Explanation:
As the cube has unit volume so energy density, $u=\frac{E}{1}=E$
where, $E=$ total energy
Energy of a photon, $E_1=h \nu$
So for n photon, $E=n h \nu$
Here, the question mentions "average energy contained in the electromagnetic waves within the same volume" which means we have to consider standing wave in a cavity (since the photons are in a cube).
Average energy density for a standing wave : $\langle u\rangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 E_0{ }^2+\frac{1}{2} \frac{B_0{ }^2}{\mu_0}\right)$
The factor $\frac{1}{2}$ comes from averaging $\sin ^2(k x)$
over the volume. $\left\langle\sin ^2(k x)\right\rangle=\frac{1}{2}$
For an $E M$ wave $E=C B$
$ \text { so } E_0=C B_0 $
Hence, $\langle u\rangle=\frac{1}{2}\left(\frac{1}{2} \varepsilon_0\left(c B_0\right)^2+\frac{B_0^2}{2 \mu_0}\right)$
$ =\frac{1}{2}\left(\frac{1}{2} \varepsilon_0 c^2 B_0^2+\frac{B_0^2}{2 \mu_0}\right) $
$ \begin{aligned} \Rightarrow\langle u\rangle=\frac{1}{2} & {\left[\frac{1}{2}\left(\frac{1}{\mu_0 c^2}\right) c^2 B_0^2+\frac{B_0^2}{2 \mu_0}\right]\left(\begin{array}{l} As, c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \\ \text { so } \varepsilon_0=\frac{1}{\mu_0 c^2} \end{array}\right.} \\ \Rightarrow\langle u\rangle & =\frac{1}{2}\left[\frac{B_0^2}{2 \mu_0}+\frac{B_0^2}{2 \mu_0}\right] \\ & \Rightarrow\langle u\rangle=\frac{B_0^2}{2 \mu_0} \end{aligned} $
Given that this is equal to the photon energy density. so $\frac{B_0^2}{2 \mu_0}=n h \nu$
Total energy of photons.
$ \begin{aligned} E & =n h \nu \\ & =35 \times 10^7 \times 6 \times 10^{-34} \times 1.15 \\ \Rightarrow E & =210 \times 10^{-12} \mathrm{~J} \end{aligned} $
For $E M$ waves : $E=\frac{B_0{ }^2}{2 \mu_0}$
So $\frac{B_0^2}{2 \mu_0}=210 \times 10^{-12}$
$ \begin{aligned} & \Rightarrow B_0^2=2 \times 4 \times \frac{22}{7} \times 10^{-7} \times 210 \times 10^{-12} \\ & \Rightarrow B_0^2=5280 \times 10^{-19} \\ & \Rightarrow B_0^2=528 \times 10^{-18} \\ & \Rightarrow B_0=\sqrt{528} \times 10^{-9} \\ & \Rightarrow B_0=22.978 \times 10^{-9} \mathrm{~T} \\ & \\ & \text { = } \alpha \times 10^{-9} \mathrm{~T} \text { (given) } \\ &So, \alpha=22.98 \text { Ans. } \end{aligned} $
[Given: The speed of light in vacuum, $c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ ]
The relation between $[E]$ and $[B]$ is
The relation between $\left[ {{\varepsilon _0}} \right]$ and $\left[ {{\mu _0}} \right]$ is
A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 $\times$ 108 ms$-$1. The final momentum of the object is