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 mica sheet is placed in front of one of the slits, so all rays from that slit travel an extra optical path.
Extra optical path introduced by a sheet of thickness $$t$$ and refractive index $$n$$ is

$$\Delta = (n-1)\,t$$

In Young’s double slit experiment, a path difference of one wavelength $$\lambda$$ shifts the fringe pattern by one fringe width (i.e. the central maximum moves to the position of the next bright band).
Hence, a shift through $$m$$ bright fringes requires a path difference

$$\Delta = m\,\lambda \quad -(1)$$

Given data:
• Refractive index of mica: $$n = 1.56$$
• Wavelength of light: $$\lambda = 450\;\text{nm} = 450 \times 10^{-9}\,\text{m}$$
• Central fringe is observed at the position of the 7-th bright fringe: $$m = 7$$

Equate the two expressions for $$\Delta$$:

$$(n-1)\,t = m\,\lambda$$

Solve for $$t$$:

$$t = \frac{m\,\lambda}{n-1}$$

Substitute the numbers:

$$t = \frac{7 \times 450 \times 10^{-9}}{1.56 - 1}$$

$$t = \frac{3150 \times 10^{-9}}{0.56}$$

$$t = 5625 \times 10^{-9}\;\text{m}$$

Hence, the thickness of the mica sheet is $$\alpha \times 10^{-9}\,\text{m}$$ with

$$\alpha = 5625$$

Answer (numerical): 5625