In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(=1.5), the central fringe shifts by 0.2 cm. The value of t is_________ cm.
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The fringe shift `S` is often expressed in terms of fringe width \( \beta = \frac{\lambda D}{d} \). The phase difference introduced is \( \Delta \phi = \frac{2\pi}{\lambda}(n-1)t \), and the shift is \( S = \frac{\beta}{2\pi} \Delta\phi \). This leads to the same formula \( S = \frac{D}{d}(n-1)t \). Notice that the shift is independent of the wavelength of light used.
In a double-slit experiment, when one slit is covered with a transparent sheet, the central fringe shifts. The extent of this shift can be calculated using the optical path difference introduced by the sheet.
The central fringe (zero-order fringe) shifts by a distance \(x\) on the screen, which is given as 0.2 cm.
The fringe shift \(x\) can be expressed as:
\[
x = \frac{t(n - 1) \cdot D}{d}
\]
\(t\) is the thickness of the sheet.
\(n\) is the refractive index of the sheet, given as 1.5.
\(D\) is the distance between the slits and the screen, given as 50 cm.
\(d\) is the distance between the slits, given as 0.1 cm.
Substituting the given values into the equation:
\[
0.2 = \frac{t(1.5 - 1) \cdot 50}{0.1}
\]
Simplify the equation:
\[
0.2 = \frac{t \cdot 0.5 \cdot 50}{0.1}
\]
\[
0.2 = \frac{25t}{0.1}
\]
Multiplying both sides by 0.1 to clear the denominator gives:
