Monochromatic light of wavelength \( 500 \, \text{nm} \) is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index \( = 1.5 \)), the central maximum is shifted to a position previously occupied by the 4\textsuperscript{th} bright fringe. The thickness of the glass plate is ___________ \( \mu \text{m} \).
Correct Answer: 4
Approach Solution - 1
The central maximum shift indicates that the additional path difference introduced by the glass plate must correspond to a phase shift equivalent to four bright fringe separations. Let's calculate this thickness.
The phase difference caused by the glass plate is given by the formula:
\[\Delta \phi = \frac{2\pi}{\lambda}(t(n-1))\]where \(t\) is the thickness of the glass plate, \(n\) is the refractive index, and \(\lambda\) is the wavelength of light.
The central maximum shifts to the position of the 4\textsuperscript{th} bright fringe, introducing an additional path difference of \(4\lambda\). Thus, we equate and solve for \(t\):
\[\frac{2\pi}{\lambda}(t(n-1)) = 4 \lambda\]Solving for \(t\):
\[t = \frac{4 \lambda^2}{2\pi (n-1)}\]Given \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}\) and \(n = 1.5\):
\[t = \frac{4 \times (500 \times 10^{-9})^2}{2\pi \times (1.5-1)}\]Now calculate:
\[t = \frac{4 \times 250,000 \times 10^{-18}}{2\pi \times 0.5} = \frac{1,000,000 \times 10^{-18}}{\pi} \approx \frac{10^{-12}}{3.1416}\]This evaluates to approximately:
\[\approx 4 \times 10^{-6} \, \text{m} = 4 \, \mu \text{m}\]Thus, the thickness of the glass plate is 4 µm, which falls within the expected range of 4 to 4.
Approach Solution -2
The optical path difference introduced by the glass plate is:
\[ \Delta x = t(\mu - 1), \]
where $t$ is the thickness of the plate and $\mu$ is its refractive index.
The fringe shift is given by:
\[ \Delta x = n\lambda, \]
where $n = 4$ (the shift corresponds to the 4th fringe) and $\lambda = 500 \, \mathrm{nm}$.
Equating: \[ t(\mu - 1) = n\lambda. \]
Substituting $\mu = 1.5$, $n = 4$, and $\lambda = 500 \, \mathrm{nm}$:
\[ t(1.5 - 1) = 4 \cdot 500. \]
Simplify: \[ t = \frac{2000}{0.5} = 4000 \, \mathrm{nm} = 4 \, \mu \mathrm{m}. \]
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