having refractive index μ=1.5 such that the angle of refraction in the plate is 60°. Calculate the smallest thickness of the glass plate which will appear dark by reflected light.

Solution:

$\lambda = 5880 A^0 = 5.880 \times 10^{-7}m$

$\mu = 1.5$

$\theta = 60^0$

$2 \mu t cos \theta = n \lambda; \text{ n = 1,2,3}$

For smallest thickness, n=1

$\begin{aligned} \therefore t_{min} &= \frac{\lambda}{ 2 \mu cos \theta} \\ &= \frac{5.880 \times 10^{-7}}{2 \times 1.5 \times cos60^0} \\ &= 3920 \times 10^{-10} m \\ &= 3920 A^0 \end{aligned}$