We are given:
- The refractive index of the medium, \( n = 1.8 \),
- The relative permeability of the medium, \( \mu_r = 2.16 \).
The refractive index \( n \) is related to the relative permittivity \( \epsilon_r \) and the relative permeability \( \mu_r \) by the formula: \[ n = \sqrt{\epsilon_r \mu_r} \]
where:
- \( \epsilon_r \) is the relative permittivity,
- \( \mu_r \) is the relative permeability.
Substitute the given values into this equation: \[ 1.8 = \sqrt{\epsilon_r \times 2.16} \] Squaring both sides: \[ 3.24 = \epsilon_r \times 2.16 \]
Now, solving for \( \epsilon_r \): \[ \epsilon_r = \frac{3.24}{2.16} \approx 1.5 \] Thus, the relative permittivity of the medium is approximately \( 1.5 \).
Conclusion: The relative permittivity of the medium is nearly 1.5, so the correct answer is Option (1).
A laser beam has intensity of $4.0\times10^{14}\ \text{W/m}^2$. The amplitude of magnetic field associated with the beam is ______ T. (Take $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N m}^2$ and $c=3\times10^8\ \text{m/s}$)
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))