68.7 nm
94.8 nm
To solve this problem, we need to consider the condition for constructive interference in thin films. For maximum transmission through the film, we require that the phase change due to the path difference between the reflected waves should result in constructive interference.
The condition for constructive interference in a thin film is given by:
2nft = (m + 0.5)λ0,
where:
Substituting the values, we have:
2 × 2.0 × t = (0 + 0.5) × 550 nm
4t = 275 nm
Solving for t gives:
t = 275 nm / 4
t = 68.75 nm
Thus, the minimum thickness of the transparent film required for maximum transmission of green light is 68.75 nm. This closely matches the provided option 68.7 nm.
Light from a point source in air falls on a spherical glass surface (refractive index, \( \mu = 1.5 \) and radius of curvature \( R = 50 \) cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is 1cm.
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: