In a Young’s double slit experiment, two slits are illuminated with light of wavelength \(800 \, \text{nm}\). The first minimum is detected at \(P\). The value of slit separation \(a\) is:
For Young’s double-slit experiments:
• Use the condition for minima or maxima to relate wavelength, slit separation, and screen distance.
• Ensure units are consistent when calculating.
Condition for Minima: Path difference for the first minimum:
\[ \Delta x = \frac{\lambda}{2}. \]
Slit Separation: From geometry:
\[ a = \frac{\lambda D}{\Delta x}. \]
Substituting values:
\[ a = \frac{800 \times 10^{-9} \times 5 \times 10^{-2}}{0.5 \times 10^{-3}} = 0.2 \, \text{mm}. \]
Final Answer: 0.2 mm
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 : \(A_2 \rightleftharpoons 2A\)