On placing a thin sheet of mica of thickness $12 \times 10^{- 7} m$ in the path of one of interferring beams in a Young's experiment.It is found that the central bright band shifts a distance equal to the width of a bright fringe.the wavelength of light used is $6 \times 10^{- 7} m$ , then find refractive index of mica.

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Solution

Given: $t = 12 \times 10^{- 7} m, λ = 6 \times 10^{- 7} m, n = 1, μ = ?$
The shifts,
$(μ - 1) t = n λ$
$μ = \frac{n λ}{t} + 1 = \frac{1 \times 6 \times 10^{- 7}}{12 \times 10^{- 7}} + 1 = 1.5$
The refractive index of mica = 1.5

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Given: t=12×10−7m,λ=6×10−7m,n=1,μ=?The shifts,(μ−1)t=nλμ=nλt+1=1×6×10−712×10−7+1=1.5The refractive index of mica = 1.5

Given: $t = 12 \times 10^{- 7} m, λ = 6 \times 10^{- 7} m, n = 1, μ = ?$
The shifts,
$(μ - 1) t = n λ$
$μ = \frac{n λ}{t} + 1 = \frac{1 \times 6 \times 10^{- 7}}{12 \times 10^{- 7}} + 1 = 1.5$
The refractive index of mica = 1.5