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For the given incident ray as shown in figure, the condition of total internal reflection of this ray the minimum refractive index of prism will be : -
A
$\dfrac{\sqrt{3} + 1}{2}$
B
$\dfrac{\sqrt{2} + 1}{2}$
C
$\sqrt{\dfrac{3}{2}}$
D
$\sqrt{\dfrac{7}{6}}$
Detailed Solution
Let the ray refract at the top face with angle of refraction r. It then meets the vertical face at an angle of incidence $90^\circ - r$, because the two faces are perpendicular.
For the minimum refractive index, the ray just suffers total internal reflection at the side face: $90^\circ - r = \theta_c$, so $r = 90^\circ - \theta_c$.
Snell's law at the top face: $1 \cdot \sin 45^\circ = \mu\sin(90^\circ - \theta_c) = \mu\cos\theta_c$
Since $\sin\theta_c = \dfrac{1}{\mu}$: $\cos\theta_c = \dfrac{\sqrt{\mu^2 - 1}}{\mu}$, so $\mu\cos\theta_c = \sqrt{\mu^2 - 1}$
$\dfrac{1}{\sqrt{2}} = \sqrt{\mu^2 - 1} \Rightarrow \mu^2 = 1 + \dfrac{1}{2} = \dfrac{3}{2}$
$\Rightarrow \mu = \sqrt{\dfrac{3}{2}}$
Alternate (objective) method: for this arrangement $\mu = \sqrt{1 + \sin^2\theta}$, so $\mu = \sqrt{1 + \sin^2 45^\circ} = \sqrt{1 + \dfrac{1}{2}} = \sqrt{\dfrac{3}{2}}$.
For the minimum refractive index, the ray just suffers total internal reflection at the side face: $90^\circ - r = \theta_c$, so $r = 90^\circ - \theta_c$.
Snell's law at the top face: $1 \cdot \sin 45^\circ = \mu\sin(90^\circ - \theta_c) = \mu\cos\theta_c$
Since $\sin\theta_c = \dfrac{1}{\mu}$: $\cos\theta_c = \dfrac{\sqrt{\mu^2 - 1}}{\mu}$, so $\mu\cos\theta_c = \sqrt{\mu^2 - 1}$
$\dfrac{1}{\sqrt{2}} = \sqrt{\mu^2 - 1} \Rightarrow \mu^2 = 1 + \dfrac{1}{2} = \dfrac{3}{2}$
$\Rightarrow \mu = \sqrt{\dfrac{3}{2}}$
Alternate (objective) method: for this arrangement $\mu = \sqrt{1 + \sin^2\theta}$, so $\mu = \sqrt{1 + \sin^2 45^\circ} = \sqrt{1 + \dfrac{1}{2}} = \sqrt{\dfrac{3}{2}}$.