Step 1: Formula for fringe width in Young’s double slit experiment.
The fringe width is given by: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) = wavelength of light, \( D \) = distance between slits and screen, \( d \) = distance between the two slits.
Step 2: Relation between fringe width and wavelength.
From the above formula: \[ \beta \propto \lambda \] i.e., the fringe width is directly proportional to the wavelength.
Step 3: Compare red and blue light.
The wavelength of red light (\( \lambda_r \)) is greater than that of blue light (\( \lambda_b \)): \[ \lambda_r > \lambda_b. \] Hence, the fringes produced by red light are wider apart (not closer) than those produced by blue light.
Step 4: Verify the truth of the statements.
- Assertion (A): False — because red fringes are farther apart, not closer. - Reason (R): True — because fringe width is directly proportional to wavelength.
\[ \boxed{(A)\ \text{is false, but (R) is true.}} \]
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 : \(A_2 \rightleftharpoons 2A\)