Condition for Overlapping Fringes:
In Young’s double slit experiment, the condition for overlapping fringes for different wavelengths is given by:
\[ n_2 \lambda_2 = n_1 \lambda_1 \]
where \( n_1 \) and \( n_2 \) are the fringe orders for wavelengths \( \lambda_1 \) and \( \lambda_2 \), respectively.
Determine the Ratio of Wavelengths:
Given:
\[ \lambda_1 = 450 \, \text{nm}, \quad \lambda_2 = 650 \, \text{nm} \]
The ratio of the wavelengths is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{450}{650} = \frac{9}{13} \]
Find the Minimum Order of Overlapping Fringes:
For the fringes to overlap, \( n_2 \lambda_2 = n_1 \lambda_1 \).
Let \( n_1 = 13 \) and \( n_2 = 9 \) (the smallest integers that satisfy the ratio):
\[ n_2 = 9 \]
Conclusion:
The minimum order of fringe produced by \( \lambda_2 \) which overlaps with the fringe produced by \( \lambda_1 \) is \( n = 9 \).
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 : \(A_2 \rightleftharpoons 2A\)