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(B) The dielectric slab is bounded by the planes $x=0, x=a$ and $y=0, y=b$. $A$ light ray incident at $y=0$ at an angle $	heta$ refracts into the die

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$\sqrt{2}$

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(B) The dielectric slab is bounded by the planes $x=0, x=a$ and $y=0, y=b$. $A$ light ray incident at $y=0$ at an angle $	heta$ refracts into the dielectric at an angle $r$ with the normal.For the ray to reach $y=b$,it must undergo Total Internal Reflection $(TIR)$ at the boundary $x=a$. The angle of incidence at the boundary $x=a$ is $(90^{\circ}-r)$.For $TIR$ to occur at $x=a$,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin C = 1/n$. Thus,$\sin(90^{\circ}-r) \geq 1/n$,which simplifies to $\cos r \geq 1/n$,or $n \geq 1/\cos r$.From Snell's law at the boundary $y=0$,$n = \sin 	heta / \sin r$. Since $	heta$ can range from $0^{\circ}$ to $90^{\circ}$,the limiting case occurs at $	heta = 90^{\circ}$,giving $n = 1/\sin r$,or $\sin r = 1/n$.Using the identity $\sin^2 r + \cos^2 r = 1$,we have $(1/n)^2 + (1/n)^2 = 1$,which leads to $2/n^2 = 1$,or $n^2 = 2$. Therefore,the minimum refractive index is $n = \sqrt{2}$.

Consider a glass cube slab of dielectric bound by the planes $x=0, x=a$; $y=0, y=b$; $z=0, z=c$; with $b > a > c$. The slab is placed in air and has a refractive index of $n$. The minimum value for $n$,such that all rays entering the dielectric at $y=0$ reach $y=b$ is

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