A light ray falls on a square glass slab on horizontal face with incident angle 45⁰. What must be the refractive index of glass, if total internal reflection occurs at the vertical face

$\begin{array}{l}\mathrm{\mu }=\frac{\sin {45}^{\circ }}{\sin \mathrm{r}}\Rightarrow \sin \mathrm{r}=\frac{1}{\sqrt{2}\cdot \mathrm{\mu }}\\ \frac{1}{\mathrm{\mu }}=\cos \mathrm{r}=\sin \left({90}^{\circ }-\mathrm{r}\right)\\ \frac{1}{\mathrm{\mu }}=\sqrt{1-{\sin }^2\mathrm{r}};\mathrm{\mu }=\sqrt{\frac{3}{2}}\end{array}$

Explantion:

At point A, the ray is incident at an angle of incidence 45°. Using Snell's law:

μ = sin(i) / sin(r)

So:

sin(r) = sin(45°) / μ = 1 / (√2 * μ) (Equation 1)

At point B:

Total internal reflection takes place. In ΔAOB,

θ + r = 90° → θ = (90° - r).

From Equation (1):

sin(90° - θ) = 1 / (√2 * μ).

Using the trigonometric identity sin(90° - θ) = cos(θ):

cos(θ) = √(μ² - 1) / μ.

Equating:

√(μ² - 1) / μ = 1 / (√2 * μ).

Simplify:

μ² - 1 = 1 / 2.

Thus:

μ² = 1 / 2 + 1 = 3 / 2.

Finally:

μ = √(3 / 2) = √3 / √2.