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.
\[ \boxed{(A)\ \text{is false, but (R) is true.}} \]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: