Three identical polaroids \(P_1\), \(P_2\), and \(P_3\) are placed one after another. The pass axis of \(P_2\) and \(P_3\) are inclined at angles of \(60^\circ\) and \(90^\circ\) with respect to the axis of \(P_1\). The source \(S\) has an intensity of \(\frac{256 \, \text{W}}{\text{m}^2}\). The intensity of light at point \(O\) is _____ \(\frac{\text{W}}{\text{m}^2}\).
For intensity through multiple polaroids:
1. Intensity After First Polaroid (\(P_1\)): - When unpolarized light passes through a polaroid, its intensity is reduced by half:
\[ I_1 = \frac{I_0}{2} = \frac{256}{2} = 128 \, \frac{\text{W}}{\text{m}^2}. \]2. Intensity After Second Polaroid (\(P_2\)): - The intensity after \(P_2\) is given by Malus’s Law:
\[ I_2 = I_1 \cos^2 60^\circ. \]- Substituting \(\cos 60^\circ = \frac{1}{2}\):
\[ I_2 = 128 \cdot \left(\frac{1}{2}\right)^2 = 128 \cdot \frac{1}{4} = 32 \, \frac{\text{W}}{\text{m}^2}. \]3. Intensity After Third Polaroid (\(P_3\)): - The intensity after \(P_3\) is again reduced according to Malus’s Law:
\[ I_3 = I_2 \cos^2 30^\circ, \]where the relative angle between \(P_2\) and \(P_3\) is \(30^\circ\) (since \(90^\circ - 60^\circ = 30^\circ\)).
- Substituting \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):
\[ I_3 = 32 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = 32 \cdot \frac{3}{4} = 24 \, \frac{\text{W}}{\text{m}^2}. \]Final Answer: \(24 \, \frac{\text{W}}{\text{m}^2}\).
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 : \(A_2 \rightleftharpoons 2A\)