When one of the slits in Young’s experiment is covered with a transparent sheet of thickness 3.6 × 10^-3 cm the central fringe shifts to a position originally occupied by the 30th bright fringe. If λ = 6000Å, find the refractive index of the sheet.

Solution

We, are considering young's double slit experiment.
Wavelength of the monochromatic source,

λ = 6000 \overset{o}{A} = 6000 \times 10^{- 10} m

Thickness of the transparent sheet, t = 3.6

\times

10^-3 cm
refractive index of the sheet,

μ

= ?

The position of the 30th bright fringe is given by,

x_{n} = n \frac{λD}{d}

i.e.,

x_{30} = 30 \frac{λD}{d}

Shift of the central fringe is,

x_{0} = 30 \frac{λD}{d}

But, we have

x_{0} = \frac{D}{d} (μ - 1) t

where, D is the distance between the screen and slit.
d is the distance between slits.

∴

30 \frac{λD}{d} = \frac{D}{d} (μ - 1) t

\Rightarrow (μ - 1) = \frac{30 λ}{t} = \frac{30 \times (6000 \times 10^{- 10})}{(3.6 \times 10^{- 5})} = 0.5

i.e.,

μ = 1 + 0.5 = 1.5

is the required refractive index of the sheet.

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