-10}}{2\times 1.5\times cos60^{\circ }}$
$=3926\times 10^{-10}m$
$=3926\mathring{A}$