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.
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: \[ t = \frac{2000}{0.5} = 4000 \, \mathrm{nm} = 4 \, \mu \mathrm{m}. \]
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
