In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index \(1.56\), the central fringe shifts to the position of \(7^{th}\) bright fringe, obtained with both slits uncovered. If the light source wavelength is \(450\) nm, the thickness of mica sheet is \(\alpha \times 10^{-9}\) m. The value of \(\alpha\) is ______.

The shift in the central fringe due to the introduction of a transparent sheet of thickness \(t\) and refractive index \(\mu\) is given by \(\Delta y = \dfrac{D}{d}(\mu - 1)t\).

The position of the \(n^{th}\) bright fringe is \(y_n = n \dfrac{D\lambda}{d}\).

Equating the two, we get:

\(\dfrac{D}{d}(\mu - 1)t = n \dfrac{D\lambda}{d}\)

\((\mu - 1)t = n\lambda\)

Substituting the given values:

\((1.56 - 1)t = 7 \times 450 \times 10^{-9}\)

\(0.56 t = 3150 \times 10^{-9}\)

\(t = \dfrac{3150}{0.56} \times 10^{-9}\)

\(t = 5625 \times 10^{-9}\) m

Comparing with \(t = \alpha \times 10^{-9}\) m, we get \(\alpha = 5625\).

Answer: \(5625\)