To find the radiation pressure on the object due to the electromagnetic wave, we start by understanding the concept of radiation pressure in physics. Radiation pressure is exerted upon any surface due to the exchange of momentum between the object and the incident electromagnetic wave. The formula to calculate radiation pressure \(P\) when the wave is completely absorbed is given by:
\(P = \frac{I}{c}\)
where:
Since the medium's refractive index is given as \(3\), the speed of light in the medium \(c_m\) is:
\(c_m = \frac{c}{n} = \frac{3 \times 10^8 \, \text{m/s}}{3} = 1 \times 10^8 \, \text{m/s}\)
Now, substitute the values into the radiation pressure formula:
\(P = \frac{6 \times 10^8 \, \text{W/m}^2}{1 \times 10^8 \, \text{m/s}} = 6 \, \text{Nm}^{-2}\)
Thus, the radiation pressure exerted on the object is \(6 \, \text{Nm}^{-2}\).
The radiation pressure \( P \) is given by:
\(P = \frac{I \cdot \mu}{c},\)
where:
- \( I \) is the intensity of the radiation,
- \( \mu \) is the absorption coefficient (fraction of radiation absorbed, here \( \mu = 1 \) for total absorption),
- \( c \) is the speed of light in a vacuum.
Substitute the given values:
- \( I = 6 \times 10^8 \, \text{W/m}^2 \),
- \( \mu = 3 \),
- \( c = 3 \times 10^8 \, \text{m/s} \).
Calculate \( P \):
\(P = \frac{I \cdot \mu}{c} = \frac{6 \times 10^8 \cdot 3}{3 \times 10^8}.\)
Simplify:
\(P = 6 \, \text{N/m}^2.\)
The Correct answer is: \(6 \, \text{Nm}^{-2}\)
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 : \(A_2 \rightleftharpoons 2A\)