Light of wavelength 6000 text{ Å} is incident | vedclass

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(D) For reflected light, the condition for destructive interference (dark fringe) is given by:$2 \mu t \cos r = n \lambda$where $\mu$ is the refractiv

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$4 	imes 10^{-7} 	ext{ m}$

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(D) For reflected light, the condition for destructive interference (dark fringe) is given by:$2 \mu t \cos r = n \lambda$where $\mu$ is the refractive index, $t$ is the thickness, $r$ is the angle of refraction, and $n$ is an integer $(n = 1, 2, 3, ...)$.For the smallest thickness, we take $n = 1$:$2 \mu t \cos r = \lambda$Given values:$\lambda = 6000 	ext{ Å} = 6000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m}$$\mu = 1.5$$r = 60^{\circ} \Rightarrow \cos 60^{\circ} = 0.5$Substituting these values into the equation:$2 	imes 1.5 	imes t 	imes 0.5 = 6 	imes 10^{-7}$$1.5 	imes t = 6 	imes 10^{-7}$$t = \frac{6 	imes 10^{-7}}{1.5} = 4 	imes 10^{-7} 	ext{ m}$

Light of wavelength $6000 \text{ Å}$ is incident on a thin glass plate of refractive index $\mu = 1.5$ such that the angle of refraction into the plate is $60^{\circ}$. Calculate the smallest thickness of the plate which will make a dark fringe by reflected beam interference.

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