A transparent block of refractive index 1.25 is surrounded by a medium of refractive index 1.0. A ray is incident on a vertical face as shown in the original figure. The maximum incidence angle θ for which the ray undergoes total internal reflection at the top surface is:

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G1 Physics · Accepted Answer

At the limiting condition for total internal reflection at the top surface, the angle of incidence there equals the critical angle θ_c. For block-to-outside transition: sinθ_c = μ_out/μ_block = 1/1.25 = 4/5. If r is the refracted angle inside the block at the vertical face, geometry gives r + θ_c = 90°. Thus, sin r = cosθ_c. Now, cosθ_c = √[1-(4/5)²] = 3/5. Snell's law at the first surface: μ_out sinθ = μ_block sinr. Therefore, sinθ = 1.25×(3/5) = 3 4 \frac{3}{4} 4 3 ​ . Hence, θ = sin⁻¹( 3 4 \…