A uniform plane wave in free space is normally incident onto a dense dielectric plate of thickness \(\lambda / 4\), having refractive index \(n\). Find the required value of \(n\) such that exactly half the incident power is reflected (and half transmitted). Remember that \(n>1\).

The reflection coefficient \(R\) can be found using the Fresnel equations for normal incidence: $$R = \frac{(n_1 - n_2)^2}{(n_1 + n_2)^2}$$ Where \(n_1\) is the refractive index of the first medium (free space, \(n_1 = 1\)) and \(n_2\) is the refractive index of the second medium (dense dielectric, \(n_2 = n\)).

Since half of the power is reflected, the reflection coefficient \(R\) is equal to 0.5. We can plug this into the equation from Step 1 and solve for \(n\): $$ 0.5 = \frac{(1 - n)^2}{(1 + n)^2} $$ To solve for \(n\), first cross multiply: $$ 2(1 + n)^2 = (1 - n)^2 $$ Now, expand the equation and simplify: $$ 2(1 + 2n + n^2) = (1 - 2n + n^2) $$ $$ 2 + 4n + 2n^2 = 1 - 2n + n^2 $$ Combine like terms: $$ n^2 - 6n + 1 = 0 $$ This is a quadratic equation which can be solved using the quadratic formula: $$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Where \(a = 1\), \(b = -6\), and \(c = 1\). Plugging in the values: $$ n = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)} $$ $$ n = \frac{6 \pm \sqrt{32}}{2} $$ There are two solutions: $$ n_1 = \frac{6 + \sqrt{32}}{2} $$ $$ n_2 = \frac{6 - \sqrt{32}}{2} $$ Since the refractive index must be greater than 1, we can discard the second solution: $$ n = \frac{6 + \sqrt{32}}{2} $$