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A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 (n1 > n2). A ray of light is incident at surface AB. What is the maximum angle of incidence a_max (in water, at AB) such that the ray exits only through surface CD (i.e., it undergoes total internal reflection at surfaces BC and AD)?

1. sin⁻¹ [ (n1/n2) * cos(sin⁻¹(n2/n1)) ]
2. sin⁻¹ [ n1 * cos(sin⁻¹(1/n2)) ]
3. sin⁻¹(n1/n2)
4. sin⁻¹(n2/n1)

Correct answer: sin⁻¹ [ (n1/n2) * cos(sin⁻¹(n2/n1)) ]

Solution

At surface BC (glass to water), TIR requires angle of incidence >= theta_c = sin⁻¹(n2/n1). The ray inside the slab hits BC at angle (90 deg - r) where r is the refraction angle at AB. For TIR: 90-r >= theta_c => r <= 90 - theta_c => sin(r) <= cos(theta_c) = cos(sin⁻¹(n2/n1)). By Snell's law at AB: n2*sin(a) = n1*sin(r), so sin(a_max) = (n1/n2)*cos(sin⁻¹(n2/n1)).

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