1. The position of bright fringes in Young's double-slit experiment is given by: \[ y_n = \frac{n \lambda D}{d}, \] where \(n\) is the fringe order, \(\lambda\) is the wavelength of light, \(D\) is the distance between the slits and screen, and \(d\) is the distance between the slits.
2. Substituting the given values: - \(n = 5, \, \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}, \, y_n = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m}, \, D = 1 \, \text{m}\).
3. Rearrange for \(d\): \[ d = \frac{n \lambda D}{y_n}. \]
4. Substituting: \[ d = \frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}} = 6 \times 10^{-6} \, \text{m} = 48 \, \mu\text{m}. \]
Thus, the distance between the slits is 48 \(\mu\text{m}\).
The fringe spacing depends on the wavelength, the slit-to-screen distance, and the slit separation. Accurate calculations require converting all quantities to SI units.
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 : \(A_2 \rightleftharpoons 2A\)