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# A young's double slit experiment uses light of wavelength $900\; nm$ A plate of thickness 't' of material of refractive index $\mu=1.5$ i placed in front of one of the slits. What should be the thickness of plate 't' so that the central fringe becomes dark.

$(a)\;4.5 \times 10^{-7}\;m \\ (b)\;9 \times 10^{-7}\;m \\ (c)\;1.5 \times 10^{-8}\; m \\ (d)\;4.5 \times 10^{-4}\;m$

Path difference produced by a plate of thickness t and refractive index $\mu$ is
$(\mu-1)t$
This path difference must be minimum equal to $\large\frac{\lambda}{2}$
$(\mu-1)t =\large\frac{\lambda}{2}$
$t= \large\frac{\lambda}{2} \times \large\frac{1}{\mu-1}$
$t= \large\frac{900 \times 10^{-9}}{2} \times \frac{1}{-5}$
$\qquad= 900 \times 10^{-9}\;m$
$\qquad= 9 \times 10^{-7}\;m$
Hence b is the correct answer.