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A glass prism whose cross-section is an isosceles triangle stands with its (horizontal) base in water; the angles that its two equal sides make with the base are each θ. An incident ray of light, above and parallel to the water surface and perpendicular to the prism’s axis, is internally reflected at the glass–water interface and subsequently re-emerges into the air. Taking the refractive indices of glass and water to be {\frac {3} {2}} and {\frac {4} {3}} , respectively, show that θ must be at least {{25.9}^{\omicron }}.

Step-by-step

With φ as defined in the Hint, Snell’s law applied to the initial entry into the glass gives sin ( {\frac {\pi } {2}} − θ) = {{n}_{g}} sin φ . Straightforward geometry then determines the angle between the incident ray and the normal to the glass–water interface at the point where the ray meets the boundary as θ+φ.
For total internal reflection to occur this must exceed {{sin}^{-1}} ({{n}_{w}}/{{n}_{g}}) . These two conditions can be combined using the formula sin(θ +φ) = sin θ cosφ+cos θ sinφ to eliminate φ and obtain {{n}^{2}_{g}}-{{n}^{2}_{w}}{\geq }{{cos}^{2}}{\theta }({{n}^{2}_{g}}+1-2{{n}_{w}}).
The substitution of the given values for the refractive indices yields the stated result.

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