When light is reflected from a surface, the reflected ray is completely polarized at the Brewster's angle. The Brewster's angle \(\theta_B\) is given by the relation: \[ \tan(\theta_B) = n \] where \(n\) is the refractive index of the material. For ray B to be completely polarized, the angle of incidence must be the Brewster's angle. For ray B, the angle of incidence is \(60^\circ\). Therefore, \[ \tan(60^\circ) = n \] We know that \(\tan(60^\circ) = \sqrt{3}\). Hence, \[ n = \sqrt{3} \approx 1.732 \]
So, the correct option is (E) : \(1.732\)
For a ray to be completely polarized after reflection, it must satisfy Brewster’s law, which states:
\(\mu = \tan \theta_B\)
where:
\(\mu\) = refractive index of glass
\(\theta_B\) = Brewster’s angle
In the given question, ray B is completely polarized at an angle of incidence of 60°. So,
\(\mu = \tan 60^\circ = \sqrt{3} = 1.732\)
Correct Option: 1.732
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
\(XPQY\) is a vertical smooth long loop having a total resistance \(R\), where \(PX\) is parallel to \(QY\) and the separation between them is \(l\). A constant magnetic field \(B\) perpendicular to the plane of the loop exists in the entire space. A rod \(CD\) of length \(L\,(L>l)\) and mass \(m\) is made to slide down from rest under gravity as shown. The terminal speed acquired by the rod is _______ m/s. 