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 substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]