As shown in the figure, a ray of light is incident at an angle of \( 45^\circ \) at the surface AB of the transparent slab. Find the value of the minimum refractive index \( n \) of the slab, when there is total internal reflection of the light ray at the vertical face BC.
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For total internal reflection, the refractive index of the material must be greater than the refractive index of the surrounding medium, and the angle of incidence must exceed the critical angle.
Step 1: Conditions for Total Internal Reflection.
Total internal reflection occurs when the angle of incidence exceeds the critical angle. The critical angle \( \theta_c \) is given by the relation:
\[
\sin \theta_c = \frac{1}{n}
\]
where \( n \) is the refractive index of the slab.
Step 2: Applying Snell's Law at the Interface AB.
At the interface AB, the angle of incidence is \( 45^\circ \). Using Snell's law:
\[
n_1 \sin \theta_1 = n_2 \sin \theta_2
\]
where \( n_1 \) is the refractive index of air (approximately 1), \( \theta_1 = 45^\circ \), and \( \theta_2 \) is the angle inside the slab. Since the light undergoes refraction at the surface, we can write:
\[
\sin 45^\circ = n \sin \theta_2
\]
Step 3: Condition for Total Internal Reflection at BC.
For total internal reflection to occur at the vertical face BC, the angle of incidence at BC must be greater than the critical angle. Thus, the angle of refraction \( \theta_2 \) inside the slab should be equal to or greater than the critical angle. Using Snell's law and the condition for total internal reflection, we have:
\[
n \sin \theta_2 = 1
\]
Step 4: Solving for \( n \).
From the equation \( \sin 45^\circ = n \sin \theta_2 \), and using \( \sin \theta_2 = \frac{1}{n} \), we get:
\[
\frac{\sqrt{2}}{2} = n \times \frac{1}{n}
\]
which simplifies to:
\[
n = \sqrt{2}
\]
Step 5: Conclusion.
Thus, the minimum refractive index \( n \) of the slab is \( \boxed{\sqrt{2}} \).
