
Remember the formula for fringe shift in Young’s double slit experiment when a thin plate is introduced. Ensure consistent units while substituting values.
Step 1: Recall the Formula for Fringe Shift
When a thin plate of thickness \(t\) and refractive index \(\mu\) is introduced in the path of one of the slits in Young’s double slit experiment, the fringe pattern shifts. The fringe shift (\(\Delta x\)) is given by:
\[ \Delta x = \frac{t (\mu - 1)}{\lambda} \beta_0 \]
where \(\lambda\) is the wavelength of light and \(\beta_0\) is the fringe width.
Step 2: Convert Units and Substitute Values
Given \(t = 10 \, \mu \text{m} = 10 \times 10^{-6} \, \text{m}\), \(\mu = 1.2\), and \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} = 5 \times 10^{-7} \, \text{m}\):
\[ \Delta x = \frac{10 \times 10^{-6} (1.2 - 1)}{5 \times 10^{-7}} \beta_0 \]
\[ \Delta x = \frac{10 \times 10^{-6} \times 0.2}{5 \times 10^{-7}} \beta_0 \]
\[ \Delta x = \frac{2 \times 10^{-6}}{5 \times 10^{-7}} \beta_0 = \frac{20 \times 10^{-7}}{5 \times 10^{-7}} \beta_0 = 4 \beta_0 \]
Step 3: Find the Value of \(x\)
The central maxima is shifted by a distance of \(x \beta_0\). We found that \(\Delta x = 4 \beta_0\). Therefore, \(x = 4\).
Conclusion: The value of \(x\) is 4.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)