Interference fringes were produced using a light of wavelength 4800 $A^{0}$ in a double slit arrangement. When sheet of uniform thickness of refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves some distance. This distance is equal to the width of 30 interference bands. The thickness (in $μ m$ ) of sheet is:

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answer is D.

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Detailed Solution

Shift of fringe pattern $= (μ - 1) \frac{t D}{d}$
$\Rightarrow \frac{30 D (4800 \times 10^{- 10})}{d} = (0.6) t \frac{D}{d}$

$\Rightarrow 30 \times 4800 \times 10^{- 10} = 0.6 t$
$\Rightarrow t = \frac{30 \times 4800 \times 10^{- 10}}{0.6} = \frac{1.44 \times 10^{- 5}}{0.6} = 24 \times 10^{- 6} m$

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